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Math Help - Related rates problem

  1. #1
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    Related rates problem

    Hey, I have this problem that I am halfway done, but I do not know how to solve the second portion.

    Question
    Ship S is 41 Km due North of ship T. Ship S sails south at a rate of 9 km/h. Ship T sails west at 10 km/h.
    a) At what rate are they approaching or separating one hour later?
    b) When do they cease to approach one another and how far apart are they at this time?

    Solution
    a)
    I figured out the distance after 1 hour: 33.52km
    I then figured out the rate of change of the seperation/approach: 0.57kmh (approaching)

    b) I do not know how to solve this part though. I could use the equation I used to find the rate of change, but setting that to zero really doesn't help in finding 3 other variables. Help?
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  2. #2
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    I think you need to find out when the hypotenuse changes from minimizing the distance as t moves west and s moves south to when the hypotenuse starts increasing the distance and then find that distance of the hypotenuse at that moment.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    I think you need to find out when the hypotenuse changes from minimizing the distance as t moves west and s moves south to when the hypotenuse starts increasing the distance and then find that distance of the hypotenuse at that moment.
    Okay. How do I do that?

    Thank you for the response!
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  4. #4
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    I haven't done related rates in awhile but when you are minimizing or maximizing you need two equations. One will be a^2+b^2=c^2 since we want to minimize c. The other might be the area of triangle not 100% on that though.
    What we know:

    \dot x =10

    \dot y =-9

    From this, I think you will have to implicit differentiate the Pythagorean identity but after that I am a little hazy on what needs to be done.
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    I haven't done related rates in awhile but when you are minimizing or maximizing you need two equations. One will be a^2+b^2=c^2 since we want to minimize c. The other might be the area of triangle not 100% on that though.
    What we know:

    \dot x =10

    \dot y =-9

    From this, I think you will have to implicit differentiate the Pythagorean identity but after that I am a little hazy on what needs to be done.
    Thanks for trying!
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  6. #6
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    Anyone know what I need to do? I think I just need to find when they are at a minimum distance from each other, that is the same as when the rate of change of distance apart would be zero.
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  7. #7
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    Quote Originally Posted by Kakariki View Post
    I could use the equation I used to find the rate of change, but setting that to zero really doesn't help in finding 3 other variables.
    Yes it does - at least, you don't need to worry about hypotenuse z because it goes to zero with dz/dt (if it's on the dz/dt side of the equals sign) or else is merely the denominator of a zero-valued fraction (if it's on that side). And you know that for y you can substitute 41 - 9/10 x.
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  8. #8
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    Quote Originally Posted by Kakariki View Post
    Hey, I have this problem that I am halfway done, but I do not know how to solve the second portion.

    Question
    Ship S is 41 Km due North of ship T. Ship S sails south at a rate of 9 km/h. Ship T sails west at 10 km/h.
    a) At what rate are they approaching or separating one hour later?
    b) When do they cease to approach one another and how far apart are they at this time?

    Solution
    a)
    I figured out the distance after 1 hour: 33.52km
    I then figured out the rate of change of the seperation/approach: 0.57kmh (approaching)

    b) I do not know how to solve this part though. I could use the equation I used to find the rate of change, but setting that to zero really doesn't help in finding 3 other variables. Help?
    Related rates problem-related-rates.jpg

    A is the starting position of ship S

    B is the starting position of ship T

    AB = 41

    x=10t

    y=41-9t

    z=\sqrt{(10t)^2+(41-9t)^2}

    \frac{dx}{dt}=10 and {dy}{dt}=-9

    x^2+y^2=z^2

    2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}

    \frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}  {z}

    Set \frac{dz}{dt}=0

    x\frac{dx}{dt}+y\frac{dy}{dt}=0

    (10t)(10)+(41-9t)(-9)=0

    Solve for t and plug into the formula for z
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