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Math Help - Linear Approximation question

  1. #1
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    Linear Approximation question

    I'm currently having a hard time with this question:

    Use differentials (or equivalently, a linear approximation) to approximate as follows: Let and find the equation of the tangent line to at a "nice" point near . Then use this to approximate .

    I'm familiar with the linear approximation formula, but this is confusing me. I tried getting (with 29.1 as a "nice" point): (cos29.1)(x-29)(sin29.1) but that is incorrect. I think it has something to do with using this number and plugging it into the equation for a tangent line? I'm really not sure. Thanks in advance for any help!
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  2. #2
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    Quote Originally Posted by fleabass123 View Post
    I'm currently having a hard time with this question:

    Use differentials (or equivalently, a linear approximation) to approximate as follows: Let and find the equation of the tangent line to at a "nice" point near . Then use this to approximate .

    I'm familiar with the linear approximation formula, but this is confusing me. I tried getting (with 29.1 as a "nice" point): (cos29.1)(x-29)(sin29.1) but that is incorrect. I think it has something to do with using this number and plugging it into the equation for a tangent line? I'm really not sure. Thanks in advance for any help!
    need to be in radians.

    30^\circ is \frac{\pi}{6} radians

    29^\circ is \frac{29\pi}{180} radians

    at x = \frac{\pi}{6} , the tangent line to the curve y = \sin{x} has a slope y' = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}

    the tangent line equation is ...

    y - \frac{1}{2} = \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{6}\right)<br />

    using x = \frac{29\pi}{180} for the approximation ...

    \sin(29^\circ) \approx y = \frac{\sqrt{3}}{2}\left(\frac{29\pi}{180} - \frac{\pi}{6}\right) + \frac{1}{2}
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  3. #3
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    Ah, I had the right idea (kind of). Thanks for showing your process. I understand how to do it now.
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