# [chain rule] Related Rates

• pine_apple
In summary, the conversation discusses solving an assignment question involving the rate at which gasoline is pumped from a tanker truck and the rate at which the gasoline level is falling. The tank is described as a cylinder and the question asks for the rate when the gasoline is at a certain depth. Different methods are explored and the conversation ends with Aero providing a clear explanation of how to solve the problem using the top surface area of the liquid.

#### pine_apple

Q: gasoline is pumped from the tank of a tanker truck at a rate of 20L/s. if the tank is a cylinder 2.5 m in diameter and 15 m long, at what rate is the level of gasoline falling when the gasoline in the tank is 0.5m deep? express in exact answer in cm/min [1L=1000 cm^3]

i did:

dv/dt = - 1200000cm^3/min

dv/dt = dv/dh*dh/dt

Vcylinder = pi r^2 l {l=h=length of the cylinder}
r = 125h/50 cm = 5h/2
V= pi(5/2h)^2l
dv/dh = pi 25/2 hl

-1200000 = pi 25/2 (50)(1500) dh/dt
dh/dt = -96/75 pi

i've got a good feeling that this is wrong. please help.

pine_apple said:

Q: gasoline is pumped from the tank of a tanker truck at a rate of 20L/s. if the tank is a cylinder 2.5 m in diameter and 15 m long, at what rate is the level of gasoline falling when the gasoline in the tank is 0.5m deep? express in exact answer in cm/min [1L=1000 cm^3]

i did:

dv/dt = - 1200000cm^3/min

dv/dt = dv/dh*dh/dt

Vcylinder = pi r^2 l {l=h=length of the cylinder}
r = 125h/50 cm = 5h/2
V= pi(5/2h)^2l
dv/dh = pi 25/2 hl

-1200000 = pi 25/2 (50)(1500) dh/dt
dh/dt = -96/75 pi

i've got a good feeling that this is wrong. please help.
A good feeling that this is wrong? I always have a bad feeling that I am wrong!:rofl:

You say "r = 125h/50 cm = 5h/2" Why would this be true? I see that the "125" is 1.25 m converted to cm but where is the 50 from? Is it the "0.5 m deep". That shouldn't be in the general formula but only applied after differentiating. And why would muliplying h by that give the radius? I take it you are assuming that the cylinder is lying on its side like it would be on a truck. In that case, as the level of gasoline falls the gasoline does not form cylinders of decreasing radius, it forms a decreasing part of the original cylinder. Set up a coordinate system with the center of the circle, of radius 1.25 m, at the origin. Then it can be written $x^2+ y^2= 1.5625. At a given depth h, the top of the gasoline is at distance 1.25- h from the center. The area, against the ends of the cylinder, of the gasoline is given by $$\int_{1.25}^{1.25-h}(1.25-h-\sqrt{1.565- x^2} dx$ and so the volume of gasoline is [tex]15\int_{1.25}^{1.25-h}(1.25-h-\sqrt{1.565- x^2} dx[/itex] Don't do that integral! Just use the "fundamental theorem of calculus" to differentiate it and set h= 0.5. HallsofIvy, thank you very much for your reply. I had realized that the gasoline doesn't form cylinders of decreasing radius too, but I couldn't figure out how to express it. We have not learned the "fundamental theorem of calculus" yet. Is it possible to calculate it in another way without using the theorem? [tex]V = 15\int_{1.25}^{1.25-h}(1.25-h-\sqrt{1.565- x^2} dx$$So taking the derivative of that would be $$15(1.25-h-\sqrt{1.565- x^2} dx$$

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but i was wondering where "dx" came from; does "d" represents the diameter?

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my fault, it should just be: $$15(1.25-h-\sqrt{1.565- x^2})$$because if we have $$F(x) = \int_{a}^{x} f(t) \; dt$$ then $$F'(x) = f(x)$$ for every $$x$$ in $$[a,b]$$.There is no $$dx$$

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Thank you.
But if you multiply those, how would it give you the volume? Doesn't the product become the area?
I don't understand why h-(1.565-x^2)^1/2 is being subtracted from 1.25-h.
Can someone please tell me why?

Consider the top surface of the liquid. Find its area as a function of the height
(remember that the cylinder is tipped on its side).

dh/dt * area of top surface = dv/dt

Plug in h = 50cm and solve for dh/dt.

Aero, thank you so much ! :D

HallsofIvy, Coutrigrad, and Aero, thank you :)

## 1. What is the chain rule?

The chain rule is a mathematical rule used to calculate the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

## 2. How is the chain rule used in related rates problems?

In related rates problems, the chain rule is used to find the rate of change of one variable with respect to another variable. This is done by taking the derivative of the related rates equation using the chain rule.

## 3. Can you give an example of a related rates problem that uses the chain rule?

One example of a related rates problem that uses the chain rule is the classic ladder sliding down a wall problem. In this problem, the chain rule is used to find the rate at which the bottom of the ladder is sliding away from the wall.

## 4. What are some common mistakes when using the chain rule in related rates problems?

Some common mistakes when using the chain rule in related rates problems include not correctly identifying the variables and their rates, not correctly applying the chain rule formula, and not using proper units in the final answer.

## 5. How can I improve my understanding and application of the chain rule in related rates problems?

To improve your understanding and application of the chain rule in related rates problems, you can practice solving various related rates problems, understand the concept of composite functions, and make sure to carefully identify and differentiate the given variables and rates in the problem. Additionally, seeking help from a teacher or tutor can also be beneficial.

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