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=

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lublana lublana · Answer 1 · 2019-08-13T08:56:20+02:00

Answer:

[tex]\frac{dr}{dt}=0.01 [/tex]cm per second

[tex]\frac{dS}{dt}=320 square centimeter per second

Step-by-step explanation:

We are given that volume of sphere

[tex]V=\frac{4}{3}\pi r^3[/tex]

Volume of sphere is increasing at a constant rate

[tex]\frac{dV}{dt}=4[/tex] cubic centimeters per second

We have to find the rate of radius at which increasing

when r= 10 cm

Differentiating w.r.t time

[tex]\frac{dV}{dt}=\frac{4}{3}\pi\cdot3r^2\frac{dr}{dt}[/tex]

[tex]4=4 r^2\frac{dr}{dt}[/tex]

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

Now ,we are given that surface area of sphere

[tex] S=4\pir^2[/tex]

Differentiate w.r.t time then we get

[tex]\frac{dS}[dt}=8\pir\frac{dr}{dt}[/tex]

[tex]\frac{dS}{dt}=8\pi\times 20\times 2[/tex]

[tex]\frac{dS}{dt}=320 [/tex]cm per second