If water is poured into a cup at a rate of 1 cubic centimeter per second, how fast is the dept of the water increasing when the water is 4 cm deep?

r = 6 cm
h = 9 cm

V = 1/3(pie) r^2 x h
V = (pie)r'h + 1/3 (pie)r^2 h'

6 / 9 = r / h

6h = 9r
r = 3 / 4.5 h

V = 1/3(pie) (3/4.5h)^2

My answer came out to 9 / 64(pie) cm/sec.

Here is what I did

from above
r = 2h/3

V = (1/3)π(r^2)h
= (1/3)π(4h^2/9)h
= (4π/27) h^3

dV/dt = (4π/9)h^2 dh/dt
when dV/dt = 1 and h = 4

1 = (4π/9)(16)dh/dt
dh/dt = 9/(64π)

(notice that Katie had r = 3h/4.5 which I found strange since that is simply 2h/3 )

what is the shape of the container?

The shape is a cone. The radius is 6 cm and the height is 9 cm

Your answer is correct if you write it as

9/(64pi).

I am curious why you went from
6h = 9r to
r = 3/4.5

why not just do
r = 6h/9 = 2h/3
instead of having decimals in your fraction.
Either we use fractions or we use decimals, but we generally don't mix them together.

I think you have a mistake. I was working on this problem at the same time.

V = 1/3 pi r^2 h
Only 'r' is squared. It appears that you squared 'h' too.
V = 1/3(pie) (3/4.5h)^2

Anyway, even with squaring h too, I don't come up with 9/64 pi.

I used r = 2/3 instead of r = 2/3 h!

Thanks so much REINY.

I really appreciate you taking the time to explain this to me.

I am teaching myself calculus, and sometimes the book just doesn't explain things well.

The r = 3h/4.5 threw me at first, until I figured out what she did (I had 2h/3).

To find the rate at which the depth of the water is increasing, we need to differentiate the volume equation with respect to time and then solve for dh/dt (the rate at which the depth is changing).

Given:
V = (1/3)πr^2h

We know that r = (3/4.5)h, which means r^2 = (9/20.25)h^2.

Substituting this into the volume equation:

V = (1/3)π(9/20.25)h^2 * h
V = (3/60.75)πh^3
V = (π/20.25)h^3

Now, differentiate both sides of the equation with respect to t (time):

dV/dt = (3π/20.25)d(h^3)/dt

Since the volume is increasing at a constant rate of 1 cm^3/sec, dV/dt = 1. Therefore:

1 = (3π/20.25)d(h^3)/dt

Now, we need to find d(h^3)/dt:

d(h^3)/dt = 20.25/(3π)

Finally, substitute this value back into the equation to find dh/dt (which represents the rate at which the depth is changing):

1 = (3π/20.25) * 20.25/(3π) * dh/dt

Therefore, the rate at which the depth of the water is increasing when the water is 4 cm deep is 1 cm/sec.