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Math rate diffrentiation

the volume V of a sphere of radius 5cm is increasing at a rate of 10cm^3/s. Find the rate of increase of its surface area A at this instant.

Answer is 4... but how?

Original post by Revision99

the volume V of a sphere of radius 5cm is increasing at a rate of 10cm^3/s. Find the rate of increase of its surface area A at this instant.

Answer is 4... but how?

Chain rule, so

\displaystyle \frac{dA}{dt}=\frac{dV}{dt} \frac{dA}{dV}

.
Maybe you can find an equation for A in terms of V or use that

\frac{dA}{dV}=\frac{dA}{dr}\frac{dr}{dV}

.

Original post by B_9710

Chain rule, so

\displaystyle \frac{dA}{dt}=\frac{dV}{dt} \frac{dA}{dV}

.
Maybe you can find an equation for A in terms of V or use that

\frac{dA}{dV}=\frac{dA}{dr}\frac{dr}{dV}

.

Are you actually allowed to use A=πr² for this question?

Original post by Dynamic_Vicz

Are you actually allowed to use A=πr² for this question?

Yes you're allowed to use the area of a sphere which is

4\pi r^2

.
Why would t you be allowed.

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