Question

V=4/3? r^2 Suppose that, for the sphere in the video, instead of being told how fast the radius is changing, we're told that the volume is increasing at a constant rate of dV/dt=4 cubic centimeters per second. How fast is the radius increasing at the instant when the radius is r=10 centimeters? dr/dt= centimeters per second.

Instead of thinking about the volume, suppose that we are interested in how the surface area of the sphere is changing. Use the surface area formula S=4? r^2 to determine how fast the surface area is changing at the instant when the radius is r=20 cm and the radius is increasing at dr/dt=2 centimeters per second. dS/dt=

How to get this answer?

Answer 1

`(dr)/(dt)=0.01`cm per second

`(dS)/(dt)=320 square  centimeter per second</strong></p><p><strong>Step-by-step explanation:</strong></p><p>We are given that volume of sphere </p><p>[tex]V=(4)/(3)\pi r^3`

Volume of sphere is increasing at a constant rate

`(dV)/(dt)=4` cubic centimeters per second

We have to find the rate of radius at which increasing

when r= 10 cm

Differentiating w.r.t time

`(dV)/(dt)=(4)/(3)\pi\cdot3r^2(dr)/(dt)`

`4=4 r^2(dr)/(dt)`

`(dr)/(dt)= (1)/(r^2)=\frac}{1}{(10)^2}=0.01` cm per second

Now ,we are given that surface area of sphere

`S=4\pir^2`

Differentiate w.r.t time then we get

`\frac{dS}[dt}=8\pir(dr)/(dt)`

`(dS)/(dt)=8\pi* 20* 2`

`(dS)/(dt)=320`cm per second