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Related rates

  1. Feb 22, 2007 #1
    a partical moves along the curve y=(17+x^3)^1/2. as it reaches the point (4,9) the y coordinate is increasing at a rate of 4 cm/s. how fast is the x-coordinate of the point changing at that instant?

    so dy/dt=4 cm/s

    i diffeientate the curve equation with respect to time to get
    dy/dt=(3x^2/2(17+x^3)^1/2)dx/dt

    i want to find dx/dt so i put 4 cm/s in for dy/dt and 4 for x (given by the point (4,9)) and then i solve for dx/dt, but i am not coming up with the right answer. please tell me what i am doing wrong.
     
  2. jcsd
  3. Feb 22, 2007 #2

    arildno

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    What did you get?
     
  4. Feb 22, 2007 #3
    dy/dt=d((17+x^3)^1/2)/d(17+x^3) . d(17+x^3)/d(x) . d(x)/dt
    4=1/2.(17+64)^1/2 . 48. dx/dt
    72/48=dx/dt
    or dx/dt=1.5 cm/s is what i expect you should have got.....
    I dont understand how to use y=9 ?
     
  5. Feb 22, 2007 #4

    arildno

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    y=9 is a necessarily given value, due to the function relation.
     
  6. Feb 22, 2007 #5
    yeah, but how does one use it? Or is it just given...
     
  7. Feb 22, 2007 #6

    arildno

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    It was a nice gesture of the exercise maker to provide the actual y-coordinate.
    besides, remember that the derivative of y wrt. to x can be written as:
    [tex]\frac{dy}{dx}=\frac{3x^{2}}{2y}[/tex]
     
  8. Feb 22, 2007 #7
    i think 9 was just given as to give the point with the x-coordinate .


    i got the right answer by the way.

    thank you for your help.
     
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