NCERT Class 12 Application of Derivatives Solutions

📘 Concept & Theory

Theory and Concept

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The rate of change of one quantity with respect to another is represented by the derivative of the first quantity with respect to the second quantity. In this question, the area \(A\) of the circle depends on its radius \(r\). Therefore, the rate of change of area with respect to radius is represented by

\[ \frac{dA}{dr}. \]

The area of a circle of radius \(r\) is

\[ A=\pi r^2. \]

Differentiating this expression with respect to \(r\) gives the instantaneous rate at which the area changes when the radius changes.

Thus, the key idea is:

\[ \text{Rate of change of area with respect to radius}=\frac{dA}{dr}. \]

This is a direct application of differentiation and is one of the fundamental ideas developed in the chapter Application of Derivatives.

🗺️ Solution Roadmap

Step-by-step Plan

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Write the formula for the area of a circle in terms of its radius.
Differentiate the area with respect to \(r\).
Substitute \(r=3\) cm into the derivative to obtain the required rate.
Substitute \(r=4\) cm into the derivative to obtain the required rate.
State the answers with appropriate units.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Express the area of the circle in terms of its radius.
The area \(A\) of a circle with radius \(r\) is
\[ A=\pi r^2. \]
Find the rate of change of area with respect to radius.
Since \(A\) is a function of \(r\), differentiate both sides with respect to \(r\):
\[\frac{dA}{dr}=\frac{d}{dr}\left(\pi r^2\right)\]
Since \(\pi\) is a constant, it can be taken outside the differentiation:
\[\frac{dA}{dr}=\pi\frac{d}{dr}\left(r^2\right)\]
Using the power rule
\[ \frac{d}{dr}(r^n)=nr^{n-1}, \]
we obtain
\[\frac{d}{dr}(r^2)=2r\]
Therefore,
\[\frac{dA}{dr}=2\pi r\]
Hence, the general rate of change of the area of the circle with respect to its radius is
\[\boxed{\frac{dA}{dr}=2\pi r}\]
(a) When \(r=3\) cm
Substitute \(r=3\) cm into
\[\frac{dA}{dr}=2\pi r\]

\[\left.\frac{dA}{dr}\right|_{r=3}=2\pi(3)\]

\[\left.\frac{dA}{dr}\right|_{r=3}=6\pi\]
Therefore, when the radius is \(3\) cm, the rate of change of the area with respect to the radius is
\[\boxed{6\pi\ \text{cm}}\]
(b) When \(r=4\) cm
Substitute \(r=4\) cm into
\[\frac{dA}{dr}=2\pi r\]

\[\left.\frac{dA}{dr}\right|_{r=4}=2\pi(4)\]

\[\left.\frac{dA}{dr}\right|_{r=4}=8\pi\]
Therefore, when the radius is \(4\) cm, the rate of change of the area with respect to the radius is
\[\boxed{8\pi\ \text{cm}}\]

💡 Answer

Final Answer

›

Radius	Rate of Change of Area
\(r=3\) cm	\(\boxed{6\pi\ \text{cm}}\)
\(r=4\) cm	\(\boxed{8\pi\ \text{cm}}\)

🎯 Exam Significance

Exam Significance

›

This problem tests the fundamental concept of rate of change using derivatives. In board examinations, questions of this type are useful for testing whether a student can correctly identify the dependent and independent variables and differentiate a standard geometrical formula.

A common examination pattern is to provide a geometrical quantity such as area, volume, circumference or surface area and ask for its rate of change with respect to another variable. The essential step is to first express the required quantity as a function of the given variable and then differentiate.

Significance for Competitive Entrance Examinations

For competitive examinations, this problem establishes the basic interpretation of a derivative as an instantaneous rate of change. Similar concepts can be extended to related-rate problems involving circles, spheres, cylinders, cones and other geometrical quantities.

Students should also recognise that the derivative

\[ \frac{dA}{dr}=2\pi r \]

is itself a function of \(r\). Therefore, the rate of change is not constant; it increases as the radius increases.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 8 points

The area of a circle is \(\displaystyle A=\pi r^2\).
The rate of change of area with respect to radius is represented by \(\displaystyle \frac{dA}{dr}\).
Differentiating \(A=\pi r^2\) gives \(\displaystyle \frac{dA}{dr}=2\pi r\).
At \(r=3\) cm, the rate of change is \(\displaystyle 6\pi\) cm.
At \(r=4\) cm, the rate of change is \(\displaystyle 8\pi\) cm.
The derivative gives the instantaneous rate of change of the area as the radius changes.
Always identify the independent variable before differentiating.
For rate-of-change questions, first form the required quantity as a function of the specified variable, then differentiate.

📘 Concept & Theory

Theory and Concept

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This is a related rates problem. In such problems, two or more quantities depend on a common variable, usually time \(t\). Here, the volume and surface area of the cube both depend on the length \(x\) of its edge, while \(x\) changes with time.

The given information is the rate at which the volume changes:

\[ \frac{dV}{dt}=8\text{ cm}^3/\text{s}. \]

We need to determine the rate at which the surface area changes:

\[ \frac{dS}{dt}. \]

For a cube of edge length \(x\),

\[ V=x^3 \]

and

\[ S=6x^2. \]

Since both \(V\) and \(S\) depend on \(x\), and \(x\) changes with time, we use the chain rule:

\[ \frac{dV}{dt}=\frac{dV}{dx}\frac{dx}{dt} \]

and

\[ \frac{dS}{dt}=\frac{dS}{dx}\frac{dx}{dt}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(x\) be the length of an edge of the cube.
Write the volume as \(V=x^3\).
Differentiate \(V=x^3\) with respect to time \(t\) using the chain rule.
Use the given value of \(\frac{dV}{dt}\) to find \(\frac{dx}{dt}\).
Write the surface area as \(S=6x^2\).
Differentiate the surface area with respect to time.
Substitute \(x=12\) cm and the previously calculated value of \(\frac{dx}{dt}\).
State the required rate of increase of surface area with its appropriate unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 25 steps

Let the edge length of the cube be \(x\).
Let \(x\) cm be the length of each edge of the cube at time \(t\).

Since the cube is changing with time, \(x\) is also a function of \(t\).
Express the volume of the cube
The volume \(V\) of a cube of edge length \(x\) is
\[V=x^3\]
The volume is increasing at the rate of \(8\text{ cm}^3/\text{s}\). Therefore,
\frac{dV}{dt}=8\text{ cm}^3/\text{s}
Differentiate the volume with respect to time
Differentiate
\[V=x^3\]
with respect to \(t\).
Since \(x\) is a function of \(t\), the chain rule gives
\[\frac{dV}{dt}=\frac{dV}{dx}\frac{dx}{dt}\]
Now,
\[\frac{dV}{dx}=\frac{d}{dx}(x^3)=3x^2\]
Hence,
\[\frac{dV}{dt}=3x^2\frac{dx}{dt}\]
Using the given rate
\[\frac{dV}{dt}=8\text{ cm}^3/\text{s},\]
we get
\[3x^2\frac{dx}{dt}=8\]
Find the rate of change of the edge length
Solving for \(\frac{dx}{dt}\),
\[\frac{dx}{dt}=\frac{8}{3x^2}\]
Therefore, the edge length of the cube is increasing at the rate
\[\boxed{\frac{dx}{dt}=\frac{8}{3x^2}\text{ cm/s}}\]
Express the surface area of the cube
A cube has six square faces. Each face has area \(x^2\). Therefore, its total surface area \(S\) is
\[S=6x^2\]
Differentiate the surface area with respect to time
Differentiate
\[S=6x^2\]
with respect to \(t\).
Applying the chain rule,
\[\frac{dS}{dt}=\frac{dS}{dx}\frac{dx}{dt}\]
Since
\[\frac{dS}{dx}=\frac{d}{dx}(6x^2)=12x,\]
we obtain
\[\frac{dS}{dt}=12x\frac{dx}{dt}\]
Substitute the value of \(\frac{dx}{dt}\)
We have already obtained
\[\frac{dx}{dt}=\frac{8}{3x^2}\]
Therefore,
\[\frac{dS}{dt}=12x\left(\frac{8}{3x^2}\right)\]
Simplifying,
\[\frac{dS}{dt}=\frac{12\times8x}{3x^2}\]

\[\frac{dS}{dt}=\frac{96x}{3x^2}\]

\[\frac{dS}{dt}=\frac{32}{x}\]
Evaluate when \(x=12\) cm
The question asks for the rate of increase of surface area when the edge length is \(12\) cm. Hence,
\[x=12\]
Substituting \(x=12\) into
\[\frac{dS}{dt}=\frac{32}{x},\]
we get
\[\left.\frac{dS}{dt}\right|_{x=12}=\frac{32}{12}\]
Simplifying,
\[\frac{32}{12}=\frac{8}{3}\]
Therefore,
\[\boxed{\frac{dS}{dt}=\frac{8}{3}\text{ cm}^2/\text{s}}\]

💡 Answer

Final Answer

›

When the edge of the cube is \(12\) cm, the surface area is increasing at the rate

\[ \boxed{\frac{8}{3}\text{ cm}^2/\text{s}}. \]

🎯 Exam Significance

Exam Significance

›

This problem is an important application of the chain rule and related rates. It tests whether a student can connect the rate of change of volume with the rate of change of the edge length and subsequently determine the rate of change of surface area.

A key examination point is that the given rate is

\[ \frac{dV}{dt}, \]

whereas the required quantity is

\[ \frac{dS}{dt}. \]

Since neither quantity is directly given as a function of time, the intermediate variable \(x\) must be introduced.

Students should also pay attention to units. Volume has units of \(\text{cm}^3\), so its rate of change has units of \(\text{cm}^3/\text{s}\). Surface area has units of \(\text{cm}^2\), so its rate of change has units of \(\text{cm}^2/\text{s}\).

Significance for Competitive Entrance Examinations

For competitive examinations, this question develops the standard related-rates framework:

\[ \frac{d(\text{quantity})}{dt} = \frac{d(\text{quantity})}{dx} \frac{dx}{dt}. \]

The same method can be extended to spheres, cylinders, cones and other geometrical figures. Competitive problems may combine this idea with algebraic simplification, multiple variables, implicit differentiation or optimization.

An efficient approach is to establish the relationship between the quantities first, differentiate with respect to time, substitute the known rate, and only then insert the specified value of the variable.

❌ Mistakes

Common Mistake to Avoid

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Do not write \[ \frac{dV}{dt}=\frac{dV}{dx} \] because these represent different rates. The correct chain-rule relationship is \[ \frac{dV}{dt} = \frac{dV}{dx}\frac{dx}{dt}. \]

Similarly, the given rate is \(8\text{ cm}^3/\text{s}\), not \(8\text{ cm}/\text{s}\). Keeping the units throughout the calculation helps prevent such errors.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 10 points

For a cube of edge length \(x\), the volume is \(\displaystyle V=x^3\).
The surface area of a cube is \(\displaystyle S=6x^2\).
When \(x\) changes with time, the chain rule must be used.
The volume rate is related to the edge-length rate by \(\displaystyle \frac{dV}{dt}=3x^2\frac{dx}{dt}\).
From the given information, \(\displaystyle \frac{dx}{dt}=\frac{8}{3x^2}\).
The surface-area rate is \(\displaystyle \frac{dS}{dt}=12x\frac{dx}{dt}\).
After substitution, \(\displaystyle \frac{dS}{dt}=\frac{32}{x}\).
At \(x=12\) cm, \(\displaystyle \frac{dS}{dt}=\frac{8}{3}\text{ cm}^2/\text{s}\).
In related-rate problems, identify the common changing variable before differentiating.
Always verify the final unit according to the physical quantity whose rate is being requested.

📘 Concept & Theory

Theory and Concept

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This is a related rates problem. The radius \(r\) of the circle changes with time, and consequently its area \(A\) also changes with time.

The given rate of change of radius is

\[ \frac{dr}{dt}=3\text{ cm/s}. \]

We need to find the rate of change of area,

\[ \frac{dA}{dt}, \]

when

\[ r=10\text{ cm}. \]

The area of a circle is

\[ A=\pi r^2. \]

Since \(r\) is a function of time, we differentiate this equation with respect to \(t\). By the chain rule,

\[ \frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(r\) denote the radius of the circle at time \(t\).
Write the area of the circle as \(A=\pi r^2\).
Differentiate the area with respect to time using the chain rule.
Substitute the given rate \(\frac{dr}{dt}=3\text{ cm/s}\).
Substitute \(r=10\) cm.
Simplify the result and state the answer with the correct unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Identify the given rate.
The radius of the circle is increasing at the rate of \(3\text{ cm/s}\). Therefore,
\[ \frac{dr}{dt}=3\text{ cm/s}. \]
Write the area of the circle.
The area \(A\) of a circle of radius \(r\) is
\[ A=\pi r^2. \]
Differentiate the area with respect to time.
Since \(r\) changes with time, differentiate both sides with respect to \(t\):
\[\frac{dA}{dt}=\frac{d}{dt}\left(\pi r^2\right)\]
Since \(\pi\) is a constant,
\[\frac{dA}{dt}=\pi\frac{d}{dt}(r^2)\]
Applying the chain rule,
\[\frac{d}{dt}(r^2)=2r\frac{dr}{dt}\]
Therefore,
\[\frac{dA}{dt}=2\pi r\frac{dr}{dt}\]
Thus, the rate of change of the area with respect to time is
\[ \boxed{\frac{dA}{dt}=2\pi r\frac{dr}{dt}}. \]
Substitute the given rate of change of radius.
We are given
\[\frac{dr}{dt}=3\text{ cm/s}\]
Hence,
\[\frac{dA}{dt}=2\pi r(3)\]

\[\frac{dA}{dt}=6\pi r\]
Substitute the radius \(r=10\) cm.
When the radius is \(10\) cm,
\[r=10\]
Therefore,
\[\left.\frac{dA}{dt}\right|_{r=10}=6\pi(10)\]

\[\frac{dA}{dt}=60\pi\]
State the result with the appropriate unit.
Since area is measured in \(\text{cm}^2\) and time is measured in seconds, the rate of change of area is measured in \(\text{cm}^2/\text{s}\).
\[\boxed{\frac{dA}{dt}=60\pi\text{ cm}^2/\text{s}}\]

💡 Answer

Final Answer

›

When the radius of the circle is \(10\) cm, its area is increasing at the rate

\[ \boxed{60\pi\text{ cm}^2/\text{s}}. \]

🎯 Exam Significance

Exam Significance

›

This question is a direct application of the concept of related rates. It tests the ability to differentiate a geometrical relation with respect to time and correctly use the chain rule.

The important relationship to remember is

\[ \frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt}. \]

Here,

\[ \frac{dA}{dr}=2\pi r, \]

so

\[ \frac{dA}{dt}=2\pi r\frac{dr}{dt}. \]

For board examinations, students should clearly show the differentiation step and substitute the given values only after obtaining the required derivative. This makes the relationship between the quantities explicit and avoids confusing \(\frac{dA}{dr}\) with \(\frac{dA}{dt}\).

Significance for Competitive Entrance Examinations

This problem forms the foundation of more advanced related-rate questions appearing in competitive mathematics. The same chain-rule technique can be applied when the radius, diameter, height, side length or another geometrical parameter changes with time.

A useful general result for a circle is

\[ \frac{dA}{dt}=2\pi r\frac{dr}{dt}. \]

Once this relationship is established, numerical values can be substituted directly.

Competitive problems may further combine related rates with algebraic manipulation, implicit differentiation or multiple changing quantities. Therefore, understanding the chain-rule structure is more important than memorising the final numerical answer.

❌ Mistakes

Common Mistakes to Avoid

›

Do not confuse \(\frac{dr}{dt}\) with \(\frac{dA}{dt}\).
Do not differentiate \(A=\pi r^2\) only with respect to \(r\) when the question asks for the rate with respect to time.
Remember the chain-rule factor \(\frac{dr}{dt}\).
Do not substitute \(r=10\) before establishing the required rate equation.
Use \(\text{cm}^2/\text{s}\), not \(\text{cm}/\text{s}\), for the rate of change of area.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The area of a circle is \(\displaystyle A=\pi r^2\).
If the radius changes with time, the area also changes with time.
The chain rule gives \(\displaystyle \frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}\).
For a circle, \(\displaystyle \frac{dA}{dr}=2\pi r\).
Therefore, \(\displaystyle \frac{dA}{dt}=2\pi r\frac{dr}{dt}\).
Given \(\displaystyle \frac{dr}{dt}=3\text{ cm/s}\) and \(r=10\) cm, the required rate is \(\displaystyle 60\pi\text{ cm}^2/\text{s}\).
The unit of a rate of change of area with respect to time is square units per unit time.
In related-rate problems, differentiate with respect to the common independent variable, usually time.

📘 Concept & Theory

Theory and Concept

›

This is a related rates problem. The edge length of the cube changes with time, and consequently the volume of the cube also changes with time.

Let \(x\) be the length of an edge of the cube. Since the edge is increasing at the rate of \(3\text{ cm/s}\),

\[ \frac{dx}{dt}=3\text{ cm/s}. \]

The volume \(V\) of a cube with edge length \(x\) is

\[ V=x^3. \]

Since \(x\) is changing with time, \(V\) also changes with time. Therefore, we differentiate the volume with respect to time using the chain rule:

\[ \frac{dV}{dt} = \frac{dV}{dx}\frac{dx}{dt}. \]

The required quantity is the rate of change of volume,

\[ \frac{dV}{dt}, \]

when

\[ x=10\text{ cm}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(x\) denote the edge length of the cube.
Write the volume as \(V=x^3\).
Use the given rate \(\frac{dx}{dt}=3\text{ cm/s}\).
Differentiate \(V=x^3\) with respect to time using the chain rule.
Substitute \(\frac{dx}{dt}=3\text{ cm/s}\).
Substitute \(x=10\) cm.
Simplify and state the rate of increase of volume with its appropriate unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Define the edge length.
Let \(x\) cm be the length of each edge of the variable cube at time \(t\).
The edge is increasing at the rate of \(3\text{ cm/s}\). Hence,
\[ \frac{dx}{dt}=3\text{ cm/s}. \]
Express the volume of the cube.
The volume \(V\) of a cube of edge length \(x\) is
\[V=x^3\]
Differentiate the volume with respect to time
Since \(x\) changes with time, differentiate both sides with respect to \(t\):
\[\frac{dV}{dt}=\frac{d}{dt}(x^3)\]
Applying the chain rule,
\[\frac{d}{dt}(x^3)=\frac{d}{dx}(x^3)\frac{dx}{dt}\]
Using the power rule,
\[\frac{d}{dx}(x^3)=3x^2\]
Therefore,
\[\frac{dV}{dt}=3x^2\frac{dx}{dt}\]
Substitute the given rate of change of the edge
We are given
\[\frac{dx}{dt}=3\text{ cm/s}\]
Hence,
\[\frac{dV}{dt}=3x^2(3)\]

\[\frac{dV}{dt}=9x^2\]
Substitute \(x=10\) cm
The question asks for the rate of increase of volume when the edge is \(10\) cm long. Therefore,
\[x=10\]
Substituting \(x=10\) into
\[\frac{dV}{dt}=9x^2,\]
we obtain
\[\left.\frac{dV}{dt}\right|_{x=10}=9(10)^2\]
Since
\[(10)^2=100,\]
we get
\[\frac{dV}{dt}=9(100)\]

\[\frac{dV}{dt}=900\]
Since volume is measured in \(\text{cm}^3\) and time is measured in seconds, the unit is \(\text{cm}^3/\text{s}\).
\[\boxed{\frac{dV}{dt}=900\text{ cm}^3/\text{s}}\]

💡 Answer

Final Answer

›

When the edge of the cube is \(10\) cm long, its volume is increasing at the rate

\[ \boxed{900\text{ cm}^3/\text{s}}. \]

🎯 Exam Significance

Exam Significance

›

This question is a standard application of the chain rule in related rates. It demonstrates how the rate of change of one geometrical dimension can be used to determine the rate of change of a derived quantity such as volume.

The important relationship is

\[ \frac{dV}{dt} = \frac{dV}{dx}\frac{dx}{dt}. \]

For a cube,

\[ V=x^3, \]

so

\[ \frac{dV}{dx}=3x^2. \]

Hence,

\[ \frac{dV}{dt}=3x^2\frac{dx}{dt}. \]

In a board examination, students should explicitly show the differentiation with respect to time and the use of the chain rule. The final answer should also include the correct unit.

Significance for Competitive Entrance Examinations

This problem reinforces the general related-rates technique used in more involved calculus problems. The essential structure is

\[ \text{required rate} = \text{derivative with respect to the changing variable} \times \text{rate of change of that variable}. \]

For a cube, the result can be written generally as

\[ \frac{dV}{dt}=3x^2\frac{dx}{dt}. \]

This formula allows rapid evaluation once the edge length and its rate of change are known.

Competitive problems may extend the same idea to the surface area of a cube or to changing dimensions of spheres, cylinders and cones. Therefore, understanding the chain-rule relationship is more useful than simply memorising the numerical answer.

❌ Mistakes

Common Mistakes to Avoid

›

Do not confuse \(\frac{dx}{dt}\) with \(\frac{dV}{dt}\).
Do not differentiate \(V=x^3\) only with respect to \(x\) when the required rate is with respect to time.
Remember to include the chain-rule factor \(\frac{dx}{dt}\).
Do not substitute \(x=10\) before establishing the general rate equation.
The rate of volume change must have units of \(\text{cm}^3/\text{s}\), not \(\text{cm}/\text{s}\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The volume of a cube of edge length \(x\) is \(\displaystyle V=x^3\).
The given rate of increase of the edge is \(\displaystyle \frac{dx}{dt}=3\text{ cm/s}\).
Since the edge changes with time, differentiate the volume with respect to time using the chain rule.
The rate of change of volume is \(\displaystyle \frac{dV}{dt}=3x^2\frac{dx}{dt}\).
Using \(\displaystyle \frac{dx}{dt}=3\), we obtain \(\displaystyle \frac{dV}{dt}=9x^2\).
At \(x=10\) cm, \(\displaystyle \frac{dV}{dt}=900\text{ cm}^3/\text{s}\).
A rate of change of volume with respect to time is expressed in cubic units per unit time.
For related-rate questions, identify the changing variable and differentiate the geometrical relation with respect to time.

📘 Concept & Theory

Theory and Concept

›

This is a related rates problem involving the changing radius and area of a circular wave. As the wave moves outward, its radius increases with time. Consequently, the area enclosed by the wave also increases with time.

Let \(r\) be the radius of the circular wave at time \(t\), and let \(A\) be the area enclosed by the wave. The speed at which the wave expands is the rate of change of its radius:

\[ \frac{dr}{dt}=5\text{ cm/s}. \]

The enclosed area is the area of a circle:

\[ A=\pi r^2. \]

Since \(r\) changes with time, \(A\) also changes with time. Therefore, we differentiate the area equation with respect to \(t\). Using the chain rule,

\[ \frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(r\) be the radius of the circular wave and \(A\) its enclosed area.
Use the given wave speed to write \(\frac{dr}{dt}=5\text{ cm/s}\).
Express the enclosed area as \(A=\pi r^2\).
Differentiate the area with respect to time using the chain rule.
Substitute \(\frac{dr}{dt}=5\text{ cm/s}\).
Substitute \(r=8\) cm.
Simplify and state the rate of increase of area with the correct unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Define the radius and its rate of change.
Let \(r\) cm be the radius of the circular wave at time \(t\).
The waves move outward at a speed of \(5\text{ cm/s}\). Therefore, the radius of the circular wave is increasing at the rate
\[ \frac{dr}{dt}=5\text{ cm/s}. \]
Express the enclosed area
The wave encloses a circular region. Therefore, its area \(A\) is
\[A=\pi r^2\]
Differentiate the area with respect to time
Since \(r\) changes with time, differentiate both sides with respect to \(t\):
\[\frac{dA}{dt}=\frac{d}{dt}\left(\pi r^2\right)\]
Since \(\pi\) is a constant,
\[\frac{dA}{dt}=\pi\frac{d}{dt}(r^2)\]
Applying the chain rule,
\[\frac{d}{dt}(r^2)=2r\frac{dr}{dt}\]
Hence,
\[\frac{dA}{dt}=2\pi r\frac{dr}{dt}\]
Substitute the rate of increase of the radius
We are given
\[\frac{dr}{dt}=5\text{ cm/s}\]
Therefore,
\[\frac{dA}{dt}=2\pi r(5)\]

\[\frac{dA}{dt}=10\pi r.\]
Substitute \(r=8\) cm
At the specified instant, the radius of the circular wave is
\[r=8\text{ cm}\]
Therefore,
\[\left.\frac{dA}{dt}\right|_{r=8}=10\pi(8)\]

\[\frac{dA}{dt}=80\pi\]
Since the area is measured in \(\text{cm}^2\) and time is measured in seconds, the unit of the rate of change of area is \(\text{cm}^2/\text{s}\).
\[\boxed{\frac{dA}{dt}=80\pi\text{ cm}^2/\text{s}}\]

💡 Answer

Final Answer

›

When the radius of the circular wave is \(8\) cm, the enclosed area is increasing at the rate

\[ \boxed{80\pi\text{ cm}^2/\text{s}}. \]

🎯 Exam Significance

Exam Significance

›

This problem is a standard application of related rates and the chain rule. It tests whether a student can translate the physical statement that the wave moves at a particular speed into the mathematical rate

\[ \frac{dr}{dt}. \]

The important distinction is that the given speed \(5\text{ cm/s}\) describes the rate of change of the radius, not the rate of change of the area. The required rate must therefore be obtained by differentiating

\[ A=\pi r^2 \]

with respect to time.

In board examinations, students should clearly write the chain-rule step

\[ \frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt} \]

before substituting numerical values.

Significance for Competitive Entrance Examinations

This question illustrates a frequently used related-rates pattern: when a geometrical dimension changes with time, any quantity depending on that dimension also changes with time.

For a circular region, the general relationship is

\[ \frac{dA}{dt} = 2\pi r\frac{dr}{dt}. \]

This can be applied directly whenever the radius and its rate of change are known.

The same reasoning extends to expanding spherical surfaces, cylindrical containers, cones and other geometrical models in which one dimension varies with time.

❌ Mistakes

Common Mistakes to Avoid

›

Do not interpret the wave speed as \(\frac{dA}{dt}\). It represents \(\frac{dr}{dt}\).
Do not forget the chain-rule factor \(\frac{dr}{dt}\).
Do not use circumference \(2\pi r\) as the area. The enclosed area is \(\pi r^2\).
Substitute \(r=8\) cm only after obtaining the general expression for \(\frac{dA}{dt}\).
The final unit must be \(\text{cm}^2/\text{s}\), because the required quantity is the rate of change of area.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The radius of the circular wave increases at the rate \(\displaystyle \frac{dr}{dt}=5\text{ cm/s}\).
The enclosed area is \(\displaystyle A=\pi r^2\).
Using the chain rule, \(\displaystyle \frac{dA}{dt}=2\pi r\frac{dr}{dt}\).
Substituting \(\displaystyle \frac{dr}{dt}=5\) gives \(\displaystyle \frac{dA}{dt}=10\pi r\).
At \(r=8\) cm, \(\displaystyle \frac{dA}{dt}=80\pi\text{ cm}^2/\text{s}\).
The speed of an expanding circular wave represents the rate of change of its radius.
The rate of change of area depends on both the current radius and the rate at which the radius changes.
Always identify exactly which quantity is changing at the given rate before applying differentiation.

📘 Concept & Theory

Theory and Concept

›

This is a related rates problem. The radius of the circle changes with time, and therefore its circumference also changes with time.

Let \(r\) be the radius of the circle at time \(t\). The given rate of increase of the radius is

\[ \frac{dr}{dt}=0.7\text{ cm/s}. \]

The circumference \(C\) of a circle of radius \(r\) is

\[ C=2\pi r. \]

Since \(r\) changes with time, we differentiate the circumference with respect to time. Using the chain rule,

\[ \frac{dC}{dt} = \frac{dC}{dr}\frac{dr}{dt}. \]

Because the circumference depends linearly on \(r\), its rate of change with respect to time will be a constant multiple of the rate of change of the radius.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(r\) denote the radius of the circle.
Write the given rate as \(\frac{dr}{dt}=0.7\text{ cm/s}\).
Write the circumference formula \(C=2\pi r\).
Differentiate the circumference with respect to time.
Substitute the given value of \(\frac{dr}{dt}\).
Simplify the result and state the rate of increase of circumference with its correct unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Identify the given rate
The radius of the circle is increasing at the rate of \(0.7\text{ cm/s}\). Therefore,
\[\frac{dr}{dt}=0.7\text{ cm/s}\]
Write the circumference of the circle
The circumference \(C\) of a circle with radius \(r\) is
\[C=2\pi r\]
Differentiate the circumference with respect to time
Since \(r\) changes with time, differentiate both sides with respect to \(t\):
\[\frac{dC}{dt}=\frac{d}{dt}(2\pi r)\]
Since \(2\pi\) is a constant,
\[\frac{dC}{dt}=2\pi\frac{dr}{dt}\]
Thus, the general rate of change of circumference is
\[\boxed{\frac{dC}{dt}=2\pi\frac{dr}{dt}}\]
Substitute the given rate
We are given
\[\frac{dr}{dt}=0.7\text{ cm/s}\]
Therefore,
\[\frac{dC}{dt}=2\pi(0.7)\]
Hence,
\[\frac{dC}{dt}=1.4\pi\text{ cm/s}\]
If we use the commonly adopted approximation
\[\pi=\frac{22}{7},\]
then
\[\frac{dC}{dt}=1.4\times\frac{22}{7}\]
Since
\[1.4=\frac{7}{5},\]
we get
\[\frac{dC}{dt}=\frac{7}{5}\times\frac{22}{7}\]

\[\frac{dC}{dt}=\frac{22}{5}\]

\[\frac{dC}{dt}=4.4\text{ cm/s}\]

💡 Answer

Final Answer

›

The circumference of the circle is increasing at the rate

\[ \boxed{1.4\pi\text{ cm/s}}. \]

Using \(\pi=\frac{22}{7}\), this is

\[ \boxed{4.4\text{ cm/s}}. \]

🎯 Exam Significance

Exam Significance

›

This problem demonstrates a particularly simple application of related rates. The circumference is directly proportional to the radius:

\[ C=2\pi r. \]

Therefore, differentiating with respect to time immediately gives

\[ \frac{dC}{dt}=2\pi\frac{dr}{dt}. \]

Unlike the area of a circle, the rate of change of circumference does not depend on the current radius. This is because the circumference is a linear function of \(r\).

For board examinations, students should clearly distinguish between the circumference \(C\), its derivative with respect to radius \(\frac{dC}{dr}\), and its rate of change with time \(\frac{dC}{dt}\).

Significance for Competitive Entrance Examinations

This question reinforces the interpretation of derivatives as rates of change and provides a simple example of the chain rule:

\[ \frac{dC}{dt} = \frac{dC}{dr}\frac{dr}{dt}. \]

Since

\[ \frac{dC}{dr}=2\pi, \]

the circumference increases at \(2\pi\) times the rate at which the radius increases.

The same reasoning can be extended to more complicated related-rate problems in which the required quantity depends nonlinearly on a changing variable.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 6 mistakes to avoid

Do not write \(\frac{dC}{dt}=2\pi\) without multiplying by \(\frac{dr}{dt}\).
Do not confuse circumference \(2\pi r\) with area \(\pi r^2\).
The correct derivative is \(\frac{dC}{dt}=2\pi\frac{dr}{dt}\).
Do not introduce a value of \(r\), because the rate of change of circumference does not depend on the current radius.
The final unit is \(\text{cm/s}\), because circumference is a length.
The expression \(4.4\text{ cm/s}\) follows from using \(\pi=\frac{22}{7}\); the exact expression is \(1.4\pi\text{ cm/s}\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The circumference of a circle is \(\displaystyle C=2\pi r\).
The given rate of increase of radius is \(\displaystyle \frac{dr}{dt}=0.7\text{ cm/s}\).
Differentiating with respect to time gives \(\displaystyle \frac{dC}{dt}=2\pi\frac{dr}{dt}\).
Substituting the given rate gives \(\displaystyle \frac{dC}{dt}=1.4\pi\text{ cm/s}\).
Using \(\displaystyle \pi=\frac{22}{7}\), the rate is \(\displaystyle 4.4\text{ cm/s}\).
The rate of change of circumference is independent of the radius at that instant.
For a linear relationship such as \(C=2\pi r\), the derivative with respect to the changing variable is constant.

📘 Concept & Theory

Theory and Concept

›

This is a related rates problem involving two changing dimensions of a rectangle. Both the length \(x\) and width \(y\) vary with time \(t\), so the perimeter and area also vary with time.

The phrase length is decreasing means that its rate of change is negative:

\[ \frac{dx}{dt}=-5\text{ cm/min}. \]

The phrase width is increasing means that its rate of change is positive:

\[ \frac{dy}{dt}=4\text{ cm/min}. \]

The perimeter of a rectangle is

\[ P=2(x+y), \]

while its area is

\[ A=xy. \]

Since both \(x\) and \(y\) change with time, we differentiate these equations with respect to \(t\). For the area, the product rule is required because \(A\) is the product of two time-dependent quantities.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given rates \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), carefully assigning the correct signs.
For part (a), write the perimeter formula and differentiate it with respect to time.
Substitute the given rates to find \(\frac{dP}{dt}\).
For part (b), write the area formula \(A=xy\).
Differentiate using the product rule.
Substitute \(x=8\), \(y=6\), \(\frac{dx}{dt}=-5\), and \(\frac{dy}{dt}=4\).
Interpret the signs of the final answers.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given —
The length \(x\) is decreasing at \(5\text{ cm/min}\), so
\[ \frac{dx}{dt}=-5\text{ cm/min}. \]

The width \(y\) is increasing at \(4\text{ cm/min}\), so
\[ \frac{dy}{dt}=4\text{ cm/min}. \]

At the instant under consideration,
\[ x=8\text{ cm},\qquad y=6\text{ cm}. \]
Rate of Change of the Perimeter
Write the perimeter of the rectangle.
The perimeter \(P\) of a rectangle with length \(x\) and width \(y\) is
\[P=2(x+y)\]
Differentiate with respect to time.
Since \(x\) and \(y\) are functions of time \(t\), differentiate both sides with respect to \(t\):
\[\frac{dP}{dt}=\frac{d}{dt}\left[2(x+y)\right]\]
Taking the constant \(2\) outside,
\[\frac{dP}{dt}=2\frac{d}{dt}(x+y)\]
Using the sum rule,
\[\frac{d}{dt}(x+y)=\frac{dx}{dt}+\frac{dy}{dt}\]
Therefore,
\[\frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)\]
Substitute the given rates.
\[\frac{dP}{dt}=2\left(-5+4\right)\]

\[\frac{dP}{dt}=2(-1)\]

\[\frac{dP}{dt}=-2\text{ cm/min}\]
Hence,
\[\boxed{\frac{dP}{dt}=-2\text{ cm/min}}\]
The negative sign indicates that the perimeter is decreasing at \(2\text{ cm/min}\).
(b) Rate of Change of the Area
Write the area of the rectangle.
The area \(A\) is
\[A=xy\]
Differentiate with respect to time.
Since both \(x\) and \(y\) depend on time, apply the product rule:
\[\frac{dA}{dt}=\frac{d}{dt}(xy)\]
By the product rule,
\[\frac{d}{dt}(xy)=x\frac{dy}{dt}+y\frac{dx}{dt}\]
Therefore,
\[\boxed{\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}}\]
Substitute the given values.
We have
\[x=8,\quad y=6,\]

\[\frac{dx}{dt}=-5,\quad \frac{dy}{dt}=4\]
Hence,
\[\frac{dA}{dt}=8(4)+6(-5)\]

\[\frac{dA}{dt}=32-30\]

\[\frac{dA}{dt}=2\]
Therefore,
\[\boxed{\frac{dA}{dt}=2\text{ cm}^2/\text{min}}\]
The positive sign indicates that the area is increasing at \(2\text{ cm}^2/\text{min}\).

💡 Answer

Final Answer

›

Quantity	Rate of Change	Interpretation
Perimeter	\(\boxed{-2\text{ cm/min}}\)	Decreasing at \(2\text{ cm/min}\)
Area	\(\boxed{2\text{ cm}^2/\text{min}}\)	Increasing at \(2\text{ cm}^2/\text{min}\)

🎯 Exam Significance

Exam Significance

›

This question is important because it combines related rates, the sum rule and the product rule. Unlike earlier problems where only one dimension changed, here both the length and width change simultaneously.

The signs of the rates are particularly important:

\[ \frac{dx}{dt}<0 \]

because the length is decreasing, whereas

\[ \frac{dy}{dt}>0 \]

because the width is increasing.

For the perimeter, the rates are added:

\[ \frac{dP}{dt} = 2\left( \frac{dx}{dt}+\frac{dy}{dt} \right). \]

For the area, the product rule must be used:

\[ \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}. \]

The negative answer for the perimeter does not indicate an error. It means that the perimeter is decreasing at that instant. Similarly, the positive area rate means that the area is increasing.

Significance for Competitive Entrance Examinations

This problem develops the ability to handle multiple simultaneously changing variables. Such problems require careful use of differentiation rules rather than direct substitution.

The area relation

\[ A=xy \]

is particularly important because differentiating a product of two time-dependent quantities requires the product rule:

\[ \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}. \]

A useful strategy for competitive problems is to differentiate the complete relation first, then substitute all numerical values. This reduces the possibility of losing a term or assigning an incorrect sign.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 7 mistakes to avoid

Since \(x\) is decreasing, use \(\displaystyle \frac{dx}{dt}=-5\), not \(+5\).
Since \(y\) is increasing, use \(\displaystyle \frac{dy}{dt}=4\).
Do not differentiate \(A=xy\) as \(\displaystyle \frac{dA}{dt}=\frac{dx}{dt}\frac{dy}{dt}\). The correct rule is the product rule.
For area, the correct expression is \(\displaystyle \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\).
Do not ignore the negative sign in \(\frac{dx}{dt}\).
Use \(\text{cm/min}\) for the rate of change of perimeter and \(\text{cm}^2/\text{min}\) for the rate of change of area.
Interpret the sign of the final derivative: negative means decreasing and positive means increasing.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 10 points

The perimeter of a rectangle is \(\displaystyle P=2(x+y)\).
The area of a rectangle is \(\displaystyle A=xy\).
A decreasing length has a negative rate: \(\displaystyle \frac{dx}{dt}=-5\text{ cm/min}\).
An increasing width has a positive rate: \(\displaystyle \frac{dy}{dt}=4\text{ cm/min}\).
The perimeter rate is \(\displaystyle \frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)\).
At the given instant, \(\displaystyle \frac{dP}{dt}=-2\text{ cm/min}\).
The area rate is \(\displaystyle \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\).
At \(x=8\) cm and \(y=6\) cm, \(\displaystyle \frac{dA}{dt}=2\text{ cm}^2/\text{min}\).
The perimeter is decreasing while the area is increasing at the specified instant.
When more than one variable changes with time, differentiate the complete relation before substituting numerical values.

📘 Concept & Theory

Theory and Concept

›

This is a related rates problem involving the volume and radius of a sphere. As air is pumped into the balloon, its volume increases. Because the balloon always remains spherical, its radius also changes with time.

Let \(r\) be the radius of the balloon at time \(t\), and \(V\) be its volume. The given rate at which air is being pumped into the balloon is the rate of increase of its volume:

\[ \frac{dV}{dt}=900\text{ cm}^3/\text{s}. \]

The volume of a sphere of radius \(r\) is

\[ V=\frac{4}{3}\pi r^3. \]

Since the radius changes with time, differentiate the volume equation with respect to time. Using the chain rule,

\[ \frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt}. \]

The required quantity is

\[ \frac{dr}{dt}, \]

evaluated when

\[ r=15\text{ cm}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(r\) be the radius of the spherical balloon.
Interpret the pumping rate as \(\frac{dV}{dt}=900\text{ cm}^3/\text{s}\).
Write the volume of the sphere as \(\displaystyle V=\frac{4}{3}\pi r^3\).
Differentiate the volume with respect to time using the chain rule.
Solve the resulting equation for \(\frac{dr}{dt}\).
Substitute \(r=15\) cm.
Simplify and state the rate of increase of the radius with the correct unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Given —
The volume of the balloon is increasing at the rate
\[ \frac{dV}{dt}=900\text{ cm}^3/\text{s}. \]

The radius at the instant under consideration is
\[ r=15\text{ cm}. \]
Write the volume of the spherical balloon
Since the balloon remains spherical, its volume is
\[V=\frac{4}{3}\pi r^3\]
Differentiate the volume with respect to time
Since both \(V\) and \(r\) change with time, differentiate both sides with respect to \(t\):
\[\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)\]
Since \(\frac{4}{3}\pi\) is a constant,
\[\frac{dV}{dt}=\frac{4}{3}\pi\frac{d}{dt}(r^3)\]
Applying the chain rule,
\[\frac{d}{dt}(r^3)=3r^2\frac{dr}{dt}\]
Therefore,
\[\frac{dV}{dt}=\frac{4}{3}\pi\left(3r^2\frac{dr}{dt}\right)\]
Cancelling \(3\), we obtain
\[\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\]
Substitute the given rate of volume change
We are given
\[\frac{dV}{dt}=900\text{ cm}^3/\text{s}\]
Therefore,
\[900=4\pi r^2\frac{dr}{dt}\]
Solving for \(\frac{dr}{dt}\),
\[\frac{dr}{dt}=\frac{900}{4\pi r^2}\]
Substitute \(r=15\) cm
At the specified instant,
\[r=15\text{ cm}\]
Hence,
\[\frac{dr}{dt}=\frac{900}{4\pi(15)^2}\]
Since
\[(15)^2=225,\]
we get
\[\frac{dr}{dt}=\frac{900}{4\pi(225)}\]
Simplifying the denominator,
\[4(225)=900\]
Therefore,
\[\frac{dr}{dt}=\frac{900}{900\pi}\]

\[\frac{dr}{dt}=\frac{1}{\pi}\]
Since the radius is measured in centimetres and time in seconds, the unit is \(\text{cm/s}\).
\[\boxed{\frac{dr}{dt}=\frac{1}{\pi}\text{ cm/s}}\]

💡 Answer

Final Answer

›

When the radius of the balloon is \(15\) cm, the radius is increasing at the rate

\[ \boxed{\frac{1}{\pi}\text{ cm/s}}. \]

🎯 Exam Significance

Exam Significance

›

This problem is an important application of related rates and the chain rule. It demonstrates how a known rate of change of volume can be used to determine the rate of change of the radius.

The central relationship is

\[ \frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt}. \]

For a sphere,

\[ V=\frac{4}{3}\pi r^3, \]

so

\[ \frac{dV}{dr}=4\pi r^2. \]

Therefore,

\[ \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}. \]

In a board examination, students should clearly show the differentiation and the substitution of the given rate. The unit of the final answer should also be stated because the required quantity is a rate of change of length.

Significance for Competitive Entrance Examinations

This problem develops the standard related-rates framework for spherical geometry. The general result

\[ \frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt} \]

shows how the rate of change of radius depends on both the volume-injection rate and the current radius.

An important observation is that, for a fixed volume-injection rate, the rate of increase of the radius decreases as the radius becomes larger because the factor \(4\pi r^2\) increases.

Similar questions can involve the surface area of a sphere, the volume of a cylinder or cone, or multiple changing geometrical quantities.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 6 mistakes to avoid

Do not interpret \(900\text{ cm}^3/\text{s}\) as the rate of change of radius. It is \(\frac{dV}{dt}\).
Use the volume formula for a sphere: \(\displaystyle V=\frac{4}{3}\pi r^3\).
Do not forget the chain-rule factor \(\frac{dr}{dt}\).
Do not substitute \(r=15\) before differentiating the volume relation.
Remember that \(\displaystyle 4(15)^2=900\), which simplifies the calculation significantly.
The final unit must be \(\text{cm/s}\), because the required quantity is the rate of change of radius.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The balloon remains spherical throughout inflation.
The volume of a sphere is \(\displaystyle V=\frac{4}{3}\pi r^3\).
The given volume rate is \(\displaystyle \frac{dV}{dt}=900\text{ cm}^3/\text{s}\).
Differentiating with respect to time gives \(\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\).
Hence, \(\displaystyle \frac{dr}{dt}=\frac{900}{4\pi r^2}\).
At \(r=15\) cm, \(\displaystyle \frac{dr}{dt}=\frac{1}{\pi}\text{ cm/s}\).
The rate of increase of radius is not constant even though the rate of increase of volume is constant.
For related-rate problems, differentiate the geometrical relationship with respect to time before inserting the numerical values.

📘 Concept & Theory

Theory and Concept

›

This question concerns the rate of change of volume with respect to radius. Since the balloon remains spherical, its volume can be expressed directly in terms of its radius \(r\).

The volume \(V\) of a sphere of radius \(r\) is

\[ V=\frac{4}{3}\pi r^3. \]

The rate at which volume changes with respect to radius is represented by

\[ \frac{dV}{dr}. \]

Therefore, we differentiate the volume formula with respect to \(r\).

It is important to distinguish between

\[ \frac{dV}{dr} \]

and

\[ \frac{dV}{dt}. \]

The former gives the rate of change of volume with respect to radius, whereas the latter gives the rate of change of volume with respect to time. Since no time-dependent rate is given in this question, the required derivative is \(\frac{dV}{dr}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the volume of the spherical balloon in terms of its radius.
Differentiate the volume with respect to \(r\).
Substitute the given radius \(r=15\) cm.
Simplify the result and state the appropriate unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Write the volume of the spherical balloon.
Since the balloon remains spherical, its volume is
\[V=\frac{4}{3}\pi r^3\]
Find the rate of change of volume with respect to radius
Differentiate both sides with respect to \(r\):
\[\frac{dV}{dr}=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)\]
Since \(\frac{4}{3}\pi\) is a constant,
\[\frac{dV}{dr}=\frac{4}{3}\pi\frac{d}{dr}(r^3)\]
Using the power rule,
\[\frac{d}{dr}(r^3)=3r^2\]
Therefore,
\[\frac{dV}{dr}=\frac{4}{3}\pi(3r^2)\]
Cancelling \(3\),
\[\frac{dV}{dr}=4\pi r^2\]
Substitute \(r=15\) cm
The question specifies that the radius is \(15\) cm. Therefore,
\[r=15\]
Substituting \(r=15\) into
\[\frac{dV}{dr}=4\pi r^2,\]
we obtain
\[\left.\frac{dV}{dr}\right|_{r=15}=4\pi(15)^2\]
Since
\[(15)^2=225,\]
we get
\[\frac{dV}{dr}=4\pi(225)\]

\[\frac{dV}{dr}=900\pi\]
Because volume is measured in \(\text{cm}^3\) and radius in cm, the unit of \(\frac{dV}{dr}\) is
\[\frac{\text{cm}^3}{\text{cm}}=\text{cm}^2\]

\[\boxed{\frac{dV}{dr}=900\pi\text{ cm}^2}\]

💡 Answer

Final Answer

›

When the radius is \(15\) cm, the rate of change of the volume with respect to the radius is

\[\boxed{900\pi\text{ cm}^2}\]

🎯 Exam Significance

Exam Significance

›

This question tests the fundamental interpretation of a derivative as the rate of change of one quantity with respect to another.

Here,

\[ \frac{dV}{dr} \]

measures how rapidly the volume changes when the radius changes. It does not represent the rate of change with respect to time.

This distinction is important in board examinations because questions in the chapter may ask for

\[ \frac{dV}{dr},\qquad \frac{dV}{dt},\qquad \frac{dr}{dt}, \]

which represent different quantities and require different information.

Significance for Competitive Entrance Examinations

The result

\[ \frac{dV}{dr}=4\pi r^2 \]

is an important derivative for spherical geometry. It connects the volume of a sphere with its radius and can also be used as part of a related-rates calculation.

If the radius itself changes with time, the chain rule gives

\[ \frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}. \]

Thus, this question provides the mathematical step that underlies more advanced spherical related-rate problems.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 5 mistakes to avoid

Do not substitute \(r=10\) cm when the question gives \(r=15\) cm.
Do not confuse \(\frac{dV}{dr}\) with \(\frac{dV}{dt}\).
Remember that \(\displaystyle \frac{d}{dr}(r^3)=3r^2\).
Do not forget to simplify \(\frac{4}{3}\times3\) to \(4\).
The unit of \(\frac{dV}{dr}\) is \(\text{cm}^2\), because it represents \(\text{cm}^3/\text{cm}\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The volume of a sphere is \(\displaystyle V=\frac{4}{3}\pi r^3\).
The rate of change of volume with respect to radius is \(\displaystyle \frac{dV}{dr}=4\pi r^2\).
At \(r=15\) cm, \(\displaystyle \frac{dV}{dr}=900\pi\text{ cm}^2\).
The given solution's substitution \(r=10\) is incorrect for the stated question.
\(\frac{dV}{dr}\) and \(\frac{dV}{dt}\) have different meanings and units.
If time is introduced, the chain rule gives \(\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\).

📘 Concept & Theory

Theory and Concept

›

This is a standard related rates problem. The ladder, wall and ground form a right-angled triangle. Since the ladder has a fixed length, the horizontal distance of its foot from the wall and its vertical height on the wall are related by the Pythagorean theorem.

Let \(x\) be the distance of the foot of the ladder from the wall and \(y\) be the height of the ladder on the wall. Since the ladder has constant length 5 m,

\[ x^2+y^2=25 \]

Both \(x\) and \(y\) change with time. Therefore, differentiating the relation with respect to time \(t\) gives a relationship between \(\dfrac{dx}{dt}\) and \(\dfrac{dy}{dt}\).

The word “away” indicates that the foot is moving farther from the wall, so \(\dfrac{dx}{dt}\) is positive. The height is decreasing, so \(\dfrac{dy}{dt}\) will be negative.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(x\) be the horizontal distance of the foot from the wall and \(y\) be the vertical height.
Use the fixed ladder length to obtain \(x^2+y^2=25\).
Differentiate the equation with respect to time.
Find \(y\) when \(x=4\) m.
Substitute \(x=4\), \(y=3\), and \(\dfrac{dx}{dt}=2\) cm/s.
Convert the units consistently and determine \(\dfrac{dy}{dt}\).
Interpret the negative sign as a decrease in height.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

📌
Given
—
- Length of ladder \(=5\) m
- Rate at which the foot moves away from the wall \(=2\) cm/s
- Distance of the foot from the wall \(=4\) m
- Required: Rate at which the height of the ladder is decreasing
Define the variables
Let \(x\) m be the distance of the foot of the ladder from the wall at time \(t\).
Let \(y\) m be the height of the ladder on the wall at time \(t\).
The length of the ladder remains constant and is 5 m.
Establish the relation using Pythagorean theorem
The ladder, wall and ground form a right-angled triangle. Therefore,
\[x^2+y^2=5^2\]
Hence,
\[x^2+y^2=25\]
Differentiate with respect to time
Since both \(x\) and \(y\) are functions of time \(t\), differentiate both sides with respect to \(t\):
\[\frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(25)\]
Applying the chain rule,
\[2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\]
Dividing throughout by \(2\),
\[x\frac{dx}{dt}+y\frac{dy}{dt}=0\]
Therefore,
\[x\frac{dx}{dt}=-y\frac{dy}{dt}\]
Hence,
\[\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\]
Find the height \(y\) when \(x=4\) m
At the instant under consideration,
\[x=4\text{ m}\]
Using
\[x^2+y^2=25,\]
we get
\[\begin{aligned}4^2+y^2&=25\\16+y^2&=25\\y^2&=25-16\\\&y^2&=9\end{aligned}\]
Since height is positive,
\[y=3\text{ m}\]
Convert the given rate into consistent units
The horizontal distance \(x\) is measured in metres, while the given rate is in centimetres per second. Therefore, convert \(2\) cm/s into m/s:
\[\frac{dx}{dt}=2\text{ cm/s}=\frac{2}{100}\text{ m/s}=\frac{1}{50}\text{ m/s}\]
Substitute the values
We have
\[\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\]
Substituting \(x=4\) m, \(y=3\) m and \(\dfrac{dx}{dt}=\dfrac{1}{50}\) m/s,
\[\frac{dy}{dt}=-\frac{4}{3}\left(\frac{1}{50}\right)\]
Therefore,
\[\frac{dy}{dt}=-\frac{4}{150}\]
Simplifying,
\[\frac{dy}{dt}=-\frac{2}{75}\text{ m/s}\]
Express the answer in cm/s
Since
\[1\text{ m}=100\text{ cm},\]
we have
\[\frac{dy}{dt}=-\frac{2}{75}\times100\text{ cm/s}\]
Hence,
\[\frac{dy}{dt}=-\frac{200}{75}\text{ cm/s}\]
Therefore,
\[\boxed{\frac{dy}{dt}=-\frac{8}{3}\text{ cm/s}}\]

💡 Answer

Final Answer

›

The height of the ladder on the wall is decreasing at the rate

\[\boxed{\frac{8}{3}\text{ cm/s}}\]

Equivalently,

\[\boxed{\frac{dy}{dt}=-\frac{8}{3}\text{ cm/s}}\]

The negative sign indicates that the height of the ladder is decreasing.

🎯 Exam Significance

Exam Significance

›

This problem is an important application of related rates and tests whether the student can translate a geometrical relationship into a differential equation. The key step is differentiating the Pythagorean relation with respect to time rather than differentiating only with respect to \(x\) or \(y\).

For board examinations, students should clearly show the formation of

\[ x^2+y^2=25 \]

followed by differentiation:

\[ x\frac{dx}{dt}+y\frac{dy}{dt}=0 \]

The calculation of the instantaneous value \(y=3\) m should also be shown before substitution. Finally, the negative sign should be interpreted explicitly as a decrease in height.

Significance for Competitive Entrance Examinations

Related-rate questions commonly test the ability to identify dependent variables, differentiate implicitly with respect to time, maintain consistent units, and interpret the sign of a rate. A quick method is to remember the directly useful relation

\[ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \]

once the fixed-length ladder relation

\[ x^2+y^2=L^2 \]

has been established.

Competitive problems may alter the ladder length, horizontal rate, position, or ask for the rate of change of a different quantity. The same differentiation framework applies.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 6 mistakes to avoid

Forgetting the time variable: Both \(x\) and \(y\) vary with time, so differentiate with respect to \(t\).
Missing the chain rule: The derivative of \(x^2\) with respect to time is \(2x\dfrac{dx}{dt}\), not merely \(2x\).
Using \(y=4\) m: The given 4 m is the horizontal distance \(x\), not the height \(y\).
Ignoring units: The distances are in metres, whereas the given rate is in cm/s. Convert units before substitution.
Dropping the negative sign: \(\dfrac{dy}{dt}<0\) indicates that the height is decreasing.
Giving the wrong quantity: The question asks for the speed of decrease of height, so the magnitude is \(\dfrac{8}{3}\) cm/s.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

A fixed-length ladder creates a right triangle whose sides vary with time.
Use the Pythagorean relation:
\[ x^2+y^2=25. \]
Differentiate with respect to time:
\[ x\frac{dx}{dt}+y\frac{dy}{dt}=0. \]
At \(x=4\) m, the corresponding height is \(y=3\) m.
The given \(2\) cm/s must be converted to \(\dfrac{1}{50}\) m/s when using metre-based quantities.
The height changes at
\[ \frac{dy}{dt}=-\frac{8}{3}\text{ cm/s}. \]
Therefore, the height is decreasing at
\[ \boxed{\frac{8}{3}\text{ cm/s}}. \]

📘 Concept & Theory

Theory and Concept

›

This is a related rates problem involving a particle moving along a curve. Since both \(x\) and \(y\) change with time \(t\), we differentiate the equation of the curve with respect to time.

The statement “the \(y\)-coordinate is changing 8 times as fast as the \(x\)-coordinate” means

\[ \frac{dy}{dt}=8\frac{dx}{dt}. \]

This relation must be substituted into the differentiated equation to determine the possible values of \(x\). The corresponding values of \(y\) are then obtained from the original equation of the curve.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the equation \(6y=x^3+2\) with respect to time \(t\).
Use the condition \(\dfrac{dy}{dt}=8\dfrac{dx}{dt}\).
Solve for the possible values of \(x\).
Substitute each value of \(x\) into the original equation to find \(y\).
Write the corresponding points on the curve.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Given —
The equation of the curve is
\[ 6y=x^3+2. \]

The \(y\)-coordinate is changing 8 times as fast as the \(x\)-coordinate:
\[ \frac{dy}{dt}=8\frac{dx}{dt}. \]
Differentiate the equation of the curve with respect to time
We have
\[6y=x^3+2\]
Differentiating both sides with respect to \(t\),
\[\frac{d}{dt}(6y)=\frac{d}{dt}(x^3+2)\]
Since \(x\) and \(y\) are functions of \(t\), applying the chain rule gives
\[6\frac{dy}{dt}=3x^2\frac{dx}{dt}\]
Dividing both sides by \(3\),
\[2\frac{dy}{dt}=x^2\frac{dx}{dt}\]
Use the given rate condition
The \(y\)-coordinate is changing 8 times as fast as the \(x\)-coordinate. Therefore,
\[\frac{dy}{dt}=8\frac{dx}{dt}\]
Substitute this into
\[6\frac{dy}{dt}=3x^2\frac{dx}{dt}\]
Thus,
\[6\left(8\frac{dx}{dt}\right)=3x^2\frac{dx}{dt}\]
Therefore,
\[48\frac{dx}{dt}=3x^2\frac{dx}{dt}\]
Assuming the particle is actually moving in the \(x\)-direction, \(\dfrac{dx}{dt}\neq0\). Hence, we can divide both sides by \(\dfrac{dx}{dt}\):
\[48=3x^2\]
Dividing by \(3\),
\[x^2=16\]
Therefore,
\[x=\pm4\]
Find the corresponding \(y\)-coordinate when \(x=4\)
The equation of the curve is
\[6y=x^3+2\]
For \(x=4\),
\[\begin{aligned} 6y&=4^3+2\\6y&=64+2\\6y&=66\\y&=11 \end{aligned} \]
Therefore, one point is
\[(4,11)\]
Find the corresponding \(y\)-coordinate when \(x=-4\)
For \(x=-4\),
\[\begin{aligned}6y&=(-4)^3+2\\6y&=-64+2\\6y&=-62\\y&=-\frac{62}{6}\\&=-\frac{31}{3}\end{aligned}\]
Hence, the second point is
\[ \left(-4,-\frac{31}{3}\right). \]

💡 Answer

Final Answer

›

The required points on the curve are

\[\boxed{(4,11)}\]

and

\[\boxed{\left(-4,-\frac{31}{3}\right)}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the application of differentiation to a moving particle and is a useful example of a related-rates question. The most important step is correctly translating the phrase “8 times as fast” into the mathematical relation

\[ \frac{dy}{dt}=8\frac{dx}{dt}. \]

For a board examination, students should clearly show the differentiation of the curve with respect to time and then substitute the given rate relationship. Both possible values of \(x\) must be considered because

\[ x^2=16 \]

gives \(x=4\) as well as \(x=-4\).

Significance for Competitive Entrance Examinations

This type of problem tests careful interpretation of rate relationships and implicit differentiation. A common trap is reversing the given rate condition. If one quantity changes \(k\) times as fast as another, the statement must be translated in the same order as given.

Here,

\[ y\text{ changes 8 times as fast as }x \quad\Longrightarrow\quad \frac{dy}{dt}=8\frac{dx}{dt}. \]

Another important point is that an equation such as

\[ x^2=16 \]

has two real solutions. Unless the problem imposes a restriction on the position of the particle, both corresponding points must be examined.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 6 mistakes to avoid

Reversing the rate relationship: Do not write \(\dfrac{dx}{dt}=8\dfrac{dy}{dt}\).
Forgetting the chain rule: The derivative of \(x^3\) with respect to time is \(3x^2\dfrac{dx}{dt}\).
Differentiating with respect to \(x\) only: Since the question concerns rates with respect to time, differentiate with respect to \(t\).
Keeping only \(x=4\): Since \(x^2=16\), both \(x=4\) and \(x=-4\) must be considered.
Incorrect substitution: For \(x=-4\), remember that \((-4)^3=-64\).
Ignoring the motion condition: Cancelling \(\dfrac{dx}{dt}\) assumes that \(\dfrac{dx}{dt}\neq0\), which is consistent with the particle's stated changing coordinates.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

For a moving particle, differentiate the curve equation with respect to time.
The given condition translates to
\[ \frac{dy}{dt}=8\frac{dx}{dt}. \]
From
\[ 6y=x^3+2, \]
differentiation gives
\[ 6\frac{dy}{dt}=3x^2\frac{dx}{dt}. \]
Substitution of the rate condition gives
\[ x^2=16. \]
Thus,
\[ x=\pm4. \]
The corresponding points are
\[ \boxed{(4,11)} \quad\text{and}\quad \boxed{\left(-4,-\frac{31}{3}\right)}. \]

📘 Concept & Theory

Theory and Concept

›

This is a standard related rates problem. The volume of an air bubble is determined by its radius. Since the radius changes with time, the volume also changes with time.

For a spherical bubble, the volume is

\[ V=\frac{4}{3}\pi r^3. \]

To find how fast the volume is changing, differentiate the volume equation with respect to time \(t\):

\[ \frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt}. \]

Thus, the rate of change of volume depends on both the instantaneous radius \(r\) and the rate at which the radius is changing \(\dfrac{dr}{dt}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the formula for the volume of a sphere.
Differentiate the volume with respect to time.
Substitute the given radius \(r=1\) cm.
Substitute the given rate \(\dfrac{dr}{dt}=\dfrac{1}{2}\) cm/s.
Simplify to obtain the rate of increase of volume.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

📌
Given
—
- Radius of the bubble:
  \[ r=1\text{ cm} \]
- Rate of increase of radius:
  \[ \frac{dr}{dt}=\frac{1}{2}\text{ cm/s} \]
- Required: Rate of increase of volume,
  \[ \frac{dV}{dt} \]
Write the volume of the spherical bubble
Since the air bubble is spherical, its volume is
\[V=\frac{4}{3}\pi r^3\]
Differentiate with respect to time
Since the radius \(r\) changes with time, differentiate both sides with respect to \(t\):
\[\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)\]
Taking the constant \(\frac{4}{3}\pi\) outside the derivative,
\[\frac{dV}{dt}=\frac{4}{3}\pi\frac{d}{dt}(r^3)\]
Applying the chain rule,
\[\frac{d}{dt}(r^3)=3r^2\frac{dr}{dt}\]
Therefore,
\[\frac{dV}{dt}=\frac{4}{3}\pi\left(3r^2\frac{dr}{dt}\right)\]
Cancelling \(3\),
\[\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\]
Substitute the given values
We are given
\[r=1\text{ cm}\]
and
\[\frac{dr}{dt}=\frac{1}{2}\text{ cm/s}\]
Hence,
\[\frac{dV}{dt}=4\pi(1)^2\left(\frac{1}{2}\right)\]
Since
\[(1)^2=1,\]
we get
\[\frac{dV}{dt}=4\pi\left(\frac{1}{2}\right)\]
Therefore,
\[\boxed{\frac{dV}{dt}=2\pi\text{ cm}^3/\text{s}}\]

💡 Answer

Final Answer

›

The volume of the air bubble is increasing at the rate

\[\boxed{2\pi\text{ cm}^3/\text{s}}\]

🎯 Exam Significance

Exam Significance

›

This problem is a direct application of the concept of related rates. It is important to understand that the required quantity is \(\dfrac{dV}{dt}\), not simply \(\dfrac{dV}{dr}\).

The essential chain-rule relationship is

\[ \frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt}. \]

Students should also remember that the volume of a sphere is

\[ V=\frac{4}{3}\pi r^3, \]

and its rate of change is

\[ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}. \]

Significance for Competitive Entrance Examinations

This question tests the ability to connect two changing quantities through the chain rule. Such questions can be solved quickly once the geometric formula and the corresponding differential relationship are identified.

A useful general result for a spherical object is

\[ \boxed{\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}}. \]

The instantaneous value of \(r\) must be substituted only after differentiation. Substituting \(r=1\) before differentiating is unnecessary and can lead to mistakes in more complicated related-rate problems.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 5 mistakes to avoid

Using the wrong formula: The bubble is spherical, so use \(V=\frac{4}{3}\pi r^3\).
Forgetting the chain rule: \(\dfrac{d}{dt}(r^3)=3r^2\dfrac{dr}{dt}\).
Confusing \(\dfrac{dV}{dr}\) with \(\dfrac{dV}{dt}\): The question asks for the rate of change with respect to time.
Forgetting the units: Since volume is measured in cm³ and time in seconds, the final unit is cm³/s.
Using \(r=1\) incorrectly: Remember that \(r^2=(1)^2=1\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The volume of a spherical bubble is
\[ V=\frac{4}{3}\pi r^3. \]
Differentiate with respect to time:
\[ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}. \]
At \(r=1\) cm,
\[ \frac{dr}{dt}=\frac{1}{2}\text{ cm/s}. \]
Therefore,
\[ \frac{dV}{dt} = 4\pi(1)^2\left(\frac{1}{2}\right) = 2\pi\text{ cm}^3/\text{s}. \]
Hence, the volume is increasing at
\[ \boxed{2\pi\text{ cm}^3/\text{s}}. \]

📘 Concept & Theory

Theory and Concept

›

Since the balloon always remains spherical, its volume is related to its radius by

\[ V=\frac{4}{3}\pi r^3. \]

However, the radius is not given directly. The diameter is given as a function of \(x\). Therefore, the solution requires a chain of relationships:

\[ x\longrightarrow d\longrightarrow r\longrightarrow V. \]

We first express the radius in terms of \(x\), substitute it into the volume formula, and then differentiate with respect to \(x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given diameter as a function of \(x\).
Use \(r=\dfrac{d}{2}\) to obtain the radius in terms of \(x\).
Substitute the radius into the formula for the volume of a sphere.
Differentiate the resulting expression with respect to \(x\).
Simplify the result carefully.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 17 steps

Given —
The diameter of the spherical balloon is

\[ d=\frac{3}{2}(2x+1). \]

Required:

\[ \frac{dV}{dx}. \]
Express the radius in terms of \(x\)
We know that the radius is half of the diameter:
\[r=\frac{d}{2}\]
Substituting the given value of \(d\),
\[r=\frac{1}{2}\left[\frac{3}{2}(2x+1)\right]\]
Therefore,
\[r=\frac{3}{4}(2x+1)\]
Write the volume of the spherical balloon
The volume of a sphere is
\[V=\frac{4}{3}\pi r^3\]
Substituting
\[r=\frac{3}{4}(2x+1),\]
we get
\[V=\frac{4}{3}\pi\left[\frac{3}{4}(2x+1)\right]^3\]
Using
\[(ab)^3=a^3b^3,\]
we obtain
\[V=\frac{4}{3}\pi\left(\frac{3}{4}\right)^3(2x+1)^3\]
Differentiate with respect to \(x\)
Differentiating both sides with respect to \(x\),
\[\frac{dV}{dx}=\frac{4}{3}\pi\left(\frac{3}{4}\right)^3\frac{d}{dx}(2x+1)^3\]
Applying the chain rule,
\[\frac{d}{dx}(2x+1)^3=3(2x+1)^2\frac{d}{dx}(2x+1)\]
Since
\[\frac{d}{dx}(2x+1)=2,\]
we get
\[\frac{d}{dx}(2x+1)^3=3(2x+1)^2(2)\]
Therefore,
\[\frac{dV}{dx}=\frac{4}{3}\pi\left(\frac{3}{4}\right)^3\left[3(2x+1)^2(2)\right]\]
Simplify the constant factor
Since
\[\left(\frac{3}{4}\right)^3=\frac{27}{64},\]
we have
\[\frac{dV}{dx}=\frac{4}{3}\pi\cdot\frac{27}{64}\cdot3\cdot2(2x+1)^2\]
Simplifying the numerical factors,
\[\frac{4}{3}\cdot\frac{27}{64}\cdot3\cdot2=\frac{27}{8}\]
Hence,
\[\boxed{\frac{dV}{dx}=\frac{27\pi}{8}(2x+1)^2}\]

💡 Answer

Final Answer

›

The rate of change of the volume of the balloon with respect to \(x\) is

\[\boxed{\frac{dV}{dx}=\frac{27\pi}{8}(2x+1)^2}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the application of differentiation and the chain rule to a geometrical quantity. The important sequence is

\[ d\rightarrow r\rightarrow V\rightarrow\frac{dV}{dx}. \]

Students should carefully distinguish between diameter and radius. Since

\[ r=\frac{d}{2}, \]

the given diameter must first be divided by \(2\) before it is used in the sphere-volume formula.

Significance for Competitive Entrance Examinations

This problem is useful for practicing nested functions and the chain rule. Instead of expanding the entire expression, it is often faster to use

\[ \frac{dV}{dx} = \frac{dV}{dr}\frac{dr}{dx}. \]

This approach reduces algebraic manipulation and lowers the possibility of losing the inner derivative

\[ \frac{d}{dx}(2x+1)=2. \]

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 5 mistakes to avoid

Using diameter as radius: Remember that \(r=\dfrac{d}{2}\).
Forgetting the chain-rule factor \(2\): The derivative of \((2x+1)^3\) contains the factor \(2\).
Changing \(2x+1\) to \(x+1\): This is algebraically incorrect.
Expanding unnecessarily: Keeping \((2x+1)^3\) in factored form makes differentiation easier.
Confusing \(\dfrac{dV}{dr}\) with \(\dfrac{dV}{dx}\): The required derivative is with respect to \(x\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The given diameter is
\[ d=\frac{3}{2}(2x+1). \]
The corresponding radius is
\[ r=\frac{3}{4}(2x+1). \]
The volume is
\[ V=\frac{4}{3}\pi r^3. \]
Using the chain rule,
\[ \frac{dV}{dx} = \frac{dV}{dr}\frac{dr}{dx}. \]
The correct rate of change is
\[ \boxed{ \frac{dV}{dx} = \frac{27\pi}{8}(2x+1)^2 }. \]
The supplied solution contains an algebraic error when it changes \((2x+1)^2\) to \((x+1)^2\).

📘 Concept & Theory

Theory and Concept

›

This is a related rates problem involving the volume of a cone. Sand is being poured continuously, so the volume of the sand cone changes with time.

The volume of a cone is

\[ V=\frac{1}{3}\pi r^2h, \]

where \(r\) is the radius of the circular base and \(h\) is the height.

The problem also gives a fixed relationship between \(r\) and \(h\):

\[ h=\frac{r}{6}. \]

Therefore, the radius can be expressed in terms of the height as

\[ r=6h. \]

Substituting this relationship into the volume formula allows us to express \(V\) entirely in terms of \(h\). We can then differentiate with respect to time and find the required rate of increase of height.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(r\) be the radius of the sand cone and \(h\) its height.
Use the given relationship \(h=\dfrac{r}{6}\) to obtain \(r=6h\).
Write the volume of the cone in terms of \(h\).
Differentiate the volume equation with respect to time.
Substitute \(\dfrac{dV}{dt}=12\text{ cm}^3/\text{s}\) and \(h=4\) cm.
Solve for \(\dfrac{dh}{dt}\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

📌
Given
—
- Rate of pouring of sand:
  \[ \frac{dV}{dt}=12\text{ cm}^3/\text{s} \]
- Relationship between height and radius:
  \[ h=\frac{r}{6} \]
- Height at the required instant:
  \[ h=4\text{ cm} \]
- Required:
  \[ \frac{dh}{dt} \]
Establish the relationship between radius and height
According to the question, the height of the sand cone is always one-sixth of the radius of its base:
\[h=\frac{r}{6}\]
Multiplying both sides by \(6\),
\[r=6h\]
Write the volume of the sand cone
The volume of a cone is
\[V=\frac{1}{3}\pi r^2h\]
Since
\[r=6h,\]
substitute this into the volume formula:
\[V=\frac{1}{3}\pi(6h)^2h\]
Squaring \(6h\),
\[V=\frac{1}{3}\pi(36h^2)h\]
Therefore,
\[V=12\pi h^3\]
Differentiate with respect to time
Since both \(V\) and \(h\) change with time, differentiate both sides with respect to \(t\):
\[\frac{dV}{dt}=\frac{d}{dt}(12\pi h^3)\]
Taking \(12\pi\) as constant,
\[\frac{dV}{dt}=12\pi\frac{d}{dt}(h^3)\]
Applying the chain rule,
\[\frac{d}{dt}(h^3)=3h^2\frac{dh}{dt}\]
Therefore,
\[\frac{dV}{dt}=12\pi\left(3h^2\frac{dh}{dt}\right).\]
Hence,
\[\frac{dV}{dt}=36\pi h^2\frac{dh}{dt}\]
Substitute the given values
We are given
\[\frac{dV}{dt}=12\text{ cm}^3/\text{s}\]
and
\[h=4\text{ cm}\]
Substituting these values,
\[12=36\pi(4)^2\frac{dh}{dt}\]
Since
\[4^2=16,\]
we get
\[12=36\pi(16)\frac{dh}{dt}\]
Therefore,
\[12=576\pi\frac{dh}{dt}\]
Solve for the rate of increase of height
Dividing both sides by \(576\pi\),
\[\frac{dh}{dt}=\frac{12}{576\pi}\]
Simplifying by dividing numerator and denominator by \(12\),
\[\frac{dh}{dt}=\frac{1}{48\pi}\text{ cm/s}\]

💡 Answer

Final Answer

›

When the height of the sand cone is \(4\) cm, its height is increasing at the rate

\[\boxed{\frac{1}{48\pi}\text{ cm/s}}\]

🎯 Exam Significance

Exam Significance

›

This problem combines the formula for the volume of a cone with the concept of related rates. The important point is that the radius and height are not independent because the shape of the sand cone remains similar as it grows.

The given relationship

\[ h=\frac{r}{6} \]

must therefore be used before differentiating. Converting the volume into a function of \(h\) gives the particularly simple equation

\[ V=12\pi h^3. \]

Significance for Competitive Entrance Examinations

This is a useful application of the chain rule and dimensional reasoning. A fast approach is to first obtain

\[ r=6h \]

and then directly write

\[ V=12\pi h^3. \]

Differentiating immediately gives

\[ \frac{dV}{dt}=36\pi h^2\frac{dh}{dt}, \]

which can then be evaluated at \(h=4\).

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 5 mistakes to avoid

Forgetting the relationship between \(r\) and \(h\): Since the cone remains similar, use \(h=\dfrac{r}{6}\).
Using the wrong volume formula: The volume of a cone is \(V=\dfrac{1}{3}\pi r^2h\).
Forgetting the chain rule: The derivative of \(h^3\) with respect to time is \(3h^2\dfrac{dh}{dt}\).
Substituting \(h=4\) before differentiating: This can lead to mistakes in more complex problems.
Incorrect algebraic simplification: Carefully simplify numerical factors to avoid errors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The volume of the sand cone is
\[ V=\frac{1}{3}\pi r^2h. \]
The given geometric relationship is
\[ h=\frac{r}{6}, \qquad r=6h. \]
Therefore,
\[ V=12\pi h^3. \]
Differentiating with respect to time gives
\[ \frac{dV}{dt}=36\pi h^2\frac{dh}{dt}. \]
At \(h=4\) cm and \(\dfrac{dV}{dt}=12\text{ cm}^3/\text{s}\),
\[ \frac{dh}{dt} = \frac{1}{48\pi}\text{ cm/s}. \]
Hence, the height of the sand cone is increasing at
\[ \boxed{\frac{1}{48\pi}\text{ cm/s}}. \]

📘 Concept & Theory

Theory and Concept

›

In economics, the marginal cost represents the rate at which the total cost changes with respect to the number of units produced.

If \(C(x)\) is the total cost of producing \(x\) units, then the marginal cost is defined as

\[ MC=\frac{dC}{dx}. \]

Therefore, to find the marginal cost at a particular production level, we first differentiate the total cost function and then substitute the required value of \(x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given total cost function \(C(x)\).
Differentiate \(C(x)\) with respect to \(x\).
Obtain the marginal cost function \(MC=\dfrac{dC}{dx}\).
Substitute \(x=17\).
Simplify the result and state the unit as rupees per unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Given —
The total cost function is

\[ C(x)=0.007x^3-0.003x^2+15x+4000. \]

Number of units produced:

\[ x=17. \]

We need to find the marginal cost:

\[ MC=\frac{dC}{dx}. \]
Write the total cost function
The total cost of producing \(x\) units is
\[C(x)=0.007x^3-0.003x^2+15x+4000\]
Differentiate the cost function
The marginal cost is the derivative of the total cost with respect to \(x\):
\[MC=\frac{dC}{dx}\]
Differentiating term by term,
\[\frac{d}{dx}(0.007x^3)=0.007(3x^2)=0.021x^2\]
Similarly,
\[\frac{d}{dx}(-0.003x^2)=-0.003(2x)=-0.006x\]
Also,
\[\frac{d}{dx}(15x)=15\]
and since \(4000\) is a constant,
\[\frac{d}{dx}(4000)=0\]
Therefore, the marginal cost function is
\[MC=\frac{dC}{dx}=0.021x^2-0.006x+15\]
Substituting \(x=17\),
\[MC=0.021(17)^2-0.006(17)+15\]
Since
\[17^2=289,\]
we get
\[MC=0.021(289)-0.006(17)+15\]
Now,
\[0.021(289)=6.069\]
and
\[0.006(17)=0.102\]
Therefore,
\[MC=6.069-0.102+15\]
Hence,
\[MC=20.967\]

💡 Answer

Final Answer

›

The marginal cost when \(17\) units are produced is

\[ \boxed{MC=20.967\text{ rupees per unit}}. \]

🎯 Exam Significance

Exam Significance

›

This problem is a direct application of differentiation in economics. The key definition to remember is

\[ \boxed{MC=\frac{dC}{dx}}. \]

The constant term in the cost function disappears during differentiation. Students should differentiate each polynomial term carefully and substitute the production level only after obtaining the marginal cost function.

Significance for Competitive Entrance Examinations

Such questions test the interpretation of derivatives as rates of change and their application to economic functions. The fastest reliable method is:

\[ C(x) \longrightarrow C'(x) \longrightarrow C'(17). \]

The constant term can immediately be ignored while differentiating because its derivative is zero.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 6 mistakes to avoid

Using \(C(17)\) instead of \(C'(17)\): Marginal cost is the derivative of total cost, not the total cost itself.
Forgetting the power rule: \(\dfrac{d}{dx}(x^3)=3x^2\).
Incorrectly differentiating the constant: \(\dfrac{d}{dx}(4000)=0\).
Using the wrong value of \(x\): The question asks for the marginal cost when \(17\) units are produced, so use \(x=17\).
Arithmetic error: \(0.021(17)^2=6.069\), while \(0.006(17)=0.102\).
Omitting units: Since cost is measured in rupees and \(x\) represents units, marginal cost is expressed in rupees per unit.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Marginal cost is the derivative of the total cost function:
\[ MC=\frac{dC}{dx}. \]
For
\[ C(x)=0.007x^3-0.003x^2+15x+4000, \]
the marginal cost function is
\[ MC=0.021x^2-0.006x+15. \]
At \(x=17\),
\[ MC=0.021(17)^2-0.006(17)+15. \]
Therefore,
\[ \boxed{MC=20.967\text{ rupees per unit}}. \]
The constant \(4000\) has no effect on marginal cost because its derivative is zero.

📘 Concept & Theory

Theory and Concept

›

In economics, marginal revenue is the rate of change of total revenue with respect to the number of units sold.

If \(R(x)\) represents the total revenue from selling \(x\) units, then the marginal revenue is

\[ MR=\frac{dR}{dx}. \]

Therefore, we first differentiate the given revenue function and then substitute the specified value \(x=7\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the total revenue function \(R(x)\).
Differentiate \(R(x)\) with respect to \(x\).
Obtain the marginal revenue function.
Substitute \(x=7\).
Simplify the result and state its unit.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
The total revenue function is

\[ R(x)=13x^2+26x+15. \]

The production or sales level is

\[ x=7. \]

We need to find

\[ MR=\frac{dR}{dx}. \]
Write the marginal revenue formula
Marginal revenue is the derivative of total revenue with respect to the number of units sold:
\[MR=\frac{dR}{dx}\]
Differentiate the revenue function
Given —
\[R(x)=13x^2+26x+15\]
Differentiating term by term,
\[\frac{dR}{dx}=\frac{d}{dx}(13x^2)+\frac{d}{dx}(26x)+\frac{d}{dx}(15)\]
Using the power rule,
\[\frac{d}{dx}(13x^2)=26x\]
Also,
\[\frac{d}{dx}(26x)=26\]
and
\[\frac{d}{dx}(15)=0\]
Therefore,
\[MR=\frac{dR}{dx}=26x+26\]
Find the marginal revenue at \(x=7\)
Substituting \(x=7\),
\[MR=26(7)+26\]
Therefore,
\[MR=182+26\]
Hence,
\[\boxed{MR=208}\]

💡 Answer

Final Answer

›

The marginal revenue when \(x=7\) is

\[\boxed{208\text{ rupees per unit}}\]

🎯 Exam Significance

Exam Significance

›

p> This question is a direct application of derivatives in economics. The key definition to remember is

\[ \boxed{MR=\frac{dR}{dx}}. \]

Students should differentiate the revenue function first and then substitute the given value of \(x\). The constant term \(15\) disappears because its derivative is zero.

Significance for Competitive Entrance Examinations

The question tests the interpretation of the derivative as an instantaneous rate of change. Once the marginal revenue function

\[ MR=26x+26 \]

is obtained, evaluation at \(x=7\) is immediate.

A useful general procedure is

\[ R(x)\longrightarrow R'(x)\longrightarrow R'(7). \]

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 5 mistakes to avoid

Using \(R(7)\) instead of \(R'(7)\): Marginal revenue is the derivative of total revenue.
Forgetting the coefficient \(13\):
\[ \frac{d}{dx}(13x^2)=26x. \]
Keeping the constant \(15\):
\[ \frac{d}{dx}(15)=0. \]
Making an arithmetic error:
\[ 26(7)=182,\qquad 182+26=208. \]
Omitting units: Since revenue is measured in rupees and \(x\) represents units, marginal revenue is expressed in rupees per unit.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Marginal revenue is the derivative of total revenue:
\[ MR=\frac{dR}{dx}. \]
For
\[ R(x)=13x^2+26x+15, \]
we get
\[ MR=26x+26. \]
At \(x=7\),
\[ MR=26(7)+26. \]
Therefore,
\[ \boxed{MR=208\text{ rupees per unit}}. \]

📘 Concept & Theory

Theory and Concept

›

The area of a circle of radius \(r\) is

\[ A=\pi r^2. \]

The question asks for the rate of change of area with respect to radius. Therefore, we need to find

\[ \frac{dA}{dr}. \]

This is different from a related-rate question involving time. Since the question specifically asks for the rate of change of area with respect to \(r\), there is no need to introduce \(t\) or calculate \(\dfrac{dr}{dt}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the area of the circle as \(A=\pi r^2\).
Differentiate with respect to \(r\).
Substitute \(r=6\).
Compare the result with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Write the area of the circle
The area of a circle with radius \(r\) is
\[A=\pi r^2\]
Differentiate with respect to \(r\)
Differentiating both sides with respect to \(r\),
\[\frac{dA}{dr}=\frac{d}{dr}(\pi r^2)\]
Since \(\pi\) is a constant,
\[\frac{dA}{dr}=\pi(2r)\]
Therefore,
\[\frac{dA}{dr}=2\pi r\]
Substitute \(r=6\)
At \(r=6\)
\[\frac{dA}{dr}=2\pi(6)\]
Hence,
\[\frac{dA}{dr}=12\pi\]

💡 Answer

Final Answer

›

The rate of change of the area with respect to the radius at \(r=6\) is

\[\boxed{\frac{dA}{dr}=12\pi}\]

Therefore, the correct answer is (B) \(12\pi\).

🎯 Exam Significance

Exam Significance

›

This question tests the direct interpretation of a derivative as a rate of change. The phrase “with respect to its radius” tells us immediately that differentiation must be performed with respect to \(r\):

\[ \frac{dA}{dr}. \]

Students should not introduce time unless the question specifically asks for a rate with respect to time.

Significance for Competitive Entrance Examinations

This is a quick derivative-based multiple-choice problem. Recognising the standard formula

\[ A=\pi r^2 \]

and differentiating directly gives

\[ \frac{dA}{dr}=2\pi r. \]

At \(r=6\), the answer follows immediately:

\[ 2\pi(6)=12\pi. \]

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 5 mistakes to avoid

Using circumference instead of area: The required quantity is the area, so start with \(A=\pi r^2\).
Forgetting the factor \(2\):
\[ \frac{d}{dr}(r^2)=2r. \]
Differentiating with respect to time: The question asks for \(\dfrac{dA}{dr}\), not \(\dfrac{dA}{dt}\).
Substituting \(r=6\) before differentiation unnecessarily: Differentiate first and then evaluate at the specified radius.
Choosing the wrong option due to arithmetic error:
\[ 2\times6=12. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Area of a circle:
\[ A=\pi r^2. \]
Rate of change of area with respect to radius:
\[ \frac{dA}{dr}=2\pi r. \]
At \(r=6\):
\[ \frac{dA}{dr}=12\pi. \]
Therefore, the correct option is
\[ \boxed{\text{(B)}\ 12\pi}. \]

📘 Concept & Theory

Theory and Concept

›

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

If \(R(x)\) is the total revenue function, then

\[ MR=\frac{dR}{dx}. \]

Therefore, we differentiate the given revenue function and then substitute \(x=15\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given total revenue function.
Differentiate it with respect to \(x\).
Obtain the marginal revenue function.
Substitute \(x=15\).
Compare the result with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Write the total revenue function
We are given
\[R(x)=3x^2+36x+5\]
Find the marginal revenue function
By definition,
\[MR=\frac{dR}{dx}\]
Differentiating the given function term by term,
\[\frac{dR}{dx}=\frac{d}{dx}(3x^2)+\frac{d}{dx}(36x)+\frac{d}{dx}(5)\]
Using the power rule,
\[\frac{d}{dx}(3x^2)=6x,\]

\[\frac{d}{dx}(36x)=36\]
and
\[\frac{d}{dx}(5)=0\]
Hence,
\[MR=6x+36\]
Find the marginal revenue at \(x=15\)
Substituting \(x=15\),
\[MR=6(15)+36\]
Therefore,
\[MR=90+36\]
Hence,
\[\boxed{MR=126}\]

💡 Answer

Final Answer

›

The marginal revenue when \(x=15\) is

\[\boxed{126\text{ rupees per unit}}\]

Therefore, the correct answer is

\[ \boxed{\text{(D)}\ 126}. \]

🎯 Exam Significance

Exam Significance

›

This question is a direct application of the definition of marginal revenue:

\[ \boxed{MR=\frac{dR}{dx}}. \]

Students must copy the revenue function accurately before differentiating. A small change in the coefficient of \(x\), as seen in the supplied solution, changes the marginal revenue and consequently the correct option.

Significance for Competitive Entrance Examinations

This is a quick application of derivatives in economics. Once the revenue function is given, the solution follows the sequence

\[ R(x)\longrightarrow R'(x)\longrightarrow R'(15). \]

Careful reading is particularly important in multiple-choice questions because changing \(36x\) to \(6x\) produces a different option.

🚫 Common Mistakes

Common Mistakes

›

Common Mistakes · 5 mistakes to avoid

Changing the given function: The coefficient of \(x\) is \(36\), not \(6\).
Using \(R(15)\) instead of \(R'(15)\): Marginal revenue is the derivative of total revenue.
Forgetting that the derivative of \(36x\) is \(36\):
\[ \frac{d}{dx}(36x)=36. \]
Forgetting the constant term disappears:
\[ \frac{d}{dx}(5)=0. \]
Arithmetic error:
\[ 6(15)+36=90+36=126. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Marginal revenue is
\[ MR=\frac{dR}{dx}. \]
For
\[ R(x)=3x^2+36x+5, \]
the marginal revenue function is
\[ MR=6x+36. \]
At \(x=15\),
\[ MR=6(15)+36=126. \]
Therefore, the correct answer is
\[ \boxed{\text{(D)}\ 126}. \]
The supplied solution is incorrect because it changes the given term \(36x\) to \(6x\).

Application of Derivatives

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

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APPLICATION OF DERIVATIVES — Learning Resources

Frequently Asked Questions

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Application of Derivatives

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

Recent posts

APPLICATION OF DERIVATIVES — Learning Resources

Frequently Asked Questions

Q 01 What is Application of Derivatives in Class 12 Mathematics?

Q 02 Why is Application of Derivatives important for Class 12 students?

Q 03 What topics are covered in NCERT Class 12 Chapter 6?

Q 04 How do I solve NCERT Application of Derivatives problems?

Q 05 What is the rate of change of a quantity?

Q 06 What is the difference between local maximum and local minimum?

Q 07 How are derivatives used in maxima and minima problems?

Q 08 Are Application of Derivatives solutions useful for JEE preparation?

Q 09 What are common mistakes in Application of Derivatives?

Q 10 How can I prepare Chapter 6 Application of Derivatives effectively?

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