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Homework Help: Determining perpendicular tangent line

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine the coordinates of the points on the graph of [tex] f(x) = \sqrt {2x+1} [/tex] where the tangent line is perpendicular to the line 3x+y+4 = 0

    2. Relevant equations

    [tex] f(x) = \sqrt {2x+1} [/tex]
    [tex] 3x+y+4 = 0 [/tex]

    3. The attempt at a solution

    For this, I was going to find the derivative of f(x) then equate that to y. This would be the point where they cross. But what does it mean for it to be perpendicular? Do I do the negative reciprocal of the slope?
  2. jcsd
  3. Oct 4, 2009 #2
    you have to first find the derivative of f(x) but why would you equate to y??
    if the tangent line of a graph f(x) is perpendicular to a line at a point then the product of the slopes is -1 (or more than a point). but in this case its just one point.
    So, first find this point.
    Last edited: Oct 4, 2009
  4. Oct 4, 2009 #3
    I don't understand.

    [tex] y= -3x -4 [/tex]
    [tex] f'(x) = \frac {1}{\sqrt{2x+1}}[/tex]
  5. Oct 4, 2009 #4
    you differentiated f(x) correctly. now differentiate y.
    you are asked to find the coordinates of the point where the tangent of the curve is perpendicular to the line y=-3x-4
    to do this, you must first understand that at this point, the product of the gradients is -1.
    so say the gradient of the curve is m and that of the line is n, then:
    this will give you a linear equation where you would easily be able to find x and then you can substitute to find y.
    you found m correctly.
  6. Oct 4, 2009 #5
    So then I take dy/dx = -3?

    Then I multiply [tex]
    f'(x) = \frac {1}{\sqrt{2x+1}}
    [/tex] by -3?
  7. Oct 4, 2009 #6
    Yes! I got the question right :). At (4, 3) it will be perpendicular.

    Though, couldn't I just not equate f'(x) to y and then solve for x? I know I will get a cubic, but there will only be one answer that fits the domain.

    Or am I completely wrong and this is the only way to do it?
  8. Oct 5, 2009 #7
    if you were to equate f'(x) to y then that would be the wrong approach, there's no point in equating a deivative of a curve to the line (no point).
    (4,3) is correct :)
    the question does say: determine the coordinates of the points (so more than one).
    note when you substitute x=4 into the equation of the curve, you get y=3. what would you get if you substituted x=4 into the equation of the line?? you see, more than one point.
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