1. A dog eats her food out of a square based rectangular prism with both lengths and width of 15cm and a height of 10cm. As soon as the box is totally empty, a machine automatically refills it at a rate of 3cm^3/s. Find the rate of change of the height of the food within the box 2 seconds after it begins filling.

Okay, I know how to do most of this question, the only thing im confused about is what to do with the time = 2s... Any help would be great, thanks.

2. Originally Posted by sugar_babee
1. A dog eats her food out of a square based rectangular prism with both lengths and width of 15cm and a height of 10cm. As soon as the box is totally empty, a machine automatically refills it at a rate of 3cm^3/s. Find the rate of change of the height of the food within the box 2 seconds after it begins filling.

Okay, I know how to do most of this question, the only thing im confused about is what to do with the time = 2s... Any help would be great, thanks.
We have a base area of $15~\text{cm} \times 10~\text{cm} = 150~\text{cm}^2$

So the volume of food in the bowl at any time after the machine starts filling it is
$V = 150h$

Thus
$\frac{dV}{dt} = 150 \frac{dh}{dt}$

We know the food is coming in at $3~cm^3/s$, so that's dV/dt.

Thus
$\frac{dh}{dt} = \frac{\frac{dV}{dt}}{150~cm^2} = \frac{3~cm^3/s}{150~cm^2} = \frac{1}{50}~cm/s$

Since the rate of height change is linear with respect to the rate of volume change, it happens (in this particular case) to be independent of the time.

-Dan

3. so the time they give doesnt have anything to do with the answer? And wouldnt the base be 15 x 15 =225cm? since the height is 10 not the width?

4. Originally Posted by sugar_babee
so the time they give doesnt have anything to do with the answer?
Correct.

You would need the time if your function for $\frac{dh}{dt}$ depended on time. Since it doesn't, you don't need the time.

Originally Posted by sugar_babee
And wouldnt the base be 15 x 15 =225cm? since the height is 10 not the width?
Whoops! Sorry about that. You are correct. The volume formula should have been
$V = 225h$

My apologies.

-Dan