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Tangent lines

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A traveler in space is moving left and right on y = x^2. He shuts off the engines and continues on the tangent line until he reaches point (4, 15). At what point on the curve should he shut off the engines to reach that point?

    3. The attempt at a solution

    Ok, the derivative is 2x. I know have to solve for the slope of the tangent line.

    Rise over run = (15-a^2) / (4 - a)

    (a, a^2) is the point to shut off. I get a^2 - 8a + 15 as the quadratic equation but the value 4 from it isn't correct when I plug it into 2x. What am I doing wrong?
     
  2. jcsd
  3. Oct 16, 2007 #2

    Hootenanny

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    What is the general form of a straight line?
     
  4. Oct 16, 2007 #3
    y = mx + b

    I know. I did that but it isn't the correct answer when looking at the graph.

    I got y = 8x - 17. It doesn't touch the graph
     
  5. Oct 16, 2007 #4

    Hootenanny

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    So we now have to equations for the gradient and our point x=a;

    [tex]m = 2a \hspace{1cm};\hspace{1cm}m = \frac{15-a^2}{4-a}[/tex]

    Can you now solve for a?
     
  6. Oct 16, 2007 #5
    I see what I did wrong now. I did the quadratic wrong because I accidentally put a wrong number in so I had to do the quadratic equation which led to a wrong number. 3 works for a and the point is (3, 9). Isn't there another way to solve this?
     
  7. Oct 16, 2007 #6

    Hootenanny

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    Not that I can think of....
     
  8. Oct 16, 2007 #7
    My teacher said there's an easy way (I'm assuming it's this way) and a hard way.
     
  9. Oct 16, 2007 #8

    Hootenanny

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    The only other option I can think of is graphing it, and with sketching skill that would be the hard way!
     
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