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Equation of a tangent, with a slope of 'e'.

  1. Apr 21, 2008 #1
    Help my guys. I'm so hopelessly lost on this question...

    Problem:

    Determine the equation of the tangent to the graph of y=ln x that has slope e.

    The answer in the book is y=ex-2.

    Please help. :smile:
     
  2. jcsd
  3. Apr 21, 2008 #2
    Where is your work? Show some and I will help.
     
  4. Apr 21, 2008 #3
    Well I just wrote y=xe as my answer but when I checked my answer it was wrong. I'm guessing that I need to do something using the derivative of y=ln x, but that is as far as know. I'm pretty weak with all this work involving Euler's constant.
     
  5. Apr 21, 2008 #4

    exk

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    y=ex can't be the answer. The general equation of a line is y=mx+b. You are given that the slope is e. What else do you know about the graph of the function and the line tangent to it at some point.

    If you think that you need to use the derivative of the function then try using it.

    and the last hint:

    What does ln(e) equal?
     
  6. Apr 21, 2008 #5
    I don't understand.:confused: Do I take the answer of y=ln e and plug it into y=ln x? If so how would I solve for x? Can someone please help me with a solution? I have an exam in 4 hours and there will be at least 3 questions like this worth(all together) 30 marks out of a possible 70.:frown: I'm in panic mode.:bugeye:

    Please. I need this.:cry:
     
  7. Apr 21, 2008 #6

    exk

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    This question has almost nothing to do with "e", it has to do with knowing how to find the equation of the line tangent to the graph. Unless you understand how to get the solution it won't help you when the question is phrased differently on the exam.

    here is the last hint:
    if the line is tangent to the curve then it shares the x,y coordinates as well as the slope.

    you shouldn't need anymore than that.
     
  8. Apr 21, 2008 #7
    But I don't have any x,y coordinates. If I did, I wouldn't be stuck.
     
  9. Apr 21, 2008 #8

    Dick

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    Science Advisor
    Homework Helper

    You haven't figured out for what value of x that the slope of the tangent is e. That's why you don't have coordinates. Start by doing that. What's the slope of the tangent to the curve y=ln(x)? For what value of x is that equal to e?
     
  10. Apr 21, 2008 #9
    I don't know how to solve it:

    y = ln x

    -sub. y = e:

    e = ln x

    Now, here is where I'm stuck. I solve for x right? But how do I do that?
     
  11. Apr 21, 2008 #10

    exk

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    y doesn't equal e.

    You said initially that you need to use the derivative of your function, try using it.

    Do you remember what the derivative of a function represents?
     
  12. Apr 21, 2008 #11
    So;

    y = ln x

    y' = 1/x

    The derivative equation gives us the slope of any tangent, at any point along the given function. What do I do from here?

    Should I sub. y' = e?

    y' = 1/x

    e = 1/x

    xe = 1

    x = 1/e ????
     
  13. Apr 21, 2008 #12

    exk

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    why can't x = 1/e? (in fact it is)

    suppose x = 1/e is correct, how do you get the y coordinate? and finally how do you get the equation of the line?
     
  14. Apr 21, 2008 #13
    To get the y-coordinate, I must sub. x = 1/e in to y = ln x. Finally, to get the equation of the tangent I would use y = m(x-p)+q and sub. in the appropriate values. Right?

    Did I get the part with the derivative right?
     
  15. Apr 21, 2008 #14

    exk

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    yes and yes.
     
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