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Math Help - Linear Approximation problem

  1. #1
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    Linear Approximation problem

    Estimate using differentials.

    sin(0.00001)
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  2. #2
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    Quote Originally Posted by Fyou88 View Post
    Estimate using differentials.

    sin(0.00001)
    f(x + h) \approx f(x) + h f'(x).

    Use f(x) = sin(x), x = 0, h = 0.00001.
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    ok thank. if it was sin(0.9999) would x = 1 and then h=0.9999?
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    Quote Originally Posted by Fyou88 View Post
    ok thank. if it was sin(0.9999) would x = 1 and then h=0.9999?
    No. If x = 1 and you require x + h = 0.9999 then h = -0.0001.
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    so it will be 0 again? and x= the value u estimating the # that is closer to?
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    Quote Originally Posted by Fyou88 View Post
    so it will be 0 again? and x= the value u estimating the # that is closer to?
    I have no idea what you are referring to.

    I have clearly answered both of your questions. I suggest you knuckle down to doing the calculations.
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  7. #7
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    how do u know that x will be = 0? anyone help please??
    Last edited by Fyou88; May 23rd 2010 at 12:42 PM.
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    Quote Originally Posted by Fyou88 View Post
    how do u know that x will be = 0? anyone help please??
    Go back and review the examples in your class notes or textbook.
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  9. #9
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    Estimate

    e^sqrt0.0001

    I already know how to do e^0.0001.
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  10. #10
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    Quote Originally Posted by Fyou88 View Post
    e^sqrt0.0001

    I already know how to do e^0.0001.
    f(x + h) \approx f(x) + h f'(x).

    Use f(x) = e^{\sqrt{x}}, x = 0, h = 0.0001
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    Thumbs up

    ok.
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  12. #12
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    so f prime (x)=e^sqrt x and e^sqrt0=1.

    so m = 1 (0,1)
    y-y1=m(x-x1)
    y-1=1(x-0)
    y=x+1

    may someone help mee??
    Last edited by Fyou88; May 23rd 2010 at 11:28 PM.
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  13. #13
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    Quote Originally Posted by Fyou88 View Post
    so f prime (x)=e^sqrt x and e^sqrt0=1.

    so m = 1 (0,1)
    y-y1=m(x-x1)
    y-1=1(x-0)
    y=x+1

    may someone help mee??
    What you have posted here makes absolutely no sense - particularly in the context of this thread (which is about applying the linear approximation).

    And if f(x) = e^{\sqrt{x}} then your derivative is wrong. The derivative is calculated using the chain rule.


    Look, the fact is that currently you're quite out of your depth with the questions you are posting - the root cause of this appears to be a poor knowledge of the pre-requisite concepts and skills. As I have previously said, you are strongly advised to go back and learn those skills properly before attempting questions that depend on knowing them. I also strongly suggest that you get remedial help from your instructor.
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  14. #14
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    Quote Originally Posted by Fyou88 View Post
    so f prime (x)=e^sqrt x and e^sqrt0=1.
    NO, you said that f(x)= e^(sqrt x). Do you know how to find the derivative of that?


    so m = 1 (0,1)
    y-y1=m(x-x1)
    y-1=1(x-0)
    y=x+1

    may someone help mee??
    mr. fantastic has been trying to help you but you are not doing what he told you to do.
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  15. #15
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    e^sqrtx*1/2sqrt x?
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