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

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Homework Help Overview

The discussion revolves around a related rates problem involving two individuals walking in different directions and the rate at which they are moving apart. The context includes the setup of a right triangle in the x-y plane, with one person walking north and the other south, while maintaining a constant eastward distance.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the problem using a right triangle and the relationship between the distances involved. There is an attempt to differentiate the equation relating the distances to find the rate of separation. Some participants question the differentiation process and the assumptions made regarding the rates of change.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the differentiation and the implications of the rates involved. There is a recognition of close approximations but no explicit consensus on the final rate of separation.

Contextual Notes

One participant notes the specific time frame of 30 minutes after the woman starts walking, which adds a temporal element to the calculations. There is also a mention of the constant distance in the eastward direction, which is a key assumption in the problem setup.

Rasine
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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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