If V=43πr3at what rate in cubic units is V increasing w

If $V = \frac{4}{3} π r^{3}$ ,at what rate in cubic units is V increasing when r = 10 and $\frac{d r}{d t} = 0.01$ ?

$π$

$4 π$

$40 π$

$4 \frac{π}{3}$

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

We have,
$V = \frac{4}{3} π r^{3} \Rightarrow \frac{d v}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow \frac{d v}{d t} = 4 π \times 100 \times 0.01 [∵ r = 10 a n d \frac{d r}{d t} = 0.01] \Rightarrow \frac{d v}{d t} = r π$
Hence, V is increasing at the rate of $4 π$ cubic units per unit of time.

Hence (b) is the correct answer.

If V=43πr3at what rate in cubic units is V increasing w

Right Answer is: B

SOLUTION

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Related Questions
A man of height 2m walks directly away from a lamp of h
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The distance moved by the particle in time t is given b
For what values of x is the rate of increase of x3−5x2+