To find rate of change, differentiate with respect to time. Then fill in the given values and evaluate.
[tex]\displaystyle V=\frac{4}{3}\pi r^3\\\\\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\\\\ \frac{dV}{dt}=4 \pi (10\,in)^2\cdot (7\,\frac{in}{min})\\\\=2800\pi\,\frac{in^3}{min}[/tex]
The rate of change of volume is 2800π in³/min ≈ 8796 in³/min.
We need to find the rate of change of the volume of a sphere whose radius increases at 7 inches per minute.
The rate of change of the volume when r = 10 inches is 8792 in^3/min
The first thing we can define is:
r = 10 in + (7 in/min)*x
Where r is the radius of the sphere, and x is the number of minutes.
The volume of the sphere will be:
V = (4/3)*3.14*r^3
Let's write this as a function of x:
V = (4/3)*3.14*( 10 in + (7 in/min)*x)^3
Now we need to differentiate it with respect to x, we will get:
V' = [3*(4/3)*3.14*( 10 in + (7 in/min)*x)^2 ]*(7 in/min)
V' = (7 in/min)*4*3.14*(10 in + (7 in/min)*x)^2
Now we need to evaluate this in r = 10 inches, remember that this happens for x = 0, then we have:
V'(0) = (7 in/min)*4*3.14*(10 in + (7 in/min)*0)^2
= (7 in/min)*4*3.14*(10 in)^2 = 8792 in^3/min
The rate of change of the volume when r = 10 inches is 8792 in^3/min
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