# calculus

A truck starts dumping sand at the rate of 17 ft^3/ min, forming a pile in the shape of a cone. The height of the pile is always 2/3 the base diameter.

After 6 minutes, what is the height of the pile?
After 6 minutes, how fast is the height increasing?
After 6 minutes, how fast is the base radius increasing?
After 6 minutes, how fast is the area of the base increasing?

V = (1/3)π r^2 h, but h = (2/3)r , so
V = (1/3)π (r^2)(2/3)r
= (2/9)π r^3

after 6 min, multiply 6 by 17 to get 102 ft^3
102 = (2/9)π r^3
r^3 = 918/(2π) = 459/π
r = 5.2689.. and
h = (2/3) 5.2689.. = 3.51126...

V = (2/9)π r^3
dV/dt = (2/27) h^2 dh/dt
17 = (2/27)(3.51126)^2 dh/dt
solve for dh/dt

from there you can get dr/dt
since 2r = 3h
2dr/dt = 3 dh/dt <--- we just found dh/dt, so find dr/dt

Area of base (A) = π r^2
dA/dt = 2π r dr/dt
you know after 6 minutes, r = 5.2689.., and you just found dr/dt

can someone explain where the 2/27 comes from?

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1. The height of the pile is always 2/3 the base diameter.
so, h = 4/3 r, and r = 3/4 h

V = (1/3)π (r^2)(4/3)r = 4/9 πr^3 = 4/9 π (3/4 h)^3 = 3/16 π h^3
dV/dt = 4/3 πr^2 dr/dt = 9/16 π h^2 dh/dt
I think you can take it from there.
I think the 2/27 is a mistake, but since the wrong value was used at the start, it is moot.

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2. jj mistake was probably thinking integral, not derivative, so typed 2/27

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