Related Rates (increasing cone radius question)

In summary, the volume of a cone is given by V = \frac{π}{3}r^{2}h and by implicit differentiation with respect to time and the chain rule, the rate of change of volume is dV/dt = \frac{π}{3}[2rh dr/dt + r^{2}dh/dt]. However, when specific relationships between r and h are given, such as h = 2r or r = h, the equation for dV/dt may be adjusted accordingly to solve for the rate of change of the radius. In the given problem, the rate of change of the radius is found to be 0.3537 ft/s and 0.1768 ft/s for
  • #1
kald13
9
0

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.
 
Physics news on Phys.org
  • #2
kald13 said:

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.
 
  • Like
Likes 1 person
  • #3
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
Dick said:
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
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
kald13 said:
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.

kald13 said:
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.
 

Related to Related Rates (increasing cone radius question)

1. How do you set up the related rates equation for an increasing cone radius?

The related rates equation for an increasing cone radius can be set up using the formula for the volume of a cone: V = (1/3)πr^2h. We can take the derivative with respect to time, t, to get dV/dt = (1/3)π(2r)(dr/dt)h + (1/3)πr^2(dh/dt). We can then plug in the given rates of change and solve for the unknown rate, in this case dr/dt.

2. How do you find the rate at which the height of the cone is changing?

To find the rate at which the height of the cone is changing, we can use the derivative of the related rates equation for the volume of a cone: dV/dt = (1/3)π(2r)(dr/dt)h + (1/3)πr^2(dh/dt). We can plug in the given rates of change and solve for the unknown rate, in this case dh/dt.

3. How do you interpret the related rates equation for an increasing cone radius?

The related rates equation for an increasing cone radius shows us the relationship between the rate at which the cone's radius is changing and the rates at which its height and volume are changing. It allows us to find the unknown rate, in this case dr/dt, by using the given rates of change for the cone's height and volume.

4. Can you use the related rates equation to find the rate at which the volume of the cone is changing?

Yes, the related rates equation can be used to find the rate at which the volume of the cone is changing. We can use the derivative of the related rates equation for the volume of a cone: dV/dt = (1/3)π(2r)(dr/dt)h + (1/3)πr^2(dh/dt). We can plug in the given rates of change and solve for the unknown rate, in this case dV/dt.

5. What is the significance of using related rates in real-life applications, such as an increasing cone radius?

Using related rates in real-life applications, such as an increasing cone radius, allows us to understand and quantify the relationships between different variables and their rates of change. This can help us make predictions and solve problems in fields such as physics, engineering, and economics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
11
Views
882
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
619
  • Calculus and Beyond Homework Help
Replies
2
Views
836
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
4K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
30
Views
3K
Back
Top