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Math Help - Distance between two points

  1. #1
    Senior Member polymerase's Avatar
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    Distance between two points

    f(x)=-x^2 for x<0 and f(x)=\sqrt{x} for x \ge 0. There is a line tangent to these two graphs at two points. Find the distance between these two points.

    Thanks!
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  2. #2
    Eater of Worlds
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    This is practically the same problem I answered before. You can use the same method to find the points, then use the distance formula to find the distance between them.
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by galactus View Post
    This is practically the same problem I answered before. You can use the same method to find the points, then use the distance formula to find the distance between them.
    I know, but when i tries it with your method, im getting extremely large values. And the other thing is, in your original post your said that q=\dfrac{p^2}{4} but when i set the derviative equal to one another im getting q=\dfrac{4}{p^2}....
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  4. #4
    Eater of Worlds
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    Then that woukd dictate I made a typo. Sorry, just typed it backwards. I will fix it.
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  5. #5
    Eater of Worlds
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    Let's do this one. I may have inadvertently made some typos in the last post. The result is correct, though.

    y=-x^{2}, \;\ y=\sqrt{x}

    Let p be a point on -x^{2}: P(p,-p^{2})

    Let q be a point on \sqrt{x}: Q(q,\sqrt{q})

    The slope of P is -2p

    The slope of Q is \frac{1}{2\sqrt{q}}

    Therefore, q=\frac{1}{16p^{2}}.......[1]

    The slope from P to Q is \frac{-p^{2}-\sqrt{q}}{p-q}

    This has to be the slope at P (or Q):

    \frac{-p^{2}-\sqrt{q}}{p-q}=-2p

    Sub in [1]:

    \frac{-p^{2}-\sqrt{\frac{1}{16p^{2}}}}{p-\frac{1}{16p^{2}}}+2p=0

    Now, solve for p and q. You can get points and find the distance between them.

    This is how I always went about these 'line tangent to two points' problems. Maybe you can find another way.
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