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Math Help - A line perpindicular to an equation

  1. #1
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    A line perpindicular to an equation

    Hi everone,

    I'm currently reviewing for a Calculas exam I have soon, and although there are a few review questions I can't seem to get, I'll stick with this one. (By the way, if it is super simple, excuse me, I really CANT do calculas.)

    Question: Determine the equation of the line perpindicular to f(x)=1/2e^x+1 at its y-intercept.

    What I did was found the dervitive, which I got to be the same thing (f'(x)=1/2e^x+1) although it could be wrong. Only problem is that I have no clue what to do after that. The answer is suppose to be y=e/2 x+e/2

    Any help is appreciated!
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  2. #2
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    Quote Originally Posted by memoIhad View Post
    Hi everone,

    I'm currently reviewing for a Calculas exam I have soon, and although there are a few review questions I can't seem to get, I'll stick with this one. (By the way, if it is super simple, excuse me, I really CANT do calculas.)

    Question: Determine the equation of the line perpindicular to f(x)=1/2e^x+1 at its y-intercept.

    What I did was found the dervitive, which I got to be the same thing (f'(x)=1/2e^x+1) although it could be wrong. Only problem is that I have no clue what to do after that. The answer is suppose to be y=e/2 x+e/2

    Any help is appreciated!
    Well, since it requires a line through the y-intercept, it would be a good idea to find that! The y-intercept of a graph is where it crosses the y-axis: where x= 0. Now assuming you meant f(x)= \frac{1}{2}e^{x+1} which is NOT what you wrote (what you wrote was f(x)= \frac{1}{2}e^x+ 1) but does have the property that its derivative is still \frac{1}{2}e^{x+1}, then since f(0)= \frac{1}{2}e, the y-intercept is (0, \frac{1}{2}e). The slope of the tangent line to the graph is the derivative evaluated at x= 0 which is again \frac{1}{2}e. Do you know how to find the slope and equation of the perpendicular line from that?
    Last edited by HallsofIvy; January 17th 2009 at 12:56 PM.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Well, since it requires a line through the y-intercept, it would be a good idea to find that! The y-intercept of a graph is where it crosses the y-axis: where x= 0. Now assuming you meant f(x)= \frac{1}{2}e^{x+1} which is NOT what you wrote (what you wrote was f(x)= \frac{1}{2}e^x+ 1) but does have the property that its derivative is still \frac{1}{2}e^{x+1}, then since [itex]f(0)= \frac{1}{2}e[/itex], the y-intercept is (0, \frac{1}{2}e). The slope of the tangent line to the graph is the derivative evaluated at x= 0 which is again [tex]\frac{1}{2}e[/itex]. Do you know how to find the slope and equation of the perpendicular line from that?
    Thanks for the quick reply. Yes, I did mean the equation you wrote . .....EDIT: nvm, I understand how you found the y-intrecept....As for the equation of the line perpindicular to that, I would assume that it would a slope that is the negative reciprocal of the original equation, so -2? So do I just plug that into the equation of the line? Which would be y-y1=m(x-x1)?
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  4. #4
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    Hi

    As written by HallsofIvy the slope of the tangent line to the graph is the derivative evaluated at x= 0 which is \frac{1}{2}e.

    This means that a direction vector of the tangent has coordinates (1,e/2).

    You have to find the coordinates of a vector perpendicular to this one.
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    Quote Originally Posted by memoIhad View Post
    Thanks for the quick reply. Yes, I did mean the equation you wrote . .....EDIT: nvm, I understand how you found the y-intrecept....As for the equation of the line perpindicular to that, I would assume that it would a slope that is the negative reciprocal of the original equation, so -2? So do I just plug that into the equation of the line? Which would be y-y1=m(x-x1)?
    The slope of the tangent line is NOT 1/2. It is, as I said, \frac{1}{2}e or \frac{e}{2}. So the slope of the perpendicular line is -\frac{2}{e}.

    Now, what is the equation of a line with slope -\frac{2}{e} passing through (0, \frac{e}{2})?
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    So, would the answer just be y=-2/e+e/2 ? that's the answer I got by plugging in the points (m=-2/e, y=e/2) The answer though, is different...am I missing something?

    Thanks a lot for being patient with me
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    Quote Originally Posted by memoIhad View Post
    So, would the answer just be y=-2/e+e/2 ? that's the answer I got by plugging in the points (m=-2/e, y=e/2) The answer though, is different...am I missing something?

    Thanks a lot for being patient with me
    Your answer is correct (apart from the observation made in the next post!)

    The answer is suppose to be y=e/2 x+e/2
    This is the equation of the tangent at the y-intercept.
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  8. #8
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    Quote Originally Posted by memoIhad View Post
    So, would the answer just be y=-2/e+e/2 ? that's the answer I got by plugging in the points (m=-2/e, y=e/2) The answer though, is different...am I missing something?
    Well, yes, you are missing something! There is no "x"!
    The equation of the line with slope -2/e, passing through (0, e/2) is
    y= (-2/e)x+ e/2.

    Thanks a lot for being patient with me
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