Finding the Rate of Change of a Cone's Height Using Related Rates

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Homework Help Overview

The problem involves determining the rate of change of the height of a conical mound of grit as it is filled at a constant volume rate. The cone maintains a constant angle of 45º between the slant side and the vertical, and the specific scenario focuses on the height when it reaches half a meter.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between the height and radius of the cone, questioning the treatment of the radius as a constant while the height changes. There is exploration of the implications of the constant angle on the rates of change of height and radius.

Discussion Status

Participants are actively engaging with the problem, attempting to derive expressions for the rates of change. Some have identified potential algebraic errors in previous calculations, while others are clarifying the relationship between the radius and height based on the fixed angle. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants note that the angle of repose remains constant at 45º, which influences the relationship between height and radius. There is also a recognition of the challenge posed by having multiple unknowns in the equations derived from the problem.

Burjam
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Homework Statement



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?

Homework Equations



V=πr2/3

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?
 
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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.
 
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.
 
Hi Burjam! :smile:
Burjam said:
I have two unknowns now.

No, you have only one unknown (h).

r is not an indpendent unknown, it is a function of h.
 
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.)
 
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:
Burjam said:
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?

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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