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Math Help - [SOLVED] Related Rates?

  1. #1
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    [SOLVED] Related Rates?

    Find the rate of change of the distance between the origin and a moving point on the graph of y = x + 1 if dx/dt = 2 cm/sec.

    Here's the solution:
    http://calcchat.tdlc.com/solutionart/calc8e/02/f/se02f01013.gif
    except I don't understand where they got that first equation D = √(x + y)

    can somebody please explain..?
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  2. #2
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    Can't see the pic but, pythag
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  3. #3
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    Quote Originally Posted by tom@ballooncalculus View Post
    Can't see the pic but, pythag
    But why would the pythagorean theorem apply here though..? I don't see a triangle in the graph..
    here's a working link to the solution if you want to look:
    http://i134.photobucket.com/albums/q...13solution.jpg

    Also, how would I do this problem if instead of y = x + 1, it's y = sin x?
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  4. #4
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    "D" represents the distance from the origin to the moving point. That's why you use the Pythagorean theorem: D=\sqrt{x^2 +y^2}

    I made you a picture:



    Since you already know that y=x^2+1, you can plug it in for "y" in the equation D=\sqrt{x^2 +y^2} -----> D=\sqrt{x^2 +(x^2+1)^2}

    From there, take the derivative of D. As you know, the derivative of that equation is \frac{dD}{dt}=\frac{1}{2} (x^4+3x^2+1)^{\frac{-1}{2}}(4x^3+6x)\frac{dx}{dt}

    Since you know that \frac{dx}{dt} = 2, you can just plug it in and multiply it by the derivative of D. That's how you're left with \frac{4x^3+6x}{\sqrt{x^4+3x^2+1}}
    Last edited by Cursed; October 24th 2009 at 11:58 AM.
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  5. #5
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    So when they say origin, they mean the vertex of the graph? ..or does it even matter where you draw the point at?
    Like if in the same problem, y = sin(x) instead of y = x + 1, would I draw a triangle the same way you did, with the point at (0, 0) instead, and then use the pythagorean theorem? or does it require a different method because it's a trig function?
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  6. #6
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    Ooops. I don't know why I did that. o_O :P

    It's supposed to be at (0,0), the origin. Here's a fixed graph:



    And yes, you would still draw a triangle from the origin in that case. Remember, you're finding the distance from the origin to a point (x,y).
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  7. #7
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    haha, ok. I understand it now. Thank you! Your graph really helped!
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