Related Rates (increasing cone radius question)

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Homework Statement



I've worked through both parts of this question twice in what I assume is the correct manner, but I'm receiving an unexpected result from part B. The question is as follows:

Sand is dumped such that the shape of the sandpile remains a cone with height equal to twice the radius.
A) If the sand is dumped at the constant rate of 20ft^3/s, find the rate at which the radius is increasing when the height reaches 6 feet.
B) Repeat for a sandpile for which the edge of the sandpile forms an angle of 45° with the horizontal.


Homework Equations



Volume of a cone [itex]V=(1/3)πr^2h[/itex]
Where r = radius
Where h = height

The Attempt at a Solution



A)
[itex]V=(1/3)πr^2h[/itex]
[itex]dV/dt=d/dt((1/3)πr^2)*h + (1/3)πr^2*(d/dt)(h)[/itex]
[itex]20ft^3/s=(2/3)πr(dr/dt)*h + (1/3)πr^2*(1)[/itex]

If the height = 2*radius, and h=6, r=3

[itex]20=(2/3)π(3)(dr/dt)*(6) + (1/3)π(3^2)[/itex]
[itex]20=(36/3)π(dr/dt) + (9/3)π[/itex]
[itex]20-3π=12π(dr/dt)[/itex]
[itex](20-3π)/12π = dr/dt[/itex]

So when the height of the cone is 6ft, the radius is expanding at about (20-3π)/12π or .2805ft/s.

B) If the edge of the sandpile forms a 45° angle with the horizontal, that means the peak of the cone is 90°, and the height is equal to the radius (since both the bisected angle at the peak and the angle at the edge of the base must be 45°)

[itex]20=(2/3)π(6)(dr/dt)*(6) + (1/3)π(6^2)[/itex]
[itex]20=(72/3)π(dr/dt) + (36/3)π[/itex]
[itex]20-12π=24π(dr/dt)[/itex]
[itex](20-12π)/24π = dr/dt[/itex]

So if the height is equal to the radius, and the height is 6ft, the radius seems to be expanding at about (20-12π)/24π or -.2347ft/s. Since the base of the cone can't be shrinking, I know I must have done something wrong, but I don't see my error.
 

Answers and Replies

  • #2
Dick
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Homework Statement



I've worked through both parts of this question twice in what I assume is the correct manner, but I'm receiving an unexpected result from part B. The question is as follows:

Sand is dumped such that the shape of the sandpile remains a cone with height equal to twice the radius.
A) If the sand is dumped at the constant rate of 20ft^3/s, find the rate at which the radius is increasing when the height reaches 6 feet.
B) Repeat for a sandpile for which the edge of the sandpile forms an angle of 45° with the horizontal.


Homework Equations



Volume of a cone [itex]V=(1/3)πr^2h[/itex]
Where r = radius
Where h = height

The Attempt at a Solution



A)
[itex]V=(1/3)πr^2h[/itex]
[itex]dV/dt=d/dt((1/3)πr^2)*h + (1/3)πr^2*(d/dt)(h)[/itex]
[itex]20ft^3/s=(2/3)πr(dr/dt)*h + (1/3)πr^2*(1)[/itex]

If the height = 2*radius, and h=6, r=3

[itex]20=(2/3)π(3)(dr/dt)*(6) + (1/3)π(3^2)[/itex]
[itex]20=(36/3)π(dr/dt) + (9/3)π[/itex]
[itex]20-3π=12π(dr/dt)[/itex]
[itex](20-3π)/12π = dr/dt[/itex]

So when the height of the cone is 6ft, the radius is expanding at about (20-3π)/12π or .2805ft/s.

B) If the edge of the sandpile forms a 45° angle with the horizontal, that means the peak of the cone is 90°, and the height is equal to the radius (since both the bisected angle at the peak and the angle at the edge of the base must be 45°)

[itex]20=(2/3)π(6)(dr/dt)*(6) + (1/3)π(6^2)[/itex]
[itex]20=(72/3)π(dr/dt) + (36/3)π[/itex]
[itex]20-12π=24π(dr/dt)[/itex]
[itex](20-12π)/24π = dr/dt[/itex]

So if the height is equal to the radius, and the height is 6ft, the radius seems to be expanding at about (20-12π)/24π or -.2347ft/s. Since the base of the cone can't be shrinking, I know I must have done something wrong, but I don't see my error.

Why are you putting dh/dt=1? In the first case h=2r, so dh/dt=2dr/dt. I can't see any reason for setting it to be 1.
 
  • #3
SteamKing
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Also, factor the (pi/3) term out. All it does is clutter your calculation and make it hard to follow.

I get a slightly different dr/dt for part 1.
 
  • #4
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Why are you putting dh/dt=1? In the first case h=2r, so dh/dt=2dr/dt. I can't see any reason for setting it to be 1.

I set d/dt[h] = 1 because I understand the power rule to state that a variable with no exponent (or more correctly, an exponent of 1) to differentiate to 1 (h^1 = 1h^0 = 1(1) = 1). Are you suggesting I convert h to terms of r prior to differentiation?

A) [itex]V = 1/3(π)(r^2)(h)[/itex]
1. [itex]dV/dt = d/dt[1/3(π)(r^2)]2r + 1/3(π)(r^2)d/dt[2r][/itex]
2. [itex]20 = 2π/3(r)dr/dt(2r) + π/3(r^2)2dr/dt[/itex]
3. [itex]20 = 2π/3(3)dr/dt(6) + 2π/3(3^2)dr/dt[/itex]
4. [itex]20 = 12πdr/dt + 6πdr/dt[/itex]
5. [itex]20 = 6πdr/dt(2+1)[/itex]
6. [itex]20/3 = 6πdr/dt[/itex]
7. [itex]6.667/6π = dr/dt[/itex]
Ans. [itex]0.3537 = dr/dt[/itex]

B) [itex]V = 1/3(π)(r^2)(h)[/itex]
1. [itex]dV/dt = d/dt[1/3(π)(r^2)]r + 1/3(π)(r^2)d/dt[r][/itex]
2. [itex]20 = 2π/3(r)dr/dt(r) + π/3(r^2)dr/dt[/itex]
3. [itex]20 = 2π/3(6)dr/dt(6) + π/3(6^2)dr/dt[/itex]
4. [itex]20 = 24πdr/dt + 12πdr/dt[/itex]
5. [itex]20 = 12πdr/dt(2 + 1)[/itex]
6. [itex]20/3 = 12πdr/dt[/itex]
7. [itex]6.667/12π = dr/dt[/itex]
Ans. [itex]0.1768 = dr/dt[/itex]

The answer for part B makes much more sense that way, thank you! I notice that the rate of the radius's expansion in part B is half that of part A, which makes sense considering the radius is twice that of A's. However, I'm not sure I understand why I would want to convert my independent variable into terms of the dependent variable? I see that in step 5 this allows me to factor out (Xπ(dr/dt)) and divide, rather than subtract (which is what led to a negative answer in my previous attempt), so does this mean I should always convert to dependent terms prior to differentiation?

I'm sorry for the simple questions. I'm taking a self-taught college course while deployed, and there's no instructor to turn to.
 
Last edited:
  • #5
SteamKing
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Let's make this simple:

The volume of a cone is V = [itex]\frac{π}{3}[/itex]r[itex]^{2}[/itex]h

By implicit differentiation w.r.t. time and the chain rule,

dV/dt = [itex]\frac{π}{3}[/itex][2rh dr/dt + r[itex]^{2}[/itex]dh/dt]

This is the general equation for dV/dt, when no relationship between r and h is known.

In your problem, two different relationships between r and h are given. In the first, h = 2r; in the second, r = h. Given these relationships, the equation for dV/dt may be adjusted:

for h = 2r, then dh/dt = 2 dr/dt,

dV/dt = [itex]\frac{π}{3}[/itex][2r(2r) dr/dt + r[itex]^{2}[/itex](2)dr/dt]

dV/dt = [itex]\frac{π}{3}[/itex][4r[itex]^{2}[/itex] + 2r[itex]^{2}[/itex]](dr/dt)

dV/dt = [itex]\frac{π}{3}[/itex][6r[itex]^{2}[/itex]](dr/dt)

With this last equation, you can now substitute the given values for h and dV/dt and solve for dr/dt. (dV/dt = 20 ft[itex]^{3}[/itex]/s and h = 6 ft, which implies r = 3 ft.)

When the relationship between r and h is altered, then the equations above can be changed according to the new relationship between variables.
 
  • #6
I set d/dt[h] = 1 because I understand the power rule to state that a variable with no exponent (or more correctly, an exponent of 1) to differentiate to 1 (h^1 = 1h^0 = 1(1) = 1).

A variable to the first power's derivative is equal to one only when differentiating with respect to that variable. So, dh/dh=1, but dh/dt doesn't necessarily.

Are you suggesting I convert h to terms of r prior to differentiation?

You don't have to, but you would need to find dh/dt in terms of r and dr/dt and substitute that at some point. Personally I think it would be easier to get h in terms of r before differentiation though.
 

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