Caculus help!!! sorry!! but i really need it!! =/?

Question

its this related rate problem and i don't know how to do it because it has two variables and i don't know if I am thinking about it properly. i also don't know if i;m applying the given information right... please help =(
here's the question:

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If the water is flowing into the tank at the rate of 10 cubic feet per minute, find the rate of chnage of the depth of the water the instant it is 8 feet deep.

this is what I have done:
Given: (dV/dt)= 10
Find: (dh/dt) when h=8 (i think h=dept)

and since V= 1/3 r^2 h pi (the Greek letter)
then it would be (dV/dt)= 2/3 pi r dr/dt h dh/dt

but then how do you find dr/dt and then how do I apply that the top is 10 feet across and that the conical tank is 12 feet deep????

thank you so much for helping =)

Answer 1

You've done the hardest part already. I suggest converting dV/dt to a single variable. Since the radius is 5 feet (10 feet across) when the depth is 12 feet, and the radius to height relation is linear, you have the relation:

r = (5/12) h

Use that plus your relation: V = (π/3) r^2 h
To get: V = (π/3) ((5/12)h)^2 h = (25 π/432) h^3

Taking the derivative: dV/dt = (25 π/432) 3 h^2 dh/dt = (25 π/144) h^2 dh/dt

Rearranging: dh/dt = (144/(25 π h^2)) dV/dt

For your problem, dV/dt = 10 ft^3/min, h = 8 ft so
dh/dt = (144/(25 π 8^2)) * 10 = 9/(10 π) ft/min.

There is a MUCH easier non-calculus way to figure this too. When the depth is 8 feet the radius is 10/3 feet. The cross-sectional area (π r^2) is 100 π/9 ft^2. Divide the cross sectional area into the flow rate and dh/dt = 10(ft^3/min)/(100 π/9 ft^2) = 9/(10π) ft/min.

Answer 2

I commend you for being close to the answer. Showing where you've come to a dead end and what confuses you is a good sign that you're trying to learn, and I respect you for that.

What you have to use is similar triangles. This is difficult to show here, but you can actually relate the radius and the height.

It would be:

[radius of conical tank] / [height of conical tank] =
[radius of water] / [height of water]

The radius of the conical tank doesn't change; it's 5 (since it's 10 feet across). The height of the conical tank doesn't change either; it's 12. Radius of the water is r, and the height of the water is h. Thus

5/12 = r/h
or, r = (5/12)h

That means in your formula V = (1/3) r^2 h, you can replace r with (5/12) h. This r has to be replaced BEFORE you differentiate, giving you

V = (1/3) [(5/12) h]^2 (h)
V = (1/3) (25/144) h^3
V = (25/432)h^3

Note that it's been quite a long time since I've done related rates but I know the answer lies in similar triangles. I hope that if I gave you the wrong answer, that I at least helped point you in the right direction.

Edit: Looks like I got a thumbs up and a similar answer below, so I guess it's correct. Basically, the fact that there is a "cone within a cone" was pivotal in being able to associate r and h, and being able to express r in terms of h. It has to do with how they are proportionate to each other.

Answer 3

You are off to a good start. You need to convert the formula for volume from two variables to one.

Let

h = height of water in tank
r = radius of the surface of the water in tank
V = volume of water in tank

Given

dV/dt = 10 ft³/min
h = 8 ft

Find

dh/dt when h = 8 ft

Notice that for the tank, the height is 12 feet and the radius is 10/2 = 5 feet. So no matter what height the tank is filled to, the ratio of the radius of the surface of the water to the height of the water will be the same.

r/h = 5/12
r = (5/12)h

V = ⅓πr²h

Substituting (5/12)h for r

V = ⅓π[(5/12)h]²h = ⅓π(25/144)h³ = 25πh³/432

dV/dh = 3*25πh²/432 = 25πh²/144

Now we have what we need to solve for dh/dt.

dh/dt = (dV/dt) / (dV/dh) = 10 / (25πh²/144)
dh/dt = 1440 / (25πh²) = 288 / (5πh²)

Plugging in the value of h = 8

dh/dt = 288 / (5πh²) = 288 / (5π(8²)) = 288 / (320π)
dh/dt = 9 / (10π) = 0.2864789 ft/min

Answer 4

V= 1/3 r^2 h pi......(1)
Since you have two variables in volume function, you need to find a restriction.

By triangle similarity, we have
r/h = 5/12
which leads to r = (5/12)h......(2)

Substitute (2) into (1) gives,
V = (25/432)h^3 pi ......(3)

Take derivative of (3),

dV/dt = pi(25/144)h^2 dh/dt = 10 ft^3/min

Plug in h = 8 ft,

dh/dt = 10(144)/(25x8^2pi) = 0.29 ft/min