A side of a cube is increasing at the rate of 3 feet per minute

solidsnake123

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A side of a cube is increasing at the rate of 3 feet per minute. Find the rate at which the area is increasing when the side is 5 feet.

Please help me. I'm totally lost.
 
Okay, so since you say you're "totally lost," we'll have to start at the very beginning. You're told that the sides of a cube are increasing at a rate of 3 feet per minute. What does that tell you? (Hint: "The derivative of (something) with respect to (something) else is 3"). If we let s = the length of one side of the cube, and t = the number of minutes elapsed, what expression can you now create? At this point, I suspect you made a typo when writing down the problem. For the rest of the problem. I'll assume you meant to write "volume" instead of "area." So, what is the formula for the volume of a cube, in terms of its side length? Or, if you meant "surface area" instead, you can do the exact same calculations, but start with formula for surface area instead.

Now, since the problem asks you to find how fast the volume is changing when the side length is 5 feet, you'll need to find the derivative of the volume with respect to the side length. But, now, remember that the side length itself is not a constant value, but is changing with respect to time. What does the Chain Rule suggest you need to do here? How does this relate to the derivative you found earlier, using the information that the side length is changing at a rate of 3 feet per minute? What does it mean that the problem wants you to find the rate the volume is changing when the side length is 5 feet?

If you get stuck again, that's okay, but please include any and all work you've done on this problem, even if you know it's wrong. Thank you.
 
Okay, so since you say you're "totally lost," we'll have to start at the very beginning. You're told that the sides of a cube are increasing at a rate of 3 feet per minute. What does that tell you? (Hint: "The derivative of (something) with respect to (something) else is 3"). If we let s = the length of one side of the cube, and t = the number of minutes elapsed, what expression can you now create? At this point, I suspect you made a typo when writing down the problem. For the rest of the problem. I'll assume you meant to write "volume" instead of "area." So, what is the formula for the volume of a cube, in terms of its side length? Or, if you meant "surface area" instead, you can do the exact same calculations, but start with formula for surface area instead.

Now, since the problem asks you to find how fast the volume is changing when the side length is 5 feet, you'll need to find the derivative of the volume with respect to the side length. But, now, remember that the side length itself is not a constant value, but is changing with respect to time. What does the Chain Rule suggest you need to do here? How does this relate to the derivative you found earlier, using the information that the side length is changing at a rate of 3 feet per minute? What does it mean that the problem wants you to find the rate the volume is changing when the side length is 5 feet?

If you get stuck again, that's okay, but please include any and all work you've done on this problem, even if you know it's wrong. Thank you.

I copied down the exact problem from my study guide. It says Area, so I assume it's surface area.

So I wrote down ds/dt=3 and A(T) = 6s^2 s for side length.
The derivative of A'(T) = 12 x s x ds/dt, right?
So would it be 12 x 5 x 3, which is 180? Is A'(t) = 180?
 
I copied down the exact problem from my study guide. It says Area, so I assume it's surface area.

So I wrote down ds/dt=3 and A(T) = 6s^2 s for side length.
The derivative of A'(T) = 12 x s x ds/dt, right?
So would it be 12 x 5 x 3, which is 180? Is A'(t) = 180?

Yes that's correct., but don't forget the units. In this case it will be square feet per minute.
 
I copied down the exact problem from my study guide. It says Area, so I assume it's surface area.

So I wrote down ds/dt=3 and A(T) = 6s^2 s for side length.
The derivative of A'(T) = 12 x s x ds/dt, right?
So would it be 12 x 5 x 3, which is 180? Is A'(t) = 180?

Yes. That's the answer I get, for the rate of change of surface area.
 
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