A calculus problem by A Former Brilliant Member

Calculus Level pending

The radius of a shpere is changing at the rate of 1 m m 1~mm per second. How fast is the volume changing ( c m 3 s e c ) \left(\dfrac{cm^3}{sec}\right) when the radius is 20 c m 20~cm . Use π = 3.1416 \pi=3.1416 . Give your answer to the nearest integer.


The answer is 503.

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Relevant wiki: Related Rates of Change - Basic

Given: d r d t = 0.1 c m s e c ; r = 20 c m \dfrac{dr}{dt}=0.1~\dfrac{cm}{sec}; r=20~cm

Formula for the volume of a sphere: V = 4 3 π r 3 V=\dfrac{4}{3} \pi r^3

Differentiate both sides with respect to t t .

d V d t = 4 3 π ( 3 r 2 ) ( d r d t ) = 4 π r 2 ( d r d t ) \dfrac{dV}{dt}=\dfrac{4}{3} \pi(3r^2) \left(\dfrac{dr}{dt} \right)=4 \pi r^2 \left(\dfrac{dr}{dt} \right)

Substitute:

d V d t = 4 ( 3.1416 ) ( 2 0 2 ) ( 0.1 ) = \dfrac{dV}{dt}=4(3.1416)(20^2)(0.1)= 503 c m 3 s e c \boxed{503~\dfrac{cm^3}{sec}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...