Answer:
Rate of change of volume when the radius is 40 inches and the height is 20 inches = 0 inch³/s
Step-by-step explanation:
We have
[tex]\texttt{Volume = }\frac{1}{3}\pi r^2h[/tex]
Taking derivative with respect to time,
[tex]V=\frac{1}{3}\pi r^2h\\\\\frac{dV}{dt}=\frac{1}{3}\pi \frac{d}{dt}\left (r^2h \right )\\\\\frac{dV}{dt}=\frac{1}{3}\pi \left ( 2r\frac{dr}{dt}\times h+r^2\frac{dh}{dt}\right)[/tex]
Given that
r = 40 inches
h = 20 inches
[tex]\frac{dr}{dt}=2inch/s\\\\\frac{dh}{dt}=-2inch/s[/tex]
Substituting
[tex]\frac{dV}{dt}=\frac{1}{3}\pi \left ( 2r\frac{dr}{dt}\times h+r^2\frac{dh}{dt}\right)\\\\\frac{dV}{dt}=\frac{1}{3}\pi \left ( 2\times 40\times 2\times 20+40^2\times (-2)\right)\\\\\frac{dV}{dt}=\frac{1}{3}\pi \left (3200-3200\right)=0inch^3/s[/tex]
Rate of change of volume when the radius is 40 inches and the height is 20 inches = 0 inch³/s
The rate at which the volume of the cone is changing when the radius is 40 inches and the height is 20 inches is 0.
Mathematically, the derivative of a function at a point is the rate of change.
The volume of a cone is given by:
[tex]V = \frac{1}{3} \pi r^{2} h[/tex] where r is the radius of cone and h is the height of the cone.
So, [tex]\frac{dV}{dt} =\frac{1}{3} \pi (r^{2} \frac{dh}{dt} +2rh\frac{dr}{dt} )[/tex]
It is given that
[tex]\frac{dr}{dt} = 2\\\\\frac{dh}{dt} =-2[/tex]
So, when r=40, h= 20
[tex]\frac{dV}{dt} =\frac{1}{3}\pi (40^{2} *(-2)+2*40*20*2)[/tex]
[tex]\frac{dV}{dt} =0[/tex]
Therefore, the rate at which the volume of the cone is changing when the radius is 40 inches and the height is 20 inches is 0.
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