Find the value of the following: The rate of change of the area of a circle with respect to its radius r at r

Exercise 6.1 · Q17

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$\frac{dA}{dr}=2\pi r=12\pi\ \text{cm}^2/\text{cm}$ at $r=6$ cm — option (B).

The idea

The phrase "rate of change of the area with respect to the radius" is a direct instruction to differentiate the area with respect to $r$ . This is a plain derivative evaluation, not a related-rates problem — no time is given, so there is no $\frac{dr}{dt}$ and no chain rule.

Set up

The area of a circle of radius $r$ is

A=\pi r^2.

Work the steps

Differentiate with respect to $r$ (treat $\pi$ as a constant):

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

Neatly, this equals the circumference — adding a thin ring of thickness $dr$ adds an area of about (circumference) $\times dr$ .

2. Substitute $r=6$ cm:

$\frac{dA}{dr}\Big|_{r=6}=2\pi(6)=12\pi.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2024Set A-11 markMCQ
$y = 2^{x^3}$
›Reveal solutionSolution

Differentiate $a^{u}$ as $a^{u}\log a \cdot u'$ with $a = 2$ , $u = x^{3}$ , then evaluate at $x = 1$ .

Step 1 — The rule for an exponential with a variable exponent.

For $y = a^{u(x)}$ with constant base $a > 0$ :
$\frac{dy}{dx} = a^{u}\,(\log a)\,\frac{du}{dx}$
(Derive it by logarithmic differentiation: $\log y = u\log a \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log a \cdot u'$ .)

Step 2 — Apply with $a = 2$ , $u = x^{3}$ .
$\frac{du}{dx} = 3x^{2}$ $\frac{dy}{dx} = 2^{x^{3}} \cdot \log 2 \cdot 3x^{2}$
Step 3 — Evaluate at $x = 1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = 2^{1^{3}} \cdot \log 2 \cdot 3(1)^{2} = 2 \cdot 3 \cdot \log 2 = 6\log 2$
Step 4 — Match the option. …
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COMEDK 2024Set 2024-E1 markMCQ
$\text { The rate of change of the volume of a sphere with respect to its surface area } \mathrm{S} \text { is }$
›Reveal solutionSolution

$\dfrac{dV}{dS}=\dfrac{dV/dr}{dS/dr}=\dfrac{r}{2}$ , and writing $r$ in terms of $S$ gives $\tfrac14\sqrt{S/\pi}$ .

For a sphere $V=\dfrac{4}{3}\pi r^{3}$ and $S=4\pi r^{2}$ .
$\frac{dV}{dr}=4\pi r^{2},\qquad \frac{dS}{dr}=8\pi r$ $\frac{dV}{dS}=\frac{dV/dr}{dS/dr}=\frac{4\pi r^{2}}{8\pi r}=\frac{r}{2}$
Express $r$ through $S$ : from $S=4\pi r^{2}$ , $\ r=\dfrac{1}{2}\sqrt{\dfrac{S}{\pi}}$ . Hence …
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KCET 2023Set A-21 markMCQ
$s = \frac{2t^3}{3} - 18t + \frac{5}{3}$
›Reveal solutionSolution

Differentiate $s$ once for velocity, set it to zero to find when the particle is at rest, differentiate again for acceleration and evaluate there.

Step 1 — The concept: derivatives as rates.

For rectilinear motion,
$v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}.$
"Comes to rest" means the velocity — not the displacement — is zero.

Step 2 — Velocity.

Given
$s = \frac{2t^{3}}{3} - 18t + \frac{5}{3},$ $v = \frac{ds}{dt} = \frac{2}{3}\cdot 3t^{2} - 18 = 2t^{2} - 18 .$
(The constant $\tfrac53$ differentiates away — it only fixes the starting position.)

Step 3 — Find the instant of rest.
$v = 0 \;\Rightarrow\; 2t^{2} - 18 = 0 \;\Rightarrow\; t^{2} = 9 \;\Rightarrow\; t = \pm 3$
Time cannot be negative, so
$t = 3\ \text{seconds}.$
Step 4 — Acceleration.
$a = \frac{dv}{dt} = \frac{d}{dt}\big(2t^{2} - 18\big) = 4t$
Step 5 — Evaluate at $t = 3$ .

$a\big|_{t=3} = 4(3) = 12$ …
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KCET 2023Set A-21 markMCQ
$5$
›Reveal solutionSolution

A related-rates problem: differentiate $A = \pi r^2$ with respect to time using the chain rule and substitute the instantaneous radius.

Step 1 — Relate the quantities.

The plate is circular, so its area is
$A = \pi r^{2}.$
Step 2 — Differentiate with respect to time (chain rule).

Both $A$ and $r$ change with $t$ , so
$\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt} = 2\pi r\,\frac{dr}{dt}$
This is the whole method of related rates: the rate we want ( $dA/dt$ ) is linked to the rate we know ( $dr/dt$ ) through the geometric relation between $A$ and $r$ .

Step 3 — Substitute the data.

We are told the radius is increasing at
$\frac{dr}{dt} = 0.05\ \text{cm/sec},$
and we want the rate at the instant when $r = 5.2$ cm (not $5$ cm — the initial $5$ cm is a distractor; the rate is asked at the later radius):
$\frac{dA}{dt} = 2\pi (5.2)(0.05)$
Step 4 — Compute. …
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KCET 2021Set A-11 markMCQ
$\theta = \dfrac{t^2}{20} + \dfrac{t}{5}$
›Reveal solutionSolution

Differentiate the angular displacement to get angular velocity, evaluate at $t=4$ to get $k$ , then multiply by 5.

Step 1 — The concept.

Angular velocity is the time-derivative of angular displacement — exactly parallel to $v = \dfrac{dx}{dt}$ in linear motion:
$\omega = \frac{d\theta}{dt}.$
This is why the question, though dressed as physics, is really a differentiation exercise.

Step 2 — Differentiate $\theta(t)$ .
$\theta = \frac{t^{2}}{20} + \frac{t}{5}$
Applying the power rule $\frac{d}{dt}(t^n) = n t^{n-1}$ term by term:
$\omega = \frac{d\theta}{dt} = \frac{2t}{20} + \frac{1}{5} = \frac{t}{10} + \frac{1}{5}.$
Step 3 — Evaluate at $t = 4$ .
$k = \omega(4) = \frac{4}{10} + \frac{1}{5} = 0.4 + 0.2 = 0.6 \ \text{rad s}^{-1}.$
Step 4 — Compute the quantity asked for.

The question asks for $5k$ , not $k$ — read it carefully:
$5k = 5 \times 0.6 = 3.$
Step 5 — Note the distractors. …
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Q.Find the value of the following: The rate of change of the area of a circle with respect to its radius $r$ at $r = 6 \text{ cm}$ is (A) $10\pi$ (B) $12\pi$ (C) $8\pi$ (D) $11\pi$

Concept understanding — Rate Of Change

Rate of Change

Average versus instantaneous rate

Rates with respect to time

Reading the sign and size

The idea

Set up

Work the steps

Method: Recognising a Plain Derivative Question Disguised as a "Rate" Question

Steps

Common Mistakes

Q.Find the value of the following: The rate of change of the area of a circle with respect to its radius rrr at r=6 cmr = 6 \text{ cm}r=6 cm is (A) 10π10\pi10π (B) 12π12\pi12π (C) 8π8\pi8π (D) 11π11\pi11π

The idea

Set up

Work the steps

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Q.Find the value of the following: The rate of change of the area of a circle with respect to its radius $r$ at $r = 6 \text{ cm}$ is (A) $10\pi$ (B) $12\pi$ (C) $8\pi$ (D) $11\pi$

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