Hello!
We can find the rate at which sand is leaking in volume per time. It is natural to measure time in minutes and volume in cubical inches for this problem.
Denote the radius of the cone (and its altitude) as `r(t)` (in inches). The rate of change of the height (and radius) is `r'(t)` and it is given to be `6` inches per minute. We also know that the volume of a (right circular)...
Hello!
We can find the rate at which sand is leaking in volume per time. It is natural to measure time in minutes and volume in cubical inches for this problem.
Denote the radius of the cone (and its altitude) as `r(t)` (in inches). The rate of change of the height (and radius) is `r'(t)` and it is given to be `6` inches per minute. We also know that the volume of a (right circular) cone is `V = 1/3 \pi r^2 h,` where `h` is the altitude. In our case it gives `V(t)=\pi/3 r^3(t).`
To find the rate of leaking, we need to take the derivative (use the Chain Rule):
`V'(t) = (\pi/3 r^3(t))' = \pi r^2(t)r'(t) = 6\pi r^2(t).`
For the moment when `r(t) = 10,` this derivative will be
`6\pi*10^2 = 600\pi approx 1885` (cubic inches per minute).
When the altitude is 10 inches, the rate at which sand is leaking out is about 1885 cubic inches per minute.
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