This is a related rate problem.

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In summary: That will give you the distance between them after 15 minutes. Differentiating that should give the rate at which they are separating after 15 minutes.In summary, this is a related rate problem involving a man and a woman walking in opposite directions at different speeds. By setting up a coordinate system, the positions of the man and woman can be represented as functions of time. Using the Pythagorean theorem, the distance between them can be calculated and differentiated to find the rate at which they are moving apart. The answer is 837/√8674 ft/s.
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rocketdiver
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This is a related rate problem.

A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts walking south at 5 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?

I'm just trying to interpret it into the numbers and drawing a triangle. But I don't get a right equations for it.


the answer is 837/ sqaure root of 8674 which is 8.99 ft/s

Thank you so much.
 
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  • #2
First be careful of your units in this problem, the speeds are given in ft/s but the times are given in minutes. Second, you need to determine where the two are at when the lady starts walking. You should get a nice number for the "y" value of his position. Third, you need to find out where they are after the fifteen minutes have passed. Fourth, the use the Pythagorean theorem as the relationship between them, differentiate wrt time, but be careful because one of the time derivatives is zero. Can you see why? Finally, substitute the values you found earlier, you should end up with the rate of change between the two as the only unknown. I worked it out and the answer you have is what I got.
 
  • #3
I work better thinking in coordinates than triangles so what I would do is set up a coordinate system with (0,0) at P, positive y-axis north, positive x-axis east.

Then the man's position, t seconds after he starts walking is (0, 4t).

The woman starts at (500,0) and walks south. Since she starts walking 5 min= 300 seconds later, at t seconds after the man starts walking she has been walking for t- 300 seconds. Her position will be (500, 5(t-300)). What is the distance from (0, 4t) to (500, 5(t-300)) as a function of t? What is the rate of change (derivative) of that function? And, as xman said, the question asks about 15 minutes after the woman started walking which is 20 minutes= 1200 seconds after the man started walking: t= 1200 s.
 

Related to This is a related rate problem.

What is a related rate problem?

A related rate problem is a type of mathematical problem that involves finding the rate of change of one quantity in relation to the rate of change of another quantity. These problems typically involve changing rates, such as speed or volume, and require the use of calculus to solve.

How do you solve a related rate problem?

To solve a related rate problem, you must first identify the quantities that are changing and the relationship between them. Then, you can use the chain rule to find the derivative of each quantity with respect to time. Finally, you can set up an equation using the given rates and solve for the unknown rate.

What are some common examples of related rate problems?

Some common examples of related rate problems include finding the rate at which the volume of a sphere is changing as its radius changes, finding the rate at which the distance between two moving objects is changing, and finding the rate at which the area of a triangle is changing as one of its sides changes.

Why are related rate problems important?

Related rate problems are important because they allow us to analyze and understand the changing relationships between quantities in real-world situations. They also require the use of calculus, which is a fundamental tool in many fields of science and engineering.

What are some tips for solving related rate problems?

Some tips for solving related rate problems include drawing a diagram to visualize the problem, using the correct units for rates, and checking your answer to make sure it makes sense in the given context. It is also helpful to practice solving various types of related rate problems to become more familiar with the process.

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