# Finding the Rate of Change of Water Height in a Filling Cylindrical Tank

• starstruck_
In summary, the conversation discusses a related rates problem involving a cylindrical tank being filled with water at a given rate. The question asks for the rate at which the height of the water is increasing. The conversation involves the use of equations and derivatives to solve the problem, with some confusion about notation and the need to cancel out constants. Ultimately, it is determined that the correct answer is 3/25pi m/min.
starstruck_

## Homework Statement

((I cannot, for the love of life, understand related rates, so please bear with me. Thank you! ))

A cylindrical tank with radius 5m is being filled with water at a rate of 3m3/min. How fast is the height of the water increasing?

I'm having trouble interpreting this question - actually, most related rates questions, having trouble differentiating between what I would call the volume and what not.

v= (pi)r2h

## The Attempt at a Solution

So, I'm going to say that I know the rate of change of the volume - which means I need to get the height in terms of the volume?

since v= (pi)r2h
h= v/((pi)r2)

h' = (dy/dx [v]((pi)r2) - dy/dx[(pi)r2](v))/((pi)2r4)
h' = (3(2pir^2) - 2pirv)/ ((pi)2r4)I'm pretty sure this is wrong since I ended up with a "v" in my numerator which I don't know the value of.
How would I go about solving this?

What is your dy/dx? You have no variable called x nor one called y in your expressions.

Orodruin said:
What is your dy/dx? You have no variable called x nor one called y in your expressions.
I probably shouldn't have written that, my bad - I just meant to use it as a way to say derivative of.

so, derivative of volume multiplied by the denominator minus the derivative of the denominator multiplied by the volume, over the denominator squared.

starstruck_ said:
I probably shouldn't have written that, my bad - I just meant to use it as a way to say derivative of.
I am sorry, but that is a really bad notation. What you want to do is to take the derivative of the expression with respect to time ##t##. That derivative operator should be written ##d/dt##, not anything else. You need to specify what you are taking the derivative with respect to. There is no general "derivative of".

Also, think about what variables in your expression actually depend on ##t##. The derivatives of the other variables with respect to ##t## are going to be zero.

Orodruin said:
I am sorry, but that is a really bad notation. What you want to do is to take the derivative of the expression with respect to time ##t##. That derivative operator should be written ##d/dt##, not anything else. You need to specify what you are taking the derivative with respect to. There is no general "derivative of".

Also, think about what variables in your expression actually depend on ##t##. The derivatives of the other variables with respect to ##t## are going to be zero.
so would it be d/dt[v](piR^2)/(pi^2R^4) ?
Does this mean my reasoning was correct? I messed up with the actual derivative part, but I've been having much more trouble understanding what to do or what reasoning is correct

Your idea of taking the derivative of the expression for h is correct, yes. I am still not completely sure what you mean by your expression. Can you insert the numbers to get the final result so that we can be sure?

I am still not completely sure what you mean by your expression. Can you insert the numbers to get the final result so that we can be sure?[/QUOTE]

I got the same answer as the textbook after plugging in my answers but if I take the expression I came up with:
d/dt[v](piR^2)/(pi^2R^4)

d/dt[v] = 3
R = 5
so: [3(pi)(25)]/[(pi)2(625)]
= 3/25(pi)

starstruck_ said:
so would it be d/dt[v](piR^2)/(pi^2R^4) ?
Does this mean my reasoning was correct? I messed up with the actual derivative part, but I've been having much more trouble understanding what to do or what reasoning is correct
Note that ##r## is a constant in this problem. So is ##\pi##!

This is the first time I've ever seen anyone differentiate ##\pi##.

You don't differentiate constants.

PeroK said:
Note that ##r## is a constant in this problem. So is ##\pi##!

This is the first time I've ever seen anyone differentiate ##\pi##.

You don't differentiate constants.

I didn't differentiate ##\pi##. It was being multiplied so I kept it as is but I did differentiate r which I wasn't supposed to.

I would suggest the following:
• You are overcomplicating your answer by not cancelling the ##\pi R^2## from your numerator with the corresponding terms in the denominator.
In this case, the units would give you dv/dt = 3 m^3/min (you really should write it like this rather than d/dt[v]), R = 5 m, R^2 = 25 m^2. Hence
$$\frac{dv}{dt}\frac{1}{\pi R^2} = \frac{3\ \mbox{m}^3/\mbox{min}}{\pi (25 \ \mbox{m}^2)} = \frac{3}{25\pi} \ \mbox{m/min}.$$

starstruck_ said:
I didn't differentiate ##\pi##. It was being multiplied so I kept it as is but I did differentiate r which I wasn't supposed to.
Okay, I see what you did now.

Also, note that you can also differentiate before you solve for ##h##, it will give you the same result. (Writing ##v'(t)## and ##h'(t)## rather than ##dv/dt## and ##dh/dt##)
$$v(t) = \pi R^2 h(t) \quad \Longrightarrow \quad v'(t) = \pi R^2 h'(t).$$
Then ##h'(t)## is just the rate you are after and so you can solve for it in terms of ##v'(t)##, which you have.

starstruck_ and PeroK

## 1. What are related rates problems?

Related rates problems are mathematical problems that involve finding the rate at which one quantity is changing with respect to another related quantity. These types of problems often involve using calculus to find the rate of change.

## 2. What are some common examples of related rates problems?

Some common examples of related rates problems include finding the rate at which the area of a circle is changing with respect to its radius, or finding the rate at which the height of a cone is changing as the radius of its base changes.

## 3. What are the key steps to solving a related rates problem?

The key steps to solving a related rates problem are: 1) Identify the quantities that are changing and the rate at which they are changing, 2) Determine how these quantities are related to each other, 3) Write an equation that represents the relationship between the changing quantities, 4) Use calculus to find the derivative of the equation with respect to time, 5) Substitute in the given values and solve for the desired rate of change.

## 4. How do you know when to use related rates in a problem?

You should use related rates when a problem involves two or more changing quantities that are related to each other, and you need to find the rate of change of one of the quantities with respect to time. This often occurs in real-world situations where multiple factors are changing simultaneously.

## 5. Are there any tips for approaching related rates problems?

Some tips for approaching related rates problems include: 1) Draw a diagram to visualize the problem, 2) Clearly label all given quantities and the desired rate of change, 3) Identify the changing quantities and their relationships, 4) Write an equation that represents the relationship between the changing quantities, 5) Take the derivative with respect to time, 6) Substitute in the given values and solve for the desired rate of change.

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