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    Hey, this question is easy but I've confused myself a bit:

    Question: find the equation of the normal to the curve y = e^x (sin2x) at the point (\pi,0)

    My solution:

    \frac{dy}{dx} = e^x(2cos2x + sin2x)

    When x = \pi, \frac{dy}{dx} = 2e^\pi

    \therefore the gradient of the normal is  -\frac{1}{2e^\pi}

    There are 2 methods now, which is correct?

    1) Using the equation of a line formula:
    y = -\frac{1}{2e^\pi}(x-\pi)

    y \div -\frac{1}{2e^\pi} = (x-\pi)

    -2e^\pi y = x -\pi

    2e^\pi y + x = \pi

    OR THIS ONE:

    2) y = -\frac{1}{2e^\pi}(x-\pi)

    y = -\frac{1}{2e^\pi}x + \frac{\pi}{2e^\pi}

    2y = -e^\pi x + \pi e^\pi

    \frac{2y}{e^\pi} = -x + \pi

    \therefore \frac{2y}{e^\pi} + x = \pi

    Which is right?! Thanks
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    Im sure some smart people on here could help?
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    But they both give different results?
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    (Original post by Mr.Pii)
    Hey, this question is easy but I've confused myself a bit:

    Question: find the equation of the normal to the curve y = e^x (sin2x) at the point (\pi,0)

    My solution:

    \frac{dy}{dx} = e^x(2cos2x + sin2x)

    When x = \pi, \frac{dy}{dx} = 2e^\pi

    \therefore the gradient of the normal is  -\frac{1}{2e^\pi}

    There are 2 methods now, which is correct?

    1) Using the equation of a line formula:
    y = -\frac{1}{2e^\pi}(x-\pi)

    y \div -\frac{1}{2e^\pi} = (x-\pi)

    -2e^\pi y = x -\pi

    2e^\pi y + x = \pi

    OR THIS ONE:

    2) y = -\frac{1}{2e^\pi}(x-\pi)

    y = -\frac{1}{2e^\pi}x + \frac{\pi}{2e^\pi}

    2y = -e^\pi x + \pi e^\pi

    \frac{2y}{e^\pi} = -x + \pi

    \therefore \frac{2y}{e^\pi} + x = \pi

    Which is right?! Thanks
    The first is definately correct. With the second, how have you got from the first line to the second line?
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    (Original post by Mr.Pii)
    Hey, this question is easy but I've confused myself a bit:

    Question: find the equation of the normal to the curve y = e^x (sin2x) at the point (\pi,0)

    My solution:

    \frac{dy}{dx} = e^x(2cos2x + sin2x)

    When x = \pi, \frac{dy}{dx} = 2e^\pi

    \therefore the gradient of the normal is  -\frac{1}{2e^\pi}

    There are 2 methods now, which is correct?

    1) Using the equation of a line formula:
    y = -\frac{1}{2e^\pi}(x-\pi)

    y \div -\frac{1}{2e^\pi} = (x-\pi)

    -2e^\pi y = x -\pi

    2e^\pi y + x = \pi

    OR THIS ONE:

    2) y = -\frac{1}{2e^\pi}(x-\pi)

    y = -\frac{1}{2e^\pi}x + \frac{\pi}{2e^\pi}

    2y = -e^\pi x + \pi e^\pi

    \frac{2y}{e^\pi} = -x + \pi

    \therefore \frac{2y}{e^\pi} + x = \pi

    Which is right?! Thanks
    You made a mistake in line 3 of (2)

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    (Original post by jdfan)
    The first is definitely correct. With the second, how have you got from the first line to the second line?
    i multiplied the -1/(2e^pi) by (x - pi) in the bracket
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    (Original post by steve2005)
    You made a mistake in line 3 of (2)

    nevermind. i have done it!

    you was a great help. i would rep but it isnt worth anything
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    (Original post by steve2005)
    You made a mistake in line 3 of (2)

    y = -\frac{1}{2}e^\pi(x-\pi)

    y = -\frac{1}{2}e^\pi x + \frac{1}{2}e^\pi \pi

    2y = -e^\pi x + e^\pi \pi

    \frac{2y}{e^\pi} = -x + \pi

    \therefore \frac{2y}{e^\pi} + x = \pi

    That looks right to me

    Can you explain why the e^\pi should be e^-pi? It's just guna annoy me otherwise. Thanks
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    (Original post by Adam92)
    y = -\frac{1}{2}e^\pi(x-\pi)

    y = -\frac{1}{2}e^\pi x + \frac{1}{2}e^\pi \pi

    2y = -e^\pi x + e^\pi \pi

    \frac{2y}{e^\pi} = -x + \pi

    \therefore \frac{2y}{e^\pi} + x = \pi

    That looks right to me

    Can you explain why the e^\pi should be e^-pi? It's just guna annoy me otherwise. Thanks
    Your first line is not correct. Look at the OP
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    (Original post by steve2005)
    Your first line is not correct. Look at the OP
    Oh ****. I never notice the obvious stuff.
 
 
 
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