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Math Help - Linear Approximation Problem?

  1. #1
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    Linear Approximation Problem?

    hi guys can you please help me with this.

    On page 431 of Physics: Calculus, 2d ed., by Eugene Hecht(Pacific Grove, CA: Brooks/Cole, 2000), in the course of derivingthe formula T=2π√L/g for the period of a dendulum oflength L, the author obtains the equation at=-gsinθfor the tangential acceleration of the bob of the pendulum. He then says, "for small angles, the value of θ in radians isvery nearly the value of sin θ; they differ by less than 2%out to about 20 degrees."

    (a) Verify the linear approximation at 0 for the sine function:

    sinx x

    (b) Use a graphing device to determine the values of x for whichsin x and x differ by less than 2%. Then verify Hecht'sstatement by converting from radians to degrees.


    I understand how to do a, where if you find the equation of the tangent line you should end up with y=x. But can someone please help me with b?
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    For part b just plot them on the same screen and estimate.
    Here is one plot of
    f(x)=x-\sin(x) \quad g(x)=.02

    their intersection is where the error is equal to 2%.

    sine.pdf

    The plot is in radians.
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  3. #3
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    wow you can actually use 2% as a data on the graph? I don't exactly understand the logic of that though. So would I have to use the amount between, say, .1 and .49? .1 radians = 5.7 degrees , .49 radians = 28.1 degrees. So 28.1-5.7 = 22 degrees? Or am I completely wrong. Thanks for your help by the way.
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by RezMan View Post
    wow you can actually use 2% as a data on the graph? I don't exactly understand the logic of that though. So would I have to use the amount between, say, .1 and .49? .1 radians = 5.7 degrees , .49 radians = 28.1 degrees. So 28.1-5.7 = 22 degrees? Or am I completely wrong. Thanks for your help by the way.
    Here is the logic behind the idea. This distance between

    x

    and

    \sin(x)

    is given by

    |x-\sin(x)|

    but we want

    |x-\sin(x)|< 2\% = .02

    So I just found the solution to this equation graphically by finding the intersection of

    f(x)=|x-\sin(x)| \text{ and } g(x)=.02
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  5. #5
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    wow that makes sense. I got confused by wondering what the 100% would be. So the answer would be .49 radians = 28.1 degrees?
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  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Yep that is the approximate answer.
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  7. #7
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    thank you so much! Your use of the graph kinda inspires me to represent everything in graph form!
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