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Math Help - Linear Approximation of 3cos^2(3) to estimate the value of |cos(3)

  1. #1
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    Linear Approximation of 3cos^2(3) to estimate the value of |cos(3)

    hey guys can you help me with this:

    Use a linear approximation of 3cos^2(3) to estimate the value of |cos(3)|

    where do I begin? would it look something like this:
    y-(3cos^2(3))= \frac{d(3cos^2(3))}{dy}(x-(3))
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  2. #2
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    Is there a particular linear approximation you had in mind?

    The "slope" piece of your line is zero (0). That's not very helpful.
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  3. #3
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    Frankly your work doesn't make a whole lot of sense to me. You want are trying to get a linear approximation at x= 3. The slope is, of course, the derivative of 3cos^2(x) which is -6cos(x)sin(x), evaluated at x= 3 which requires calculating cos(3) and sin(3) anyway.

    TKHunny is suggesting you use a linear approximation at x= \pi rather than at x= 3. There, as TKHunny says, the slope is 0 so your approximation is just the value at x= \pi which is 3.
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  4. #4
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    Well this question is the third part of the initial question, which is:
    Given the function y=xcos^2(x)
    a) find the differential of the function:
    which I found to be: cos^2(x)-2xcos(x)sin(x)'

    b) Give the linearization of the function at x=pi. Which I found to be: y=x

    and then there's that question
    c) Use a linear approximation of 3cos^2(3) to estimate the value of |cos(3)|
    which I don't really understand how to do it.
    I guess I have to find value of x that allow 3cos^2(3) to equal |cos(3)|?
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  5. #5
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    Very good.

    a) OK
    b) OK
    c) Well, what is the approximation? You have A(x) = x, the linearization of y(x) close to pi. x = 3 is pretty close to pi. Use this to create an approximation of y(3). Then use algebra to produce the approximation of |cos(3)|.
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  6. #6
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    you know I should understand what that means but I don't. Could you please detail it a little more
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  7. #7
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    Quote Originally Posted by RezMan View Post
    you know I should understand what that means but I don't. Could you please detail it a little more
    So you know the linear approximation is

    y \approx L(x)=x

    So just evaluate this at

    x=3 \implies y(3) \approx L(3)=3
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  8. #8
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    this is what my teacher says:
    "So in this case, SUPPOSE you had y = 2x - 5 as a linear approximation for y = f(x) near x = 3 (that is not the case, i'm just saying pretend that is your linear approximation).

    then we could say f(3)= 3 cos^2(3) is APPROXIMATELY EQUAL to y = 2(3) - 5 = 1

    Since 3 cos^2(3) is APPROXIMATELY EQUAL to 1, we conclude that cos^2(3) is APPROXIMATELY EQUAL to 1/3, so by taking square roots we find |cos(3)| is APPROXIMATELY EQUAL to sqrt (1/3)."


    So 3cos^2(3) = 3, because y=x. So maybe cos^2(3) =1? so 3(1)=3. So sqrt(1)=|cos(3)| ?
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  9. #9
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    she says yes. Case closed. Thanks guys!
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