Word Problem, How Fast is the Distance Changing: Calc 1 : Related Rates

A

akarbarz

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At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 km/hr and ship B is sailing east at 25 km/hr. How fast is the distance between them changing at 4:00 PM? I have come to an answer but am not sure it is correct or if I set up my equation properly any guidance would be great, thanks. Please let me know if you need to see more work.

I began by drawing a picture of a triangle. Then I took the derivative of the pythagorean theorem.
I ended up with the equation 2(dx/dt)(100+100) + 2yy' = 2zz'
(is 100 +100) correct or should it be (100-100)?
After plugging all values in I got 2(25)(200) + (2)(35)(140) / [(2)(sqrt(59600))]
My final answer was 40.5519.
 
Last edited:
akarbarz said:
At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 km/hr and ship B is sailing east at 25 km/hr. How fast is the distance between them changing at 4:00 PM? I have come to an answer but am not sure it is correct or if I set up my equation properly any guidance would be great, thanks. Please let me know if you need to see more work.

I began by drawing a picture of a triangle. Then I took the derivative of the pythagorean theorem.
I ended up with the equation 2(dx/dt)(100+100) + 2yy' = 2zz'
(is 100 +100) correct or should it be (100-100)?
After plugging all values in I got 2(25)(200) + (2)(35)(140) / [(2)(sqrt(59600))]
My final answer was 40.5519.
Click to expand...
I can't quite follow your reasoning, but I do get the same value. ;)
 
akarbarz said:
At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 km/hr and ship B is sailing east at 25 km/hr. How fast is the distance between them changing at 4:00 PM?

I began by drawing a picture of a triangle. Then I took the derivative of the pythagorean theorem.

I ended up with the equation 2(dx/dt)(100+100) + 2yy' = 2zz'

(is 100 +100) correct or should it be (100-100)?

After plugging all values in I got [2(25)(200) + (2)(35)(140)] / [2*sqrt(59600)]

My final answer was 40.5519.
Click to expand...
I'm not familiar with your method, but your answer matches mine.

Placing ship A at the origin (at noon), a function for the distance between the ships in terms of hours (t) since noon is:

f(t) = 5*sqrt(74t^2 + 200t + 400)

f"(4) = 40.55 kph (rounded)
 
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