the volume of a sphere is increasing at a rate of 2cm^3/sec. Find the rate of change of its surface area when its volume is 256pi/3 cm^3.

Answer:

The rate of change of the surface area of the sphere is 0.99 cm²/s.  

Step-by-step explanation:

The surface area (A) of a sphere is given by:

[tex] A = 4\pi r^{2} [/tex]                      

If we derivate the above equation with respect to the time we have:

[tex] \frac{dA}{dt} = 4\pi (2r)\frac{dr}{dt} [/tex]

[tex] \frac{dA}{dt} = 8\pi r\frac{dr}{dt} [/tex]    (1)

Where:          

r: is the radius    

We need to find [tex]\frac{dr}{dt}[/tex] and r.

From the volume we can find the radius:

[tex] V = \frac{4}{3}\pi r^{3} = 256 \frac{\pi}{3} cm^{3} [/tex]  

[tex] r = \sqrt[3]{\frac{3V}{4\pi}} = \sqrt[3]{\frac{3*256*\frac{\pi}{3}}{4\pi}} = 4 cm [/tex]                

And by derivating the volume of the sphere with respect to the time we can calculate [tex]\frac{dr}{dt}[/tex]:  

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

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

[tex] \frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^{2}} = \frac{2}{4\pi(4)^{2}} = 0.0099 cm/s [/tex]      

Now, we can calculate the rate of change of the surface area (equation 1):

[tex] \frac{dA}{dt} = 8\pi r\frac{dr}{dt} = 8\pi*4*0.0099 = 0.99 cm^{2}/s [/tex]

Therefore, the rate of change of the surface area of the sphere is 0.99 cm²/s.  

I hope it helps you!          

an someone help please and thx:) !

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