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Posted by Joel on October 29, 2002 at 19:59:14:

In Reply to: Related Rates Problems posted by C. Slayden on October 29, 2002 at 18:06:30:

: I have been working through my pages of homework and I have come across a few related rate problems that seem to have confused me. So I wanted to see if someone could help me with these. Thank you in advance.

: 1. Find the rate of change of the distance between the origin and a moving point on the graph of y=sin(x) if dx/dt=2 centimeters per second.

: I figured you had to use the distance formula from here but I am not sure at how to.

I don't think you can get a numerical solution to this without also knowing x. Given only dx/dt you can get the answer in terms of x.
Call the distance you are looking for s. You can find s in terms of x, and then relate ds/dt to dx/dt using:
ds/dt = ds/dx * dx/dt

Pythagoras tells you that
s = (x^2 + sin^2(x))^(1/2)
so ds/dx = (x + sin(x)cos(x))/((x^2 + sin^2(x))^(1/2))
and ds/dt is 2 times that.

: 2. The radius r of a sphere is increasing at a rate of 2 inches per minutes. Find the rate of change of the volume when r=6 inches and r=24 inches.

: I worked it out and I recieved a hilatious answer. For r=6 inches I got 288(pi) and for r=24 I got 4608(pi). Here is my work. I differentiated the volume for a sphere and got dv/dt=4(pi)r^2(dr/dt) and then pluged in the numbers 4(pi)6^2(2)=dv/dt=288(pi)and 4(pi)24^2(2)=dv/dt=4608(pi). right? wrong?

: 3.All edges of a cube are expanding at a rate of 3 centimeters per second. How fast is the volume changing when each edge is 1 centimeter?

: My work. Took the volume for a cube and differentiated it dv/dt=dl/dt(dw/tw)(dh/dt) and just plugged in the 1 for each and recieved a answer of 1. I have no clue if I did that right.

: 4.The formula for a volume of a cone is v=(1/3)(pi)r^2(h). Find the rate of change of the volume if dr/dt is 2 inches per minute and h=3r when r=6 inches.

: My work. Replaced h for 3r in the formula and differentiated. Recieved dv/dt=3(pi)dr/dt(r^2+2r) plugged the numbers in again and recieved 288(pi)

: 5.A conical tank (with vertex down) is 10 feet acros the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rete of change of the depth of the water when the water is 8 feet deep.

: For some reason couldn't figure this one at in any shape or form. Again, thank you in advance

: C. Slayden

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