Answer:
0.194 ft/min
Step-by-step explanation:
Differentiating the volume equation with respect to time, we get ...
V' = (1/3)π(2r·r'·h +r²·h')
We need to find the height when the volume is 293 ft³:
239 ft³ = (π/3)(55 ft)²h
h = 3·239 ft³/(3025π ft²) = 717/(3025π) ft ≈ .07545 ft
Using volumes in ft³, areas in ft², and rates in ft/min, we can fill in the given information to get ...
694 = (π/3)(2·55·9·717/(3025π) + 3025h')
694 = 4302/55 +(π/3)3025h' . . . . simplify a bit
33868/55 = (π/3)3025h' . . . . . . . . subtract 4302/55
Now, divide by the coefficient of h.
101604/(166375π) = h' ≈ 0.194 . . . . ft/min
The rate of change of height is about 0.194 ft/min.
Step-by-step explanation:
We have volume of cone is
[tex]V=\frac{1}{3}\pi r^2h[/tex]
We need to find the rate of change of the height when the radius of the cone is 55 feet and the volume is 239 cubic feet.
Substituting
[tex]239=\frac{1}{3}\pi \times 55^2\times h\\\\h=0.275 feet[/tex]
Given
[tex]\frac{dV}{dt}=694ft^3/min\\\\\frac{dr}{dt}=9ft/min[/tex]
Finding rate of change of volume
[tex]V=\frac{1}{3}\pi r^2h\\\\\frac{dV}{dt}=\frac{1}{3}\pi (2r\frac{dr}{dt}h+r^2\frac{dh}{dt})\\\\694=\frac{1}{3}\pi (2\times 55\times 9\times 0.275+55^2\times \frac{dh}{dt})\\\\\frac{dh}{dt}=-0.108ft/min[/tex]
Rate of change of height is -0.108 ft per minute.
