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Related rates differentiation problem

  1. Jul 27, 2012 #1
    1. At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/hr and ship B is sailing north at 25 km/hr. How fast is the distance between the ships changing at 4:00 pm?



    2. None



    3. I have the distance as 150 km. I have the variables [itex]\frac{dx}{dt} = 35[/itex] and [itex]\frac{dy}{dt} = 25[/itex] and [itex]\frac{dZ}{dt} = ?[/itex] I am just wondering on how to set up the equation.
     
    Last edited: Jul 27, 2012
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  3. Jul 27, 2012 #2

    eumyang

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    What's the relationship between x, y, and Z? Note that Z is the hypotenuse of a right triangle.
     
  4. Jul 27, 2012 #3
    x is ship A's position, y is ship B's position and Z is the rate of change in distance.
     
  5. Jul 27, 2012 #4
    Realize that the distance equation is the Pythagorean equation, so try setting up a triangle with what you know, and see if you can figure out how to get the hypotenuse of the triangle from what you are given.
     
  6. Jul 27, 2012 #5
    Ok, thanks for the help.
     
  7. Jul 27, 2012 #6

    eumyang

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    Oops, I misread the part that stated that A started west of B and is traveling east. Suppose we call the point where ship B started point O. |AO| = 150. As time goes on, A gets closer to O. What expression would give us the distance |AO| in terms of x?

    Also, Z is the distance between A and B. It is not the rate of change in distance.
     
  8. Jul 27, 2012 #7
    The equation would be: [itex]\frac{dz}{dt}[/itex] = [itex]\frac{(2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}}{2z}[/itex]
     
  9. Jul 27, 2012 #8

    eumyang

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    No. First, the distance |AO| would be 150 - x, not 150 + x. 2nd, show us the equation BEFORE taking the derivative.
     
  10. Jul 27, 2012 #9
    The equation before taking the derivative is (150 - x)[itex]^{2}[/itex] + y[itex]^{2}[/itex] = z[itex]^{2}[/itex]
     
  11. Jul 27, 2012 #10

    eumyang

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    I think I see what you did earlier. On the side with the x and y, take the derivative separately, and put the dx/dt and dy/dt next to each term. Earlier, you wrote [itex](2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}[/itex] (ignoring the 150+x error), which makes no sense. The numerator should be written as
    (derivative of (150 - x)2)(dx/dt) + (derivative of y2)(dy/dt).
     
  12. Jul 27, 2012 #11
    It would be [itex]\frac{dz}{dt}[/itex] = [itex]\frac{(150 - x)\frac{dx}{dt} + y \frac{dy}{dt}}{z}[/itex]
     
  13. Jul 27, 2012 #12

    eumyang

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    Nope. You didn't take the derivative of (150 - x)2 correctly. Nor the derivative of y2.
     
  14. Jul 27, 2012 #13
    The derivative would be [itex]\frac{2(150-x)\frac{dx}{dt} + 2y\frac{dy}{dt}}{2z}[/itex]
     
  15. Jul 27, 2012 #14

    eumyang

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    Close. You need a negative in front of the 2(150 - x)(dx/dt). (You need to use the chain rule.) Also, to simplify things, factor out the 2 in the numerator and cancel with the 2 in the denominator.

    Find the distance ship A traveled (x) in 4 hours. Find the distance ship B traveled (y) in 4 hours. Find Z. Then plug everything in the dZ/dt equation.
     
  16. Jul 27, 2012 #15
    What I got is [itex]\frac{4300}{10\sqrt{101}}[/itex] km/hr
     
  17. Jul 27, 2012 #16
    Which is about 42.79 km/hr
     
  18. Jul 27, 2012 #17

    eumyang

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    That's not what I got. Can you please show all of your work here, instead of just writing the final answer?
     
    Last edited: Jul 27, 2012
  19. Jul 27, 2012 #18
    35 * 4 = 140 for x, 25 * 4 = 100 for y; [itex]\frac{-2(150-140)*35+2*100*25}{\sqrt{(150-140)^{2}+ 100^{2}}}[/itex] = [itex]\frac{-2*10*35 + 5000}{\sqrt{10^{2} + 100^{2}}}[/itex] = [itex]\frac{-700 + 5000}{\sqrt{10100}}[/itex] = [itex]\frac{4300}{10\sqrt{101}}[/itex]
     
  20. Jul 27, 2012 #19

    eumyang

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    Looks like you removed the 2 in the denominator, but never factored out or canceled the 2's in the numerator. Let's just forget about canceling out the 2. You should have written:
    [tex]\frac{-2(150-140) \cdot 35+2 \cdot 100 \cdot 25}{2\sqrt{(150-140)^{2}+ 100^{2}}}[/tex]
    (Also, don't use "*" for multiplication in LaTeX. Use "\cdot" instead.)
     
  21. Jul 27, 2012 #20
    I see, my mistake. I must have missed the 2 in the denominator.
     
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