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

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The discussion centers on a related rates problem involving a man and a woman walking in perpendicular directions. The man walks north at 6 ft/s, while the woman starts walking south at 2 ft/s after a delay of five minutes. A right triangle is established to relate their distances, leading to the equation dd/dt = dy/dt, where dy/dt is the combined speed of both individuals. After 30 minutes of the woman walking, the rate at which they are moving apart is calculated, with clarification on differentiating the distance formula. The conversation also briefly touches on a separate question regarding a pyramid-shaped tank, emphasizing the importance of keeping discussions focused.
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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 becuase 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 becuase 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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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