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Math Help - Exam question involving e and lnx

  1. #1
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    Exam question involving e and lnx

    Hey i really need quick help with this exam question so I will be able to explain it in class later on today I'll need the explanation of how its done so i can use it for revision:


    The curve C has equation y=x^2 lnx, x>0. The point A, with x-coordinate e, lies on C. Find an equation of the tangent to C at A.
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  2. #2
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    Quote Originally Posted by Big_Joe View Post
    Hey i really need quick help with this exam question so I will be able to explain it in class later on today I'll need the explanation of how its done so i can use it for revision:


    The curve C has equation y=x^2 lnx, x>0. The point A, with x-coordinate e, lies on C. Find an equation of the tangent to C at A.
    Is that  y = x^2 ln|x| or  y = x^{2.ln|x|} ?
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  3. #3
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    The first equation. Sorry not sure how to use the script.
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  4. #4
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    Quote Originally Posted by Big_Joe View Post
    The first equation. Sorry not sure how to use the script.
     y = x^2 ln|x|

    Gradient of the tangent by differentiating this using product rule:

     \frac{dy}{dx} = 2x.ln|x| + x

    We have x = e

     \frac{dy}{dx} = 2e ln|e| + e

    ln|e| = 1 ( e = exp(1), and we know that ln(x) and exp(x) are inverse functions, so (ln(exp(x)) = x)

     \frac{dy}{dx} = 2e+e = 3e .

    Need the y-coord.

     = y = e^2 ln|e| = e^2

    Put it into the equation of a straight line:

     y_{tangent} = \frac{dy}{dx}x + C

    You have dy/dx, and you have the point (e, e^2), so you can find C.
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  5. #5
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    Thanks a lot, that's really helpful!
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  6. #6
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    Quote Originally Posted by Big_Joe View Post
    Hey i really need quick help with this exam question so I will be able to explain it in class later on today I'll need the explanation of how its done so i can use it for revision:


    The curve C has equation y=x^2 lnx, x>0. The point A, with x-coordinate e, lies on C. Find an equation of the tangent to C at A.
    The point is (e,e^2) , since \ln(e)=1

    \frac{dy}{dx}=2x \ln(x)+x

    So the slope when x=e is:

    m=2e+e=3e.

    So you have the coordinates of the point and the slope of the tangent at the point, you should now be able to write the equation of the tangent at the point.

    (careless on my part not noticing there was already an adequate answer)

    CB
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