Answer:
The height of cone is decreasing at a rate of 0.085131 inch per second.
Step-by-step explanation:
We are given the following information in the question:
The radius of a cone is decreasing at a constant rate.
[tex]\displaystyle\frac{dr}{dt} = -7\text{ inch per second}[/tex]
The volume is decreasing at a constant rate.
[tex]\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}[/tex]
Instant radius = 99 inch
Instant Volume = 525 cubic inches
We have to find the rate of change of height with respect to time.
Volume of cone =
[tex]V = \displaystyle\frac{1}{3}\pi r^2 h[/tex]
Instant volume =
[tex]525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}[/tex]
Differentiating with respect to t,
[tex]\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)[/tex]
Putting all the values, we get,
[tex]\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131[/tex]
Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.
