00:01
We have a conical funnel and that usually means a cone with a vertex at the bottom.
00:06
Okay, so something that looks like this.
00:12
So we're told that it is 15 centimeters in diameter and 15 centimeters deep.
00:19
So let's draw out a cone.
00:24
So that means this height is 15 centimeters.
00:29
And the diameter is also 15, which means the radius is 7 .5.
00:37
Okay, so we have this cone that's a little bit.
00:39
Originally filled with liquid.
00:42
But liquid is running out at a rate of 5 centimeters cubed per minute.
00:47
So that means the volume is changing.
00:51
So dvd t is equal to negative five centimeters cubed per minute.
00:59
And it's negative because we're losing volume.
01:04
So the question is at what rate is the depth of the liquid changing when the depth is 8 centimeters.
01:13
So meaning we're looking for, so we want d, h, d t, evaluated at h is equal to 8.
01:28
Okay, so the first thing we're going to do is we're going to write down an equation representing volume.
01:35
So volume of a cone is equal to 1 over 3 pi, r, square, times h.
01:52
Okay, and the next thing we're going to do is take the derivative with respect to t on both sides.
01:57
So the left side, that'll be dvd t.
02:02
And then the right side, we have our constants, pi over three.
02:07
And then here we need to use a product tool.
02:09
So this will be two times r, times drdt times h, plus r square times dh d.
02:22
Okay, so since we're looking for dhd, we want to somehow express drdt in terms of dhdt.
02:31
And we can do that by looking at similar triangles.
02:37
So if we write radius over height, then that's going to be r over h is equal to 7 .5 over 15...