gravel is being dumped from a conveyor belt at a rate of 15 ft^3/hr and its coarseness is such that it forms a pile in the shape of an inverted right cone whose height is three times its base radius. How fast is the height of the pile increasing when the pile has a height of 10ft
height ------- h
base radius --- h/3
V = (1/3)π r^2 h
= (1/3)π (h^2/9)(h) = (π/27) h^3
dV/dt = (π/9) h^2 dh/dt
for the given data:
15 = π/9 (100)dh/dt
dh/dt = 135/(100π) ft/hr
How did you get "dh/dt = 135/(100π) ft/hr?"
That's the only part I'm confused about.
Thanks for your help btw.
from
15 = π/9 (100)dh/dt
I multiplied both sides by 9/(100π)
which left me with dh/dt on the one side
and 15(9/(100π) ) or 135/(100π) on the other side
oh okay, I ended up getting .43ft/hr. I'm 99% sure that's the correct answer.
Thanks for clearing that up for me, I'm really sucking at related rates, so I'm doing all the problems I can.
Why did the pile go to therapy? Because it had low self-esteem and couldn't stop comparing itself to mountains!
Now, let's solve the problem. We are given that the rate at which gravel is being dumped is 15 ft³/hr. Let's call the base radius of the pile "r" and the height of the pile "h." We are also told that the height of the pile is three times its base radius: h = 3r.
To find how fast the height of the pile is increasing, we need to determine the derivative dh/dt, using the given information. We know that dh/dt = (dh/dr) * (dr/dt), applying the chain rule.
Since h = 3r, we can differentiate both sides of the equation with respect to time (t) to find:
dh/dt = 3(dr/dt).
Now, we need to find dr/dt, the rate at which the base radius is changing. To do this, we'll use the relationship between the rate of volume and the rate of change of the base radius.
The volume V of a cone can be expressed as follows: V = (1/3) * π * r² * h.
We're given that the rate of change of volume with respect to time (dV/dt) is 15 ft³/hr. Thus, we can differentiate both sides of the equation with respect to time to find:
dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r² * dh/dt.
Since h = 3r, we substitute this value and the given values (dV/dt = 15 ft³/hr) into the equation and rearrange to solve for dr/dt:
15 = (1/3) * π * (2r * dr/dt) * (3r) + (1/3) * π * r² * (3dr/dt).
Simplifying the equation:
15 = 6πr²(dr/dt) + 3πr²(dr/dt).
Combine like terms:
15 = 9πr²(dr/dt).
Now, we can solve for dr/dt by rearranging the equation:
dr/dt = 15 / (9πr²).
Given that the pile has a height of 10 ft (h = 10), we can determine the base radius (r = 10/3) and substitute this value into the equation:
dr/dt = 15 / (9π(10/3)²).
Simplifying further:
dr/dt = 45 / (900π).
Finally, we substitute this value of dr/dt into the equation: dh/dt = 3(dr/dt):
dh/dt = 3 * (45 / (900π)).
Simplifying the expression:
dh/dt = 135 / (900π).
So, when the pile has a height of 10 ft, the height of the pile is increasing at a rate of 135 / (900π) ft/hr.
To find the rate at which the height of the pile is increasing, we'll need to work with related rates. Let's assign some variables to the given information:
Let h be the height of the pile (in ft).
Let r be the base radius of the pile (in ft).
From the problem, we know:
The rate at which gravel is being dumped is 15 ft^3/hr.
The height of the pile (h) is three times its base radius (r), so h = 3r.
We need to find dh/dt, which represents the rate at which the height (h) is changing when the pile has a height of 10 ft. In other words, we need to find the value of dh/dt when h = 10 ft.
To find dh/dt, we can relate the variables h, r, and V (volume of the pile) using the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Differentiating both sides of the equation with respect to time t gives us:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Now, we can substitute the given values and solve for dh/dt:
- dV/dt is given as 15 ft^3/hr.
- π is a constant (approximately 3.14159).
- h = 10 ft (since we need to find the rate when h = 10 ft).
We also know from the problem that h = 3r, so we can substitute this relationship into the equation:
15 = (1/3) * 3.14159 * (2r * dr/dt * 10 + r^2 * dh/dt)
At this point, we have a relationship between r, dr/dt, and dh/dt. We need to solve for dh/dt.
Simplifying the equation further:
5 = π * (2r * dr/dt * 10 + r^2 * dh/dt)
Now, we can substitute h = 3r (since we know h = 10 ft and solve for dh/dt):
5 = π * (2r * dr/dt * 10 + r^2 * (3 * dr/dt))
5 = π * (20r * dr/dt + 3r^2 * dr/dt)
To solve for dh/dt, we need to isolate the term that contains dh/dt:
5 = π * (20r + 3r^2) * dr/dt
Now divide both sides by π * (20r + 3r^2):
5 / (π * (20r + 3r^2)) = dr/dt
Therefore, the rate at which the height of the pile is changing (dh/dt) is equal to 5 / (π * (20r + 3r^2)).
To find the value of dh/dt when the pile has a height of 10 ft, substitute h = 10 ft:
h = 3r
10 = 3r
r = 10/3 ft
Substitute this value of r into the equation:
dh/dt = 5 / (π * (20(10/3) + 3(10/3)^2))
= 5 / (π * (200/3 + 300/9))
= 5 / (π * (600/9 + 300/9))
= 5 / (π * (900/9))
= 5 / (π * 100)
= 0.0159 ft/hr
Therefore, when the pile has a height of 10 ft, the height of the pile is increasing at a rate of approximately 0.0159 ft/hr.