If $V = \frac{4}{3} π r^{3}$ , at what rate in cubic units is $V$ increasing when $r = 10$ and $\frac{d r}{d t} = 0.1$ ?

π

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4 π

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40 π

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\frac{4 π}{3}

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Solution

The correct option is B $40 π$
Given, $r = 10$ and $\frac{d r}{d t} = 0.1$
We have,
$V = \frac{4}{3} π r^{3}$
$\Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
$\frac{d V}{d t} |_{r = 10} = 4 π (100) (0.1) = 40 π$ cubic units per sec

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Similar questions

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

V = \frac{4}{3} π r^{3}

\frac{d r}{d t} = 0.01

V = \frac{4}{3} π r^{3}

r = 10

\frac{d r}{d t} = 0.01 ?

f (x) = 2 s i n^{3} x - 3 s i n^{2} x + 12 s i n x + 5 \forall x \in (0, \frac{π}{2})

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