Question

A snowball is melting at the rate of 2 inches per hour. How fast is the surface area of the snowball changing at the instant the snowball has a radius of 3 inches

i tried (d/dt) = 4pir^2 (dr/dt)

its multiple choice and -72pi isnt there so im just trying to figure this out

Answer 1

SA = 4πr^2

d(SA)/dt = 8πr dr/dt
subbing in our values
d(SA)/dt = 8π(3)(-2) = -48π

you differentiated volume, it asked to find the rate of change of the surface area. So you need to differentiate surface area

Answer 2

it's -48pi. ur welcome flvs fam

Answer 3

[AI-generated answer — Clown Bot]

Well, I hate to break it to you, but it seems like this snowball is melting faster than a popsicle in the Sahara! Let's get to the math, shall we?

First, we need to find the rate at which the radius is changing. Let's call that "dr/dt". In this case, we are given that the snowball is melting at a rate of 2 inches per hour. So, dr/dt = -2 inches per hour (since the radius is getting smaller).

Next, we need to find the surface area of the snowball, which is given by the formula A = 4πr^2. Since we want to find how fast the surface area is changing, we need to take the derivative with respect to time, which gives us dA/dt = 8πr (dr/dt).

Now plug in the given radius of 3 inches and the rate of radius change of -2 inches per hour into the equation:

dA/dt = 8π(3)(-2) = -48π

So, at the instant the snowball has a radius of 3 inches, the surface area is changing at a rate of -48π square inches per hour.

I apologize for not having -72π as one of the options, but hey, at least now you know how quickly that snowball is vanishing!

Answer 4

[AI-generated answer — Explain Bot]

To find the rate of change of the surface area of the snowball, you can use the formula for the surface area of a sphere, which is A = 4πr^2, where A represents the surface area and r represents the radius.

To start, you correctly identified that you need to differentiate the equation with respect to time, t, using the chain rule. So, the correct formula is:

(dA/dt) = 4πr^2 (dr/dt)

In this case, you are given that the snowball is melting at a rate of 2 inches per hour (dr/dt = -2, since the radius is decreasing). Moreover, you are also given that the radius of the snowball is 3 inches at the instant of interest (r = 3). Plugging these values into the formula, we get:

(dA/dt) = 4π(3)^2 (-2)
= -72π

Hence, the correct answer is -72π. Make sure to double-check the options to ensure you haven't missed it. If it is not provided as an option, it could be a case of rounding or approximation used in the options.