# Finding the Rate of Height Change for a Changing Volume in a Trough

• kuahji
In summary, the conversation is about finding the rate at which the height of a trough is changing at 2m. The volume of the trough is changing at a rate of 0.2 m/s and the formula for volume is V=0.5bhl. However, the original formula for dV/dt is incorrect as it assumes that only h and V are changing, when in fact b is also changing. To find a correct formula, the relationship between h and b needs to be considered, taking into account that both variables are functions.

#### kuahji

There is a trough that is 10 meters long, 6 meters wide, and is in the shape of an equalateral triangle. The volume is changing at a rate of 0.2 m/s a second. The goal is to find the rate at which the height is changing at 2m.

So I initially set the problem up as V=0.5bhl (where b=width, h=height, and l=length).

Then I took the derivative of both sides & found
dv/dt=0.5bl(dh/dt)

From here I plugged in the numbers 0.2=0.5(10)(6)(dh/dt)
& found dh/dt=1/150.

However, I don't have a place to plug height in for 2m. So would it be correct to reason dh/dt is independent from the height? Or do I have something incorrect in my setup?

You have a mistake.
Your formula for dV/dt is incorrect, since it assumes that only h and V are changing. If you draw a picture of the trough with some water in it, you should see that b is not constant.

Your formula for volume V is correct, but it doesn't show the relationship between b and h, where b is the width of the water across its top edge, and h is the depth of the water. The only quantity that isn't changing in this problem is the length of the water, 10 meters.

You are given that the cross section of the trough is an equilateral triangle, which tells you that the interior angles of the triangle are all equal. Use that knowledge to get a relationship between the height h of water, and b, the width across the top edge of the water. Then write your formulas for volume V and dV/dt.

Remember that when the volume is changing, the height and width are also changing. The size of the trough will be the same, but since the amount of liquid is changing the height of the liquid in the trough is a function, and so is the width. If you take some liquid out of the trough the width is going to be smaller, right? Only the length stays constant.

So you need to consider that you have more than one function on the right hand side of your volume equation when differentiating. Try expressing the width as a function of height to leave yourself with only one function on the right hand side. This time however it will not be a linear function and when you take the derivative remember to use implicit differentiation.

Well it looks like Mark beat me to it. What he said!

>.< I forgot about that. Thanks it makes sense now. :)

## What is a related rates problem?

A related rates problem is a type of mathematical problem that involves finding the rate of change of a quantity with respect to time, given the rates of change of other related quantities. It is commonly used in calculus and physics to model real-world situations.

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

Some common examples of related rates problems include finding the rate at which the volume of a balloon is changing as it is being inflated, determining the rate at which the distance between two people walking is changing, and calculating the rate at which the depth of water in a conical tank is changing as it is being filled.

## What is the general approach to solving a related rates problem?

The general approach to solving a related rates problem involves identifying the relevant quantities and their rates of change, setting up an equation that relates these quantities, differentiating both sides of the equation with respect to time, and finally solving for the desired rate of change.

## What are some common challenges in solving related rates problems?

One common challenge in solving related rates problems is properly identifying the relevant quantities and their rates of change. Another challenge is setting up the correct equation that relates these quantities. Additionally, the use of implicit differentiation and understanding how to apply the chain rule can also be challenging for some individuals.

## What are some tips for successfully solving related rates problems?

Some tips for successfully solving related rates problems include carefully reading and understanding the problem, drawing a diagram to visualize the situation, and clearly defining the variables and their rates of change. It can also be helpful to use units consistently and to check your answer for reasonability.