# Cone Related Rates Problem

1. Nov 6, 2013

### Burjam

1. The problem statement, all variables and given/known data

Grit, which is spread on roads in winter, is stored in mounds which are the shape of a cone. As grit is added to the top of a mound at 2 cubic meters per minute, the angle between the slant side of the cone and the vertical remains 45º. How fast is the height of the mound increasing when it is half a meter high?

2. Relevant equations

V=πr2/3

3. The attempt at a solution

So I need to solve for dh/dt. I know dV/dt=2 and I know the height, but not the radius. So I draw a right triangle. Since the angle is 45º, r=h. So r=0.5. Now time to take d/dt of each side.

dV/dt=d/dt[πr2/3]
2=1/3πr2*dh/dt

I treated r as a constant and h as a function of time here. I applied the product rule and the chain rule.

dh/dt=6/πr2

Substitue 0.5 for r and I get

dh/dt=24/π

The correct answer is 8/π. Where did I go wrong?

2. Nov 6, 2013

### SteamKing

Staff Emeritus
You can't treat r as a constant. As grit is added to the cone, both h and r are changing w.r.t. time.

If r stays constant and h increases, then the angle of repose of the grit (45 degrees) will increase, which it cannot do. The angle stays constant.

3. Nov 6, 2013

### Burjam

Here's the derivative I get now:

2=1/3πr2*dh/dt + 2/3πrh*dr/dt

This is great and all, but what am I supposed to sub in for dr/dt to solve the equation? I have two unknowns now.

4. Nov 6, 2013

### tiny-tim

Hi Burjam!
No, you have only one unknown (h).

r is not an indpendent unknown, it is a function of h.

5. Nov 6, 2013

### SteamKing

Staff Emeritus
You know the angle of repose of the grit (45 degrees). For each cm the height increases, how many cm does the radius increase? (Hint: draw a diagram.)

6. Nov 6, 2013

### Burjam

The radius will increase at the same rate as the height because the angle is 45º and constant. I actually drew a diagram initially to get r so this is partially from that. So dr/dt=dh/dt. From here I can solve it:

2=1/3πr2*dh/dt + 2/3πrh*dh/dt
2=dh/dt(1/3πr2 + 2/3πrh)
dh/dt=9/(πr2 + πrh)

When I plug in 0.5 for r and h I get:

dh/dt=18/π

Still not the right answer. What's wrong?

Last edited: Nov 6, 2013
7. Nov 6, 2013

### MuIotaTau

It seems to me that there's an algebra issue going between the two bolded lines. I found that using $r = h$ made it fairly simple, personally.

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