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Math Help - Related rates word problem. What steps do you follow to solve this?

  1. #1
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    Cool Related rates word problem. What steps do you follow to solve this?

    "At t = 0, a single-engine military jet is flying due east at 12 mi/min. At the same altitude and 208 mi directly ahead of it, still at t = 0, is a commercial jet, flying due north at 8 mi/min. When are the two planes closest to each other? What is the minimum distance between them?"

    I already have the answer, verified by the text, but had to do some unkosher gymnastics to arrive at the answer.

    Thanks
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  2. #2
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    Re: Related rates word problem. What steps do you follow to solve this?

    Could you describe/show the method you used?
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    Re: Related rates word problem. What steps do you follow to solve this?

    Hi Mark from Florida. I drew a diagram first. With this diagram, I set up the planes at their nearest position. The jet is at (x0, 0) and the commercial jet is at (208, y1). I squared the distance formula to avoid radicals. d^2=(x1-x0)^2 + (y1-y0)^2. This gives me d^2=(208-x)^2 + y^2. Differentiation yields 2d dd/dt = 2 (208-x)(dx/dt) + 2y dy/dt. After inputting the knowns (dd/dt is 0, as this is a minimum problem; dx/dt is 12, dy/dt is 8), I simplified and found a formula relating x and y (a ratio). Then I simply made a chart with that ratio per minute. However, I could have just done that all along without the differentiation. What did I miss in the problem solving?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Related rates word problem. What steps do you follow to solve this?

    Here's how I would set this up, which isn't to say it's the best method to say the least!

    I would orient my xy-plane such that the positive y-axis is due north and the positive x-axis is due east. The units are in miles. Put the commercial jet at the origin, and the military jet at (-208,0).

    To find the parametric equations of motion:

    Let t be the parameter representing time measured in minutes.

    Military jet:

    \frac{dx}{dt}=12

    x(t)=12t-208

    Commercial jet:

    \frac{dy}{dt}=8

    y(t)=8t

    Square of distance D between the jets:

    D^2(t)=x^2(t)+y^2(t)=(12t-208)^2+(8t)^2=208(t^2-24t+208)

    Differentiate with respect to t and equate to zero:

    2D(t)D'(t)=416(t-12)=0

    Critical value at t=12

    The first derivative test shows the extremum here is a minimum.

    Now, to find the minimum distance, we compute:

    D(12)=\sqrt{208(12^2-24(12)+208)}=32\sqrt{13}

    Do these results agree with yours?
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    Re: Related rates word problem. What steps do you follow to solve this?

    Mark: thanks again. I don't have my text with me to check your answer. This makes sense, although I have to admit I would'nt have thought of this approach (using x(t), y(t)) on my own. Thanks again.
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