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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 4

Lesson 5: Solving related rates problems

# Related rates: water pouring into a cone

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.E (LO)
,
CHA‑3.E.1 (EK)
As you pour water into a cone, how does the rate of change of the depth of the water relate to the rate of change in volume. Created by Sal Khan.

## Want to join the conversation?

• At how are the diameter and height having the same ratios. What does he mean by "these are lines"?
• Imo is easier if u notice the ratio between the Diameter and the height in the bigger cone. for every 4 cm of D, there is 4 cm of height, so the ratio D/h = 4/4 = 1/1. Since D = 2r u can say then the ratio 2r/h = 1/1 ---> r= h/2
• how is r h/2? I still don't get it
• The proportion of height and diameter is the same in this problem. So h = d. The diameter is 2 times the radius.

``h = dh = 2rh/2 = r``

Now we have r in terms of h.
• At I don't understand why Sal uses "h" as height when he has real value for that height which is 2.
• We also need to know the height of the water at times near "now" so that we can answer the question of how quickly the height of the water is rising. Therefore, we do need to define h as a function of t so that we can calculate its derivative.
• Why can't you make the volume formula h = V/.3(area of base) and then take the derivative of that?

d/dt [V(.3(pi(h/2)^2)^-1] = d/dt [h]

then solve? The answer doesn't seem to match, but I can't figure out why. Is it because the h in the Volume formula is not implicitly a function of time?
• Is it as simple as you changed 1/3 in the Volume formula to 0.3? I followed that process, left it as 1/3 and was still able to get 1/pi as dh/dt:

h = (12/pi)vh^-2. Take derivative of both sides, use product rule:

dh/dt = (12/pi)(h^-2 dV/dt + V*(-2h^-3) dH/dT). - Substitute h=2, V=pi/3, dV/dt = 1 & simplify:

dh/dt = (12/pi)(1/4 - pi/6 dh/dt) - I multiplied both sides by pi/12 to move it over:

pi/12 dh/dt = 1/4 - pi/6 dh/dt - get rid of fractions by *12 both sides:

pi dh/dt = 3 - 2pi dh/dt - Move the right term over (add 2pi dh/dt):

3pi dh/dt = 3 - divide both sides by 3pi:

dh/dt = 1/pi
• How would we derive a function to determine the rate at which the increase of the height of the water is decreasing with respect to time?
• The first derivative of the variable h with respect to time (dh/dt, or h' ) shows how the height changes with time. (ie. where is the height at any time). The second derivative of the variable h with respect to time (h'' ) would show how fast the rate the of the height is changing with respect to time. (ie. how fast is the rate of the height changing at any time). Solve for the second derivative.
• Why can he take out the constants at ? I'm guessing it won't change the answer, but I don't get how
• the constant rule states that we can take the constant out because a derivative of a constant is 0
• Sal always surprises me with his deep understanding of math, i would have never thought about relating diameter and height and somehow getting r^2 in terms of height. Anyways, is the reason we want the radius in terms of height is that so when we differentiate the equation we only have one unknown (dh/dt) instead of two (dr/dh and dh/dt)?
• Exactly. Differentiation and integration are very easy to do with one variable but highly complicated with multiple variables.
• I have a general question about related rates. I am trying to solve a problem two ways and keep getting two different answers.

The volume of a cone of radius r and height h is given by V = 1/3 pi r^2 h. If the radius and the height both increase at a constant rate of 1/2 cm per second, at what rate in cubic cm per sec, is the volume increasing when the height is 9 cm and the radius is 6 cm.

I tried letting r = 2/3 h and doing a substitution. Then when I differentiated with respect to time, I got an equation for dV/dt involving dh/dt. This process didn't give me the correct answer though.

Better yet, if r/h = 6/9 = 2/3, how could dr/dt = dh/dt = 1/2???? Isn't r = 2/3 h and therefore dr/dt = 2/3???

Nikki
Calculus instructor
• The reason you are having problems is that you are assuming that r and h have a constant ratio. They do not. The height and radius are 9 and 6 at the point in time that you are looking at. However, two seconds before they would have been 8 and 5. The ratio changes as their measures change.

To solve this problem, you do not need to get rid of one of the variables. You have rates of change for both. Differentiate using the product rule and substitute in both dr/dt and dh/dt.
• I still don't get why do we do h(t). Is it like that in every situation? When do you know when to do it?

Thanks