How Fast Are the Man and Woman Moving Apart in the Related Rates Problem?

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SUMMARY

The discussion focuses on solving a related rates problem involving a man and a woman moving apart. The man walks north at 6 ft/s, while the woman starts walking south at 2 ft/s after a 5-minute delay, beginning from a point 500 ft east of the man. The correct approach to find the rate at which they are moving apart involves differentiating the equation \(d^2 = x^2 + y^2\) and applying the chain rule, yielding \(dd/dt = (y * dy/dt) / d\). After 30 minutes of the woman walking, the rate of separation is confirmed to be 8 ft/s.

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A man starts walking north at 6 ft/s from a point P. Five minutes later, a woman starts walking south at 2 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 30 min after the woman starts walking?

so i was ableto set up a right triangle using the x-y plane (y being north and x being east)

so i was trying to relate the components with distance^2=x^2+y^2
then where i differientate i get dd/dt=dx/dt+dy/dt

i know dy/dt=8 and i want to know dd/dt

there dx/dt=0 because there is no change in x...x is always 500 ft

so would dd/dt=dy/dt=8 ?
 
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Rasine said:
A man starts walking north at 6 ft/s from a point P. Five minutes later, a woman starts walking south at 2 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 30 min after the woman starts walking?

so i was ableto set up a right triangle using the x-y plane (y being north and x being east)

so i was trying to relate the components with distance^2=x^2+y^2
then where i differientate i get dd/dt=dx/dt+dy/dt
Then you differentiated wrong! If dd^2= x^2+ y^2 then, differentiating, 2dd d(dd)/dt= 2x (dx/dt)+ 2y (dy/dt). Since dx/dt= 0,
2dd d(dd)/dt= 2y(dy/dt) so dd d(dd)/dt= 2y dy/dt. Now, 30 minutes after the woman starts walking (35 minutes after the man starts walking) what are dd and y?

i know dy/dt=8 and i want to know dd/dt

there dx/dt=0 because there is no change in x...x is always 500 ft

so would dd/dt=dy/dt=8 ?
Very close but not exactly!
 
Pyramid shaped tank...?

Does anyone know how to do this problem on related rates?:

The base of a pyramid-shaped tank is a square with sides of length 8 meters, and the vertex of the pyramid is 8 meters above the base. The tank is filled to a depth of 2 meters, and water is flowing into the tank at the rate of 4 cubic meters per minute. Find the rate of change of the depth of water in the tank. (Hint: the volume of a pyramid is given by V=1/3bh, where b is the base area and h is the height of the pyramid)
 
Please, please, please! Do not "hijack" someone else's thread for a different question. Start your own thread- it's not that difficult!

The base of a pyramid-shaped tank is a square with sides of length 8 meters, and the vertex of the pyramid is 8 meters above the base. The tank is filled to a depth of 2 meters, and water is flowing into the tank at the rate of 4 cubic meters per minute. Find the rate of change of the depth of water in the tank. (Hint: the volume of a pyramid is given by V=1/3bh, where b is the base area and h is the height of the pyramid)
Okay, can you determine how h is related to b for different heights of the water?
 

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