Name: Rate of Change of Function: Example
Uploaded: 2019-07-19
Duration: 3 min 57 s
Description: In this video, we will use the rate of change to find the speed of the lower end of the ladder when it is sliding on the ground. Watch this video to know more.

Question

A circular ink blot grows at the rate of

2 {cm}^{2} / \sec

. Find the rate at which the radius is increasing after

2 \frac{6}{11}

seconds.

Accepted Answer

We know that, the area of the circle $A = π r^{2}$ , where $r =$ radius

Given that, rate of change in area $\frac{d A}{d t} =$ $2 {cm}^{2} / sec$

Differentiate $A$ with respect to time, then we get

$\frac{d A}{d t} = \frac{d}{d t} (π r^{2}) = 2 π r \cdot \frac{d r}{d t}$

$\Rightarrow 2 = 2 π (r) \cdot \frac{d r}{d t}$

$\Rightarrow 1 = π (r) \cdot \frac{d r}{d t} \to (1)$

At $t = 0$ , we have area $A = 0$ and area is increasing at $2 {cm}^{2} / sec$ .

Hence, area after $\frac{28}{11}$ seconds will be $2 \times \frac{28}{11} = \frac{56}{11}$

Thus, new area $A = \frac{56}{11} {cm}^{2}$

$\Rightarrow π r^{2} = \frac{56}{11}$

$\Rightarrow r = \sqrt{\frac{56}{11 π}}$

So, from $(1)$ , we have

$1 = π \sqrt{\frac{56}{11 π}} \cdot \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t} = \sqrt{\frac{11}{56 \times \frac{22}{7}}}$

$\Rightarrow \frac{d r}{d t} = \sqrt{\frac{1}{8 \times 2}}$

$\Rightarrow \frac{d r}{d t} = \sqrt{\frac{1}{8 \times 2}}$

$\Rightarrow \frac{d r}{d t} = \sqrt{\frac{1}{16}}$

$\Rightarrow \frac{d r}{d t} = \frac{1}{4} = 0.25 cm / sec$

Hence, the rate at which the radius is increasing $\frac{d r}{d t} = 0.25 cm / sec$ .

A circular ink blot grows at the rate of $2 {cm}^{2} / \sec$ . Find the rate at which the radius is increasing after $2 \frac{6}{11}$ seconds.

Important Questions on Physical Interpretation of Derivative

Learn and Prepare for any exam you want !

A circular ink blot grows at the rate of 2 cm2/sec. Find the rate at which the radius is increasing after 2611 seconds.

Important Questions on Physical Interpretation of Derivative

Learn and Prepare for any exam you want !

A circular ink blot grows at the rate of $2 {cm}^{2} / \sec$ . Find the rate at which the radius is increasing after $2 \frac{6}{11}$ seconds.