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Homework Help: Related rates help

  1. Apr 27, 2006 #1
    Code (Text):

    A particle is moving along the graph of y=x^1/3.  
    Suppose x is increasing at the rate of 3 cm/s.  
    At what rate is the angle of inclination, theta, changing when x=8?  
    [Hint: when x=8, theta approx. 0.24 rad]
     
    I'm stuck on problem. I know x=8, y=2 and the hyp=2.87. They want the derivative of theta so am I supposed to take the derivative of theta=0.24?
     
  2. jcsd
  3. Apr 27, 2006 #2
    No, you want to find some function that defines theta, and take the derivative of that.
     
  4. Apr 27, 2006 #3

    HallsofIvy

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    Well, the derivative of a constant (0.24) is 0 so that wouldn't do a whole lot of good would it? :wink: It's a lot more helpful to find the derivative of a function rather than a number! Draw a picture, showing the "angle of inclination" and write theta as a function of x. Then find the derivative of that function with respect to x. (Do you remember that the derivative of a function is the tangent of the angle the tangent line to the graph makes with the x-axis?)
     
  5. Apr 27, 2006 #4
    yo.gif
    Is this correct? Or am I completely off my rocker?:frown:
     
  6. Apr 27, 2006 #5

    Tom Mattson

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    You're off your rocker. :biggrin:

    Try following these steps:

    1.) Draw a graph of f(x)=x1/3 vs. x.

    2.) Consider an arbitrary point on the curve: (x,y)=(x,x1/3)

    3.) Draw a line segment from the origin to (x,y) and drop a vertical line segment down from (x,y) to the x-axis. You now have a right triangle.

    4.) The angle made by the line segment (the one connecting (0,0) to (x,y)) and the x-axis is the angle of inclination. Write down an expression for it in terms of x.

    Try that and let's see what you come up with.
     
  7. Apr 27, 2006 #6
    yup.gif
    Is this right?
     
  8. Apr 27, 2006 #7

    Tom Mattson

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    So far so good, but you're supposed to plug in x=8 and get a number for [itex]d\theta/dt[/itex].
     
  9. Apr 27, 2006 #8
    Sweet! Thanks to all for the help!
     
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