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Curves and parallel lines

  1. Mar 8, 2005 #1
    I'm kinda stuck on this question... my text gives me a simple example, but it's far from enlightening. Just wondered if I could get some help!

    "At what point on the curve y = 1 + 2e^x - 3x is the tangent line parallel to the line 3x -y =5?"

    Would I just need to simply compare the slopes?

    I found the derivative of the curve: y'= 2e^x - 3

    So I suppose you just set 2e^x = 3... then you get x = ln(3)/2 ?

  2. jcsd
  3. Mar 8, 2005 #2
    parallel means same slope, so u found [tex]\frac{dy}{dx} = 2e^x - 3 [/tex]
    [tex]y=-5+3x, \frac{dy}{dx} = 3[/tex]
    [tex]3 = 2e^x - 3[/tex]
    [tex]3 = e^x[/tex]
    [tex]ln(3) = x[/tex]
  4. Mar 8, 2005 #3
    Oh, so I had to compute the derivatives for both slopes, then compare them? Interesting... thanks! I have to digest this info now.
  5. Mar 8, 2005 #4


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    zenity --

    Your original approach was CORRECT. You just made the careless error shown above. Method is to determine slope of the line (compute dy/dx of the line OR determine by inspection (like you did) from the line's {y = mx + b} equation that the slope is "3") and then to equate this value to the curve's tangent slope (found by differentiating the curve's equation like you did).

  6. Mar 8, 2005 #5
    so the point would be... ( ln(3), 7-3ln(3) ) ?

    I'm just a bit confused graphically.
    Last edited: Mar 8, 2005
  7. Mar 8, 2005 #6


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    Your answer is CORRECT. If helpful, sketch the graph to better understand the math. (Or use a graphing calculator.)

    Last edited: Mar 8, 2005
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