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Math Help - Fine Tangent Line

  1. #1
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    Fine Tangent Line

    Hi! I am new to this forum I am looking forward to discuss math problems with everyone here ;]

    1)a. Find the tangent line to the curve f(x) = 2^sin(x) at the point x = 0

    my answer is y = 1+ln2[2^sin(x)]cos(x)[x]

    I don't know what I did wrong

    b. Use the line in part (a) to estimate the value of f(.5)
    my value = 1.424 I got it wrong also ><

    I used this formula: f(x) = f(a) + f'(a)(x-a)

    c. What is the error in the approximation, accurate to three decimal places?
    my answer is -.454 and it's also wrong ><

    I used this formula: E(x) = F(x) - L(x)



    can someone please explain to me what I did wrong???
    I can't figure it out

    Thanks in Advance ;D
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ryuu View Post
    1)a. Find the tangent line to the curve f(x) = 2^sin(x) at the point x = 0

    my answer is y = 1+ln2[2^sin(x)]cos(x)[x]

    I don't know what I did wrong
    f(x) = 2^{sin(x)}

    Thus
    f^{\prime}(x) = ln(2) \cdot cos(x)~2^{sin(x)}

    At x = 0:
    f^{\prime}(0) = ln(2) \cdot cos(0)~2^{sin(0)} = ln(2)

    So you are looking for a line with a slope of ln(2) that passes through the point (0, 1).
    y = mx + b

    1 = ln(2) \cdot 0 + b

    Thus
    y = ln(2)x + 1

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    f(x) = 2^{sin(x)}

    Thus
    f^{\prime}(x) = ln(2) \cdot cos(x)~2^{sin(x)}

    At x = 0:
    f^{\prime}(0) = ln(2) \cdot cos(0)~2^{sin(0)} = ln(2)

    So you are looking for a line with a slope of ln(2) that passes through the point (0, 1).
    y = mx + b

    1 = ln(2) \cdot 0 + b

    Thus
    y = ln(2)x + 1

    -Dan

    How did you get point (0,1) can't it be (0,0) ??
    other than that thank you for the explanation

    OH OKAY you plug in the 0 to the original equation and get 1

    okay okay THANK YOU =]

    now part B and C I still need help ><
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  4. #4
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    Quote Originally Posted by Ryuu View Post
    Hi! I am new to this forum I am looking forward to discuss math problems with everyone here ;]

    1)a. Find the tangent line to the curve f(x) = 2^sin(x) at the point x = 0

    my answer is y = 1+ln2[2^sin(x)]cos(x)[x]

    I don't know what I did wrong

    b. Use the line in part (a) to estimate the value of f(.5)
    my value = 1.424 I got it wrong also ><

    I used this formula: f(x) = f(a) + f'(a)(x-a)

    c. What is the error in the approximation, accurate to three decimal places?
    my answer is -.454 and it's also wrong ><

    I used this formula: E(x) = F(x) - L(x)



    can someone please explain to me what I did wrong???
    I can't figure it out

    Thanks in Advance ;D
    Hello,

    to a) your calculations are OK - you only mixed the calculation of the slope with the claculation of the equation of the tangent:

    the slope of the tangent is m =\ln(2) \cdot (2^{\sin(0)}) \cdot \cos(0) = \ln(2) \cdot 2^0 \cdot 1 = \ln(2)

    Thus the equation of the tangent is:

    t:\ y=\ln(2) \cdot x + 1

    to b) Plug in x = 0.5 into the equation of the tangent (that means you use the tangent line as an approximation of the function):

    f(0.5) \approx \ln(2) \cdot 0.5 + 1 \approx 1.34657...


    to c) I'm not certain what you should do here. The nearly exact value of

    f(0.5) = 1.394188409
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  5. #5
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    Quote Originally Posted by earboth View Post
    Hello,

    to a) your calculations are OK - you only mixed the calculation of the slope with the claculation of the equation of the tangent:

    the slope of the tangent is m =\ln(2) \cdot (2^{\sin(0)}) \cdot \cos(0) = \ln(2) \cdot 2^0 \cdot 1 = \ln(2)

    Thus the equation of the tangent is:

    t:\ y=\ln(2) \cdot x + 1

    to b) Plug in x = 0.5 into the equation of the tangent (that means you use the tangent line as an approximation of the function):

    f(0.5) \approx \ln(2) \cdot 0.5 + 1 \approx 1.34657...


    to c) I'm not certain what you should do here. The nearly exact value of

    f(0.5) = 1.394188409


    Okay! I got part B now, which mean i don't need to use the formula ><
    I still don't get part C ><

    how did you get that exact value? and doesn't mean you have to subtract to get the error in approximation?


    Thanks
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