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Math Help - confusing calculus 1 question

  1. #1
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    Red face confusing calculus 1 question

    it says if f(x) = sin (x), this means f '(x) = cos (x).
    Using this, find a nonzero solution to x^3 - sin (x/10000) = 0
    *Hint: use a tangent line, no calculator required*

    how does the tangent line connect to any of this?
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  2. #2
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    Sine function

    Hello liltiger22
    Quote Originally Posted by liltiger22 View Post
    it says if f(x) = sin (x), this means f '(x) = cos (x).
    Using this, find a nonzero solution to x^3 - sin (x/10000) = 0
    *Hint: use a tangent line, no calculator required*

    how does the tangent line connect to any of this?
    I'm not sure about the method, but there is a solution very close to x = 0.01.

    This is because, for small \theta, \sin \theta \approx \theta. So when x = 0.01

    \sin \frac{x}{10000} \approx \frac{0.01}{10000} = 10^{-6}= x^3

    And 10^{^-6} is extremely small, so the approximation is a very good one!

    I hope this helps

    Grandad
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  3. #3
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    First, a definition: A tangent line is a line which locally touches a curve at exactly one point.

    Let's do an easier example before we tackle the problem at hand:
    Find the equation of the line tangent to  f(x) = x^4 + 13x - 3 at  x=2 .

    We have:
     f(x) = x^4 + 13x - 3
     f'(x) = 4x^3 + 13

    Note that  f'(x) , the derivative of  f(x) , is the equation for the slope of the tangent line to  f(x) = x^4 + 13x - 3 .

    Thus,  f'(2) is the slope of the line tangent to f(x) at  x=2 .
    We observe through plug-n-chug that  f'(2) = 45 . Hence, the slope of the line tangent to  f(x) at  x=2 is 45.

    Good, so we know that the slope of the tangent line is 45. Since we want to write the equation of the tangent line, let's find a point on said tangent line.
    By definition (of tangent line), the tangent line shares at least one point with our original equation  f(x) = x^4 + 13x - 3 . [Note: We say "at least one point" and not "exactly one point" in the line above because we did not specify an "at" condition, i.e., we did not say, "... point with our original equation  f(x) = x^4 + 13x - 3 at x=2 (or x= 3, etc).] Since we are interested in the line tangent to  f(x) = x^4 + 13x - 3 at  x=2 , we have an x-coordinate value, 2, for one of the points on the tangent line. We plug this value into our original equation and observe:  f(2) = 39 . Thus,  (2, 39) is the point shared by  f(x) and the line tangent to  f(x) at  x=2 .

    We now have the slope of the tangent line and a point on said tangent line.

    The rest is easy: we use the point-slope equation  y - y_{1} = m(x - x_{1}) . We have  x_{1} = 2 ,  y_{1} = 39 , and  m = f'(2) = 45 . We plug these suckers in:
     y - 39 = 45(x - 2) \implies y = 45x - 51. This can equivalently be written:  y = 3(15x - 17) .

    Now, let's try to solve your problem.
    -------------
    it says if f(x) = sin (x), this means f '(x) = cos (x).
    Using this, find a nonzero solution to x^3 - sin (x/10000) = 0
    *Hint: use a tangent line, no calculator required*
    -------
    We have:
     f(x) = x^3 - \sin{\frac{x}{10000}} = 0

    A nonzero solution to this equation is asking us to find a value for  x \neq 0 such that  x^3 = \sin{\frac{x}{10000}} .

    Note that for small values of  \theta (i.e., values of  \theta close to 0),  \sin{\theta} \approx{\theta} .

    Now, since we want to find  x \neq 0 such that  x^3 = \sin{\frac{x}{10000}} , let's use the principal described in the above line. Namely, let  x=0.001 be an approximation to 0, so that  (0.001)^3 = x^3 = \sin{(\frac{x}{10000})} \approx \frac{0.001}{10000} = 0.0000001

    So we have  x^3 \approx 0.0000001 .
    Last edited by abender; January 27th 2009 at 11:16 AM.
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