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
the height is 2/3 of the diameter not the radius though
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.
where did you get the 2/27 from when solving for dh/dt
To find the rate at which the height of the sandpile is changing when the height is 6 ft, we can use related rates.
Let's denote the height of the cone h, the radius of the base r, and the volume of the cone V.
The volume of a cone is given by the formula: V = (1/3)πr^2h.
We are given that the height of the pile is always 2/3 the base diameter. The base diameter of a cone is twice the radius, so the height is (2/3)(2r) = (4/3)r.
Now, we need to relate the variables to each other. We want to find dh/dt, the rate of change of the height with respect to time.
We are given that the rate at which sand is being dumped is 17 ft^3/ min. This corresponds to dV/dt, the rate of change of the volume with respect to time.
We can differentiate the formula for the volume of the cone with respect to time:
dV/dt = (1/3)π [2rh(dr/dt) + r^2(dh/dt)].
We know that r = (1/2)h, since the height is 2/3 the base diameter. Differentiating r with respect to t gives us dr/dt = (1/2)(dh/dt).
Substituting these values into the equation for dV/dt, we get:
17 = (1/3)π [2(2/3)rh(1/2)(dh/dt) + (1/2)h^2(dh/dt)].
Simplifying:
17 = (1/3)π [(4/3)(1/2)rh(dh/dt) + (1/2)h^2(dh/dt)].
17 = (1/3)π [(2/3)rh(dh/dt) + (1/2)h^2(dh/dt)].
To solve for dh/dt, we can isolate it on one side of the equation:
dh/dt = (3/π) [17 - (2/3)rh(dh/dt)] / [(2/3)h^2].
Now, we can plug in the given values to find the rate of change of the height at h = 6 ft. We have r = (1/2)h = 3 ft.
dh/dt = (3/π) [17 - (2/3)(3)(6)(dh/dt)] / [(2/3)(6)^2].
Simplifying further:
dh/dt = (3/π) [17 - 12(dh/dt)] / 72.
Multiplying both sides by 72π:
72π(dh/dt) = 3(17) - 12(dh/dt).
Combining like terms:
72π(dh/dt) + 12(dh/dt) = 51.
(84π)(dh/dt) = 51.
Finally, solving for dh/dt:
dh/dt = 51 / (84π).
Therefore, the rate at which the height of the sandpile is changing when the height is 6 ft is approximately 0.061 ft/min.