The area of a circle increases at a rate of 6 cm^2/s

Question

The area of a circle increases at a rate of 6 cm^2/s

a. How fast is the radius changing when the radius is 3 ​cm?

b. How fast is the radius changing when the circumference is 4 cm?

Answer 1

The first step to these types of problems is to write the implicit formula. (in this case the area of a circle)

A = π*r²

Then you take the derivative with respect to time of each side of the equation.

d/dt(A) = d/dt(π*r²)

The left side is simple dA/dt the right side is not that easy.

Remember to use the chain rule.

If it was the derivative with respect to r it would clearly be 2πr, but we are deriving with respect to t.

When using the chain rule first take the derivative ignoring the inside

then multiply by the derivative of the inside.

The derivative as mentioned earlier is 2πr, the derivative of the inside is d/dt(r) which is clearly dr/dt

dr/dt is the rate the radius changes with respect to time.

The full equation is dA/dt = 2πr * dr/dt.

Remember dr/dt is just a number, it's another variable.

Substituting the information given in the problem we get;

6 = 2 * π * 3 * dr/dt

Solving we get dr/dt = 1/π= 0.318 cm/s

Part b is similar, remember that you are given the circumference not the radius.

Answer 2

In this situation, you're needing to find a way to express a change in a circle's area in terms of radius. We can do this using implicit differentiation.

First, we need to differentiate the area formula of a circle with respect to time. This will look very similar to derivatives that you have likely done previously in your calculus studies.

So if the formula is A = π*r ², then we would find the derivative of each side, but making sure we do it in terms of time. On the left side, the derivative of A is 1, but we need to use dA/dt because of how we are differentiating. On the right side, we would get 2π*r , making sure we also put in dr/dt because the change in radius is with respect to time, as well.

Put it all together, and you get:

dA/dt = 2π*r(dr/dt)

Now, let's look at the information we've been given in the problem. We know the rate at which our area is changing - 6 cm²/s - and that's our dA/dt.

For part a, it asks us to find dr/dt when r = 3 cm. Plugging everything in, we're able to find:

dA/dt = 2π*r(dr/dt)

6 = 2*π*3(dr/dt)

6 = 6π(dr/dt)

1/π = dr/dt

So our radius is changing at a rate of 1/π cm/s.

For part b, we are told that the circumference is 4 cm. Since we don't have a formula comparing circumference to volume, we need to get the radius from the given circumference before we start our calculations.

So, since the formula for circumference is C = 2*π*r, we can plug in our values and find:

4 = 2*π*r

4/2π = r

2/π = r

We can now plug that into our differentiated volume formula and get:

dA/dt = 2π*r(dr/dt)

6 = 2*π*(2/π)(dr/dt)

6 = 4(dr/dt)

3/2 = dr/dt

So, our radius is changing in this situation at a rate of 3/2 cm / s.