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Math Help - Related rates into regards to applications of the derivative.

  1. #1
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    Related rates into regards to applications of the derivative.

    Stuck on this one.
    An airplane leaves a field at noon and flies east at 100 km/h and a second airplane leaves the same field at 1pm and flies south at 150 km/h. How fast are the airplanes separating at 2pm.

    So if I do a right triangle.
    My Vx right would be 200 and x would be 2 for hours//
    My Vy(south) would be 150 and y 1 hour.
    My Resultant of Vz is what I am looking for and z is 2.23606.
    I am to find Vz and I come up with 245 km/h and the answer is 170 km/h.

    Can someone see what I did wrong?? Thanks
    Joanne
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  2. #2
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    You need to pay attention to the sign +/- of the derivatives you're using. Velocities are vectors and vectors have directions. Did you draw a diagram and set up a coordinate system?
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  3. #3
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    Hi, Our lessons did not include that and none of the questions I have answered, if - or + have to included.
    Were just coming up with the number only. no direction for this question.
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    Quote Originally Posted by bradycat View Post
    Hi, Our lessons did not include that and none of the questions I have answered, if - or + have to included.
    Were just coming up with the number only. no direction for this question.
    Oh, ok. Maybe you can avoid using vectors, and just look at it as a triangle. To find the position of the plane directed east, you need \frac{dx}{dt}t, where t is in hours. The distance of the second plane will be one hour behind, so it's position will be \frac{dy}{dt}(t-1). The distance function could be written:

    s(t)=\sqrt{\left(\frac{dx}{dt}t\right)^2+\left[\frac{dy}{dt}(t-1)\right]^2}

    You're given the velocities:

    \frac{dx}{dt}=100

    \frac{dy}{dt}=150

    So you plug those into the formula, and expand to obtain:

    s(t)=\sqrt{32,500t^2-45,000t+22,500}

    Use the chainrule to differentiate this function in order to find the rate at which the distance between the planes is increasing:

    \frac{ds}{dt}= \frac{2(32,500)t-45,000}{2\sqrt{32,500t^2-45,000t+22,500}}

    This derivative is equal to 170 at t=2.
    Last edited by adkinsjr; February 11th 2011 at 09:51 PM.
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  5. #5
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    Ok got it, I figured this out, and then did it another way how we should be doing it.
    First by the Pyth to find z in kms, then doing a derivative of the x^2+ y^2=z^2.
    THANK YOU for your help.
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