# Re: Related Rates Problems

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Posted by Joel on October 30, 2002 at 00:37:33:

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

Nothing wrong with your answers to #2. Why do you think they're off? Just look at what's happening to the volume. At 6 in radius the volume is 288pi cu in. At 8 in radius the volume is 683pi. At 24 in radius the volume is way up to 18432pi. The rates of increase that you came up with are not out of line with that, are they?

: 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.

: 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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