A tank of water is in the shape of a cone and is leaking water at a rate of 35 cm^3/sec. The base radius of the tank is 1 meter and the height of the tank is 2.5 meters. When the depth of the water is 1.25 meters at what rate is the (a) depth changing and (b) the radius of the top of the water changing?

I figured out the values, but I don't understand how to find dr/dt and dh/dt.

Question

A tank of water is in the shape of a cone and is leaking water at a rate of 35 cm^3/sec.  The base radius of the tank is 1 meter and the height of the tank is 2.5 meters.  When the depth of the water is 1.25 meters at what rate is the (a) depth changing and (b) the radius of the top of the water changing?

I figured out the values, but I don't understand how to find dr/dt and dh/dt.

anonymous · Answer

RATE=D/T

ashleynguyenx3 · Answer

I drew a picture and used proportions|dw:1350278686551:dw|

anonymous · Answer

so then based on what you wrote above I need to figure out exactly where to start. Can you tell me the equation of the depth in relation to time?

ashleynguyenx3 · Answer

so the volume of the cone is v=(1/3)pi(r^2)h
then I just found the derivative with respect to t

ashleynguyenx3 · Answer

so then i plug in my values which are
v'=35cm^3/sec
h=1.25
r=0.5

ashleynguyenx3 · Answer

$$v'=2/3\pi rr'h+1/3\pi r^2 h'$$

anonymous · Answer

sorry I keep getting responses at other questions

ashleynguyenx3 · Answer

Oh, it's okay

ashleynguyenx3 · Answer

I just don't get how to get r' and h' cause I have all the other values already

anonymous · Answer

One second to help a stranger on the internet I am going to go and get my old math 31 book from a year ago to remember some stuff

ashleynguyenx3 · Answer

Alright :)

anonymous · Answer

My math 31 textbook cheats a little in its example if you want I will show it to you. I believe it should give you the correct answer and full marks anyways if anything for this problem it actually ends up be a little bit longer

ashleynguyenx3 · Answer

Okay, sure.

anonymous · Answer

okay if we go back to the original volume formula of v = (1/3)*pi*r^2*h we can replace r with an equation in the form of h. The key relationship for this switch is that "the ratio of height and radius of the water in the tank will remain constant throughout the fill"

anonymous · Answer

This relationship is given by
r/h = 1/2.5
solve for h to give you
r = (1/2.5)*h
now sub this in for r in the original equation for volume giving you
V = (1/3)*pi*(h/2.5)^2*h
V = (1/3)*pi*((h^2)/(2.5^2))*h
V = (1/3)*pi*((h^3)/(2.5^2))

anonymous · Answer

or
V = (4/75)*pi*h^3

ashleynguyenx3 · Answer

So then you'd find the derivative of that to solve for h'
And once you solve for h', you'd have all the values.
Am I correct? Or?

anonymous · Answer

now apply d/dt to the equation you found to get
V' = (4/75)*pi*3*h^2*(dh/dt)
yes I think you are correct as you have the V' and the height in question so therefore you can solve for dh/dt

ashleynguyenx3 · Answer

And we have r too right?

ashleynguyenx3 · Answer

I think, I got it from here. Thanks! :D

anonymous · Answer

This is what I was talking about earlier. In this part r was replaced as a function of h to get rid of it so that we did not run into it or its counter part dr/dt . In part b we are looking for dr/dt therefore we will need a different equation then the one used in part a. Since we now have dh/dt you could use the original volume expression or you could sub the other way around finding h as a function of r. Okay then I am heading to bed goodnight and you are welcome and please remember to medal me if I helped

ashleynguyenx3 · Answer

Yeah, I will. Good night. :)

anonymous · Answer

one thing to calculate r at that moment use the ratio formula we talked about earlier. I don't think I mentioned that and I do not want you to get stuck at that part.