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?
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.
jj mistake was probably thinking integral, not derivative, so typed 2/27
To understand where the coefficient 2/27 comes from in the equation (2/27)h^2(dh/dt) = 17, let's break down the problem.
We know that the volume of a cone is given by V = (1/3)πr^2h, where r is the base radius and h is the height. In this case, the height is always 2/3 the base diameter, so we can substitute h = (2/3)r.
Now, we want to find how the height of the pile is changing with respect to time, or dh/dt. And we are also given that the rate of sand dumping is 17 ft^3/min.
Since V = (1/3)πr^2h, we can find the rate of change of volume with respect to time, dV/dt, by taking the derivative of V with respect to t:
dV/dt = (1/3)π[(2r)(dr/dt)h + r^2(dh/dt)]
Now, we can substitute h = (2/3)r:
dV/dt = (1/3)π[(2r)(dr/dt)(2/3)r + r^2(dh/dt)]
= (2/9)πr^2 dr/dt + (1/3)πr^2(dh/dt)
Since we are given that dV/dt = 17, we can further simplify:
17 = (2/9)πr^2 dr/dt + (1/3)πr^2(dh/dt)
Now, we want to solve for dh/dt, so let's isolate it:
(2/9)πr^2(dh/dt) = 17 - (2/9)πr^2 dr/dt
(dh/dt) = (27/2πr^2)(17 - (2/9)πr^2 dr/dt)
(dh/dt) = (2/27)(27/π)(17 - (2/9)πr^2 dr/dt)
(dh/dt) = (2/27)(17 - (2/9)πr^2 dr/dt)
Therefore, the coefficient 2/27 comes from the derivation of the volume formula and the substitution for h in terms of r, where r represents the base radius and dh/dt represents the rate of change of the height with respect to time.