The radius of a sphere is measured at the rate of 2 cm/sec.Find the rate increase of the volume when the radius is 6cm?

Question

Charles Anctil · Accepted Answer

# 288pi" cc/sec."# Let #v and r# be the volume and radius of a sphere, resp. #:. v=4/3pir^3#. #:. (dv)/dt=4/3pid/dt(r^3)=4/3pi*d/(dr)(r^3)*(dr)/dt#. #:. (dv)/dt=4/3pi*3r^2*(dr)/dt=4pir^2(dr)/dt............(ast)#, In #(ast), (dr)/dt=2, r=6#. #;. (dv)/dt=4pi*36*2=288pi" cc/sec."#

Christopher Agresta · Answer

undefined To find the rate of increase of the volume when the radius is 6 cm, we first need to express the volume of the sphere as a function of its radius. The volume $ V $ of a sphere with radius $ r $ is given by the formula:

$$ V = \frac{4}{3} \pi r^3 $$

Next, we'll take the derivative of the volume function with respect to time $ t $, since the rate of change of the volume depends on the rate of change of the radius. This derivative will give us the rate of change of the volume with respect to time.

$$ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} $$

Given that the radius $ r $ is increasing at a rate of $ 2 $ cm/sec, we have $ \frac{dr}{dt} = 2 $ cm/sec. Now, we'll find $ \frac{dV}{dr} $ by taking the derivative of the volume function $ V $ with respect to the radius $ r $:

$$ \frac{dV}{dr} = 4 \pi r^2 $$

Substituting the given rate of increase of the radius ($ \frac{dr}{dt} = 2 $) and the expression for $ \frac{dV}{dr} $, we can find $ \frac{dV}{dt} $ when $ r = 6 $ cm:

$$ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4 \pi (6)^2 \cdot 2 $$

$$ = 4 \pi (36) \cdot 2 $$

$$ = 288 \pi $$

So, the rate of increase of the volume when the radius is $ 6 $ cm is $ 288 \pi $ cubic centimeters per second.