# Math Help - Calculus - Related Rate...?

1. ## Calculus - Related Rate...?

#1 Water is poured into a rectangular box with a square base (4 cm x 4 cm x 12 cm) at a rate of 50 cm^3/min. What rate does the surface of the water rise?

V = h(l^2)
V = 12(l^2)
dV/dt = 24 = 24l(dh/dt)
50 = dV/dt = 24(4)(dh/dt)
50 = 96(dh/dt)
50/96 cm/min = dh/dt

#2 A cone is filled with water at a rate of 6m^3/min.

If the reservoir is 3 m deep, and has a 4 m radius at the top, what is the rate at which the surface level of the water is rising at the instant the depth of the water is 2 m?

2. Originally Posted by AlphaRock
#1 Water is poured into a rectangular box with a square base (4 cm x 4 cm x 12 cm) at a rate of 50 cm^3/min. What rate does the surface of the water rise?

V = h(l^2)
V = 12(l^2)
dV/dt = 24 = 24l(dh/dt)
50 = dV/dt = 24(4)(dh/dt)
50 = 96(dh/dt)
50/96 cm/min = dh/dt
apparently they gave you the dimensions of the box, unless i am misinterpreting what that "4 cm x 4 cm x 12 cm" means.

in that case, we have

V = 16h

that's it. so h = V/16

now differentiate both sides with respect to time and plug in dV/dt

#2 A cone is filled with water at a rate of 6m^3/min.

If the reservoir is 3 m deep, and has a 4 m radius at the top, what is the rate at which the surface level of the water is rising at the instant the depth of the water is 2 m?

the reservoir is the cone?

that's not the answer i got. what equation did you use?

3. Originally Posted by Jhevon
apparently they gave you the dimensions of the box, unless i am misinterpreting what that "4 cm x 4 cm x 12 cm" means.

in that case, we have

V = 16h

that's it. so h = V/16

now differentiate both sides with respect to time and plug in dV/dt

the reservoir is the cone?

that's not the answer i got. what equation did you use?

I used V=(1/3)pi(r^2)h

I got 8/3pi = dh/dt for the cone. If this is wrong, can you show me how to do it right?

Can somebody show me how to do the rectangular box question? Because I'm stuck on that one still.

4. Originally Posted by AlphaRock
Thanks.

I'm stuck on the cone problem. Can somebody show me how to do it?

I used V = (1/3)pi(r^2)h
see post #3 here, then try the problem again and post your result

5. Thanks Jhevon.

That was fast!

I edited my post after realizing I got 8/3pi.

I'm still not sure on how to get the rectangular box question.

6. Originally Posted by AlphaRock
#1 Water is poured into a rectangular box with a square base (4 cm x 4 cm x 12 cm) at a rate of 50 cm^3/min. What rate does the surface of the water rise?

V = h(l^2)
V = 12(l^2)
dV/dt = 24 = 24l(dh/dt)
50 = dV/dt = 24(4)(dh/dt)
50 = 96(dh/dt)
50/96 cm/min = dh/dt
you have made height a constant and the square base variable ... it's the other way around. think about what dimension is changing as water is poured into the box. the height of the water is variable, and the square base is a constant.

V = 16h

now find take the time derivative and solve for dh/dt.

7. Since we are dealing with a box with straight sides, you do not even need calculus.

We need the height when V=50. Think of that as a rectangular box inside the box that holds 50 cm^3

$16h=50$

Solve for h and that is the height per minute.

You can get the same result through calc though.

$V=16h$

$\frac{dV}{dt}=16\cdot\frac{dh}{dt}$

Since dV/dt=50, solve for dh/dt.

It's the same thing as the first. The height change remains constant because we have a nice shape we're dealing with.

8. Thanks guys. You really helped me out.

Did you guys get 8/3pi for the cone question?

9. Originally Posted by AlphaRock
Thanks guys. You really helped me out.

Did you guys get 8/3pi for the cone question?
no.

$\frac{r}{h} = \frac{4}{3}$

$r = \frac{4h}{3}$

$V = \frac{\pi}{3} r^2 h$

$V = \frac{\pi}{3} \left(\frac{4h}{3}\right)^2 h$

$V = \frac{16\pi}{27} h^3$

$\frac{dV}{dt} = \frac{16\pi}{9} h^2 \cdot \frac{dh}{dt}$

... sub in your known values and re-calculate $\frac{dh}{dt}$

10. Originally Posted by skeeter
no.

$\frac{r}{h} = \frac{4}{3}$

$r = \frac{4h}{3}$

$V = \frac{\pi}{3} r^2 h$

$V = \frac{\pi}{3} \left(\frac{4h}{3}\right)^2 h$

$V = \frac{16\pi}{27} h^3$

$\frac{dV}{dt} = \frac{16\pi}{9} h^2 \cdot \frac{dh}{dt}$

... sub in your known values and re-calculate $\frac{dh}{dt}$
6 = (16pi/9)(2^2)(dh/dt)
6 = (16pi/9)(4)(dh/dt)
6 = (64pi/9)(dh/dt)
6/(64pi/9) = (64pi/9)(dh/dt)/(64pi/9)
54/64pi = dh/dt
27/32pi = dh/dt

My final answer. I hope this is right...

11. Originally Posted by AlphaRock
6 = (16pi/9)(2^2)(dh/dt)
6 = (16pi/9)(4)(dh/dt)
6 = (64pi/9)(dh/dt)
6/(64pi/9) = (64pi/9)(dh/dt)/(64pi/9)
54/(64pi) = dh/dt

My final answer. I hope this is right...
units?

12. Originally Posted by skeeter
units?
Okay, this is my FINAL, FINAL answer:

27/(32pi) m/min

?