At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when the surface area is 4 square meters and the radius is increasing at the rate of 1/6 meters per minute?

Question

At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when the surface area is 4 square meters and the radius is increasing at the rate of 1/6 meters per minute?

1 Answer

Impact of this question

5475 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-25T17:23:07

We know the volume of a sphere relative to its radius is given by:
$V (r) = \frac{4}{3} π r^{3}$ $V(r) = 4/3 pi r^3$

We are given that
Surface area at the time in question is $4 m^{2}$ $4 m^2$
which implies since $S = 4 π r^{2}$ $S = 4 pi r^2$ that the radius at that time is
$r = \frac{1}{\sqrt{π}}$ $r = 1/sqrt(pi)$

We are asked for $\frac{d V}{d t}$ $(dV)/(dt)$

$\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t}$ $(d V)/(dt) = (d V)/(dr) * (dr)/(dt)$

$\frac{d V}{d r} = 4 π r^{2}$ $(d V)/(dr) = 4 pi r^2$ (using the derivative ofour formula for the Volume)
and we are told that $\frac{d r}{d t} = \frac{1}{6} \frac{m}{sec}$ $(dr)/(dt) = 1/6 m/sec$

So at the time indicated
$\frac{d V}{d t} = 4 π r^{2} \cdot \frac{1}{6}$ $(d V)/(dt) = 4 pi r^2 * 1/6$

and with $r = \frac{1}{\sqrt{π}}$ $r= 1/sqrt(pi)$

$\frac{d V}{d t} = \frac{2}{3} \frac{m}{sec}$ $(d V)/(dt) = 2/3 m/sec$

Science

Math

Humanities

... and beyond

At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when the surface area is 4 square meters and the radius is increasing at the rate of 1/6 meters per minute?

1 Answer

Related questions

Impact of this question