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Math Help - Linear Approximations

  1. #1
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    Exclamation Linear Approximations

    Hello, Everyone!

    Any help would be greatly appreciated

    Use linear approximation, i.e. the tangent line, to approximate as follows:
    Let . The equation of the tangent line to at can be written in the form where is: ________
    and where is:______
    Using this, we find our approximation for is _________
    NOTE: For this part, give your answer to at least 9 significant figures or use fractions to give the exact answer.


    Im lost
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xcelxp View Post
    Hello, Everyone!

    Any help would be greatly appreciated

    Use linear approximation, i.e. the tangent line, to approximate as follows:
    Let . The equation of the tangent line to at can be written in the form where is: ________
    and where is:______
    Using this, we find our approximation for is _________
    NOTE: For this part, give your answer to at least 9 significant figures or use fractions to give the exact answer.


    Im lost
    for linear approximation, we use the formula:

    f(x) ~= f(a) + f ' (a)(x - a)

    where x is the value that you want to find f(x) for and a is a value close to x that you know the value of f(x) for.

    note, that in your question they said use the line y = mx + b, this is the same thing, here b = f(a) and m = f ' (a) and (x - a) is our x

    so let f(x) = sqrt(x)
    we want to find sqrt(81.2), that is we want to find f(81.2)

    do we know f(81.2)? no, but we know f(81) and 81 is very close to 81.2

    so here our x = 81.2 and a = 81

    so f(a) = f(81) = sqrt(81) = 9

    f ' (x) = (1/2)x^(-1/2)
    so f ' (a) = f ' (81) = (1/2)(81)^(-1/2) = (1/2)*(1/9) = 1/18

    now we can just plug in the values into our equation:

    using f(x) ~= f(a) + f ' (a)(x - a)

    => sqrt(81.2) = f(81.2) ~= f(81) + f ' (81)(81.2 - 81)
    => sqrt(81.2) = f(81.2) ~= 9 + (1/18)(0.2)
    => sqrt(81.2) = f(81.2) ~= 9 + 0.01111111111111111
    => sqrt(81.2) = 9.01111111

    or for exact answer:

    sqrt(81.2) ~= 9 + (1/18)(2/10) = 9 + 1/90 = 811/90
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  3. #3
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    I am working out the problem, just one more question. Shouldnt b =9? This doesnt seem to be the right answer. Everything else worked out fine. Thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xcelxp View Post
    I am working out the problem, just one more question. Shouldnt b =9? This doesnt seem to be the right answer. Everything else worked out fine. Thanks!
    let's find the equation of the tangent line at x = 81

    f(x) = sqrt(x)
    => f ' (x) = (1/2)x^(-1/2)
    => the slope at 81 = f ' (81) = (1/2)(81)^(-1/2) = (1/2)(1/9) = 1/18
    this is our m

    now, when x = 81, y = sqrt(81) = 9

    using the point slope form

    y - y1 = m(x - x1)
    => y - 9 = (1/18)(x - 81)
    => y = (1/18)(x - 81) + 9
    => y = (1/18)x - 81/18 + 9
    => y = (1/18)x + 9/2

    so 9/2 is our b
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