# Find the derivative of a balloon's circumference

• JosephTraverso2
In summary: Problems involving derivatives are definitely in the bailiwick of calculus, regardless of whether the class you are taking is called "precalculus."
JosephTraverso2
A balloon's volume is increasing at a rate of dV/dt. Express the rate of change of the circumference with respect to time (dc/dt) in terms of the volume and radius.

## Homework Equations

Vsphere = (4/3)(π)(r^3)
C = (2)(π)(r)

## The Attempt at a Solution

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My strategy was to come up with two separate equations of dr/dt and substitute leading to an equation relating dv/dt and dc/dt

dV/dt = (4)(π)(r^2) (dr/dt)
dc/dt = (2)(π)(dr/dt)

(dc/dt) (1/2π) = dr/dt
Substitute...

(4)(π)(r^2) (dc/dt) (1/2π) = dv/dt

Canceling like terms

(2r^2)(dc/dt) = dv/dt

(1/2r^2) (dv/dt) = (dc/dt)

Now I was curious and wanted to graph the function just for fun. I know dv/dt is some number so I assumed it is 3. This yields
3/2r^2 = dc/dt
If I were to graph this function on a graphing utility would I be getting the circumference on the y-axis and time on the x? I ask this because the function is decreasing but the circumference of the balloon is increasing. Please help.

JosephTraverso2 said:
Where did you get this idea? dC/dt is positive which means that C increases as time increases. The r2 in the denominator means that C is increasing at a decreasing rate as the balloon grows larger, not that C is decreasing.
If you were to graph this, you need to integrate first to find C as a function of time. You would plot C on the vertical axis and t on the horizontal axis.

Last edited:
Clear and concise answer much thanks. But I'm currently in a pre-calculus class and we have not gone over integration yet. If it is not too much trouble can you please show how to do it to obtain circumference as a function of time? Thanks once again

JosephTraverso2 said:
Clear and concise answer much thanks. But I'm currently in a pre-calculus class and we have not gone over integration yet. If it is not too much trouble can you please show how to do it to obtain circumference as a function of time? Thanks once again
Is the statement of the problem exactly as was given to you? If so, then you cannot figure out what C as a function of time looks like because you need to know what dV/dt is, i.e. how the volume is increasing. You assumed as an example that it is 3, but is it really? Furthermore, is dV/dt constant or does itself depend on time? So if dV/dt is not given, you cannot do anything more than you have already done.

JosephTraverso2 said:
But I'm currently in a pre-calculus class and we have not gone over integration yet.
Thread moved. Problems involving derivatives are definitely in the bailiwick of calculus, regardless of whether the class you are taking is called "precalculus."
Please post problems involving derivatives and related rates in the Calculus & Beyond Homework section - thanks...

## 1. What is the definition of the derivative of a balloon's circumference?

The derivative of a balloon's circumference is the rate of change of its circumference with respect to its radius. It measures how much the balloon's circumference changes for a small change in its radius.

## 2. How do you find the derivative of a balloon's circumference?

To find the derivative of a balloon's circumference, you can use the formula for the derivative of a circle's circumference, which is 2π. This means that the derivative of a balloon's circumference is always 2π.

## 3. Why is finding the derivative of a balloon's circumference important?

Finding the derivative of a balloon's circumference is important because it allows us to understand how the circumference of a balloon changes as its size changes. This can be useful in various applications, such as designing balloons for different purposes or understanding the behavior of materials used in balloons.

## 4. Can the derivative of a balloon's circumference be negative?

No, the derivative of a balloon's circumference cannot be negative. Since the derivative measures the rate of change, it only gives a positive value for the change in circumference for a positive change in radius. Therefore, the derivative of a balloon's circumference is always positive.

## 5. How does the shape of a balloon affect its derivative of circumference?

The shape of a balloon does not affect its derivative of circumference. As long as the balloon has a circular base, the derivative will always be 2π. This is because the derivative is determined by the relationship between the circumference and the radius, which remains constant for a circular shape.

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