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.

Answer 1

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?

Answer 2

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

Answer 3

the height is 2/3 of the diameter not the radius though

Answer 4

arghhh, my bad

so just change it to h = (2/3)(2r) = (4/3)r
and make the corresponding changes in the work above
The steps will remain the same.

Answer 5

where did you get the 2/27 from when solving for dh/dt