# Related Rates (increasing cone radius question)

## 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 $V=(1/3)πr^2h$
Where h = height

## The Attempt at a Solution

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

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

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

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°)

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

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.

Dick
Homework Helper

## 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 $V=(1/3)πr^2h$
Where h = height

## The Attempt at a Solution

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

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

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

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°)

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

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.

• 1 person
SteamKing
Staff Emeritus
Homework Helper
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.

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

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

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:
SteamKing
Staff Emeritus
Homework Helper
Let's make this simple:

The volume of a cone is V = $\frac{π}{3}$r$^{2}$h

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

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

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

dV/dt = $\frac{π}{3}$[6r$^{2}$](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$^{3}$/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.

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.