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Draw a tangent that is perpendicular to the line

  1. Sep 30, 2005 #1
    I have a curve [itex]-\sqrt {2x^3 }[/itex] on which I'm supposed to draw a tangent that is perpendicular to the line [itex]\[y = \frac{4}{3}x + \frac{1}{3}[/itex].

    I know that this tangent must have a "steepness" of -3/4 in order to make it perpendicular, but how do I now find the point on the graph?
     
  2. jcsd
  3. Sep 30, 2005 #2

    EnumaElish

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    Solve for the x from the curve formula that would make its slope equal to that of the tangent.
     
  4. Sep 30, 2005 #3

    Diane_

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    Given that the tangent needs to be perpendicular to the line y = 4/3x + 1/3, he's right that the slope must be -3/4. Negative reciprocal for the normal line.

    TSN79 - What you need is the point on the curve where the first derivative is -3/4. Just take the derivative, plug in that value, and solve for x.

    Note: I'm presuming you're taking or have taken calculus, as I don't see any other way to do this.
     
  5. Sep 30, 2005 #4

    EnumaElish

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    He could do it geometrically I suppose, but that may or may not be exact.
     
  6. Oct 1, 2005 #5
    Take the derivative of what? And plug in what value? -3/4? I tried with [itex]
    - \sqrt {2x^3 } [/itex] but didn't really get anywhere...
     
  7. Oct 1, 2005 #6

    EnumaElish

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    Take the derivative of [itex] - \sqrt {2x^3 } [/itex] then equate it to -3/4 and solve for x.
     
  8. Oct 1, 2005 #7
    Ah, now where getting somewhere, thanks ya all!
     
  9. Oct 1, 2005 #8

    HallsofIvy

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    Did you really need to be told that? You were told that the line had to be tangent to the curve given by [itex]y=-\sqrt {2x^3 }[/itex]. Didn't you connect "tangent" with "derivative"- after you had been told the find the derivative- the only tangent mentioned was to [itex]y= -\sqrt {2x^3 }[/itex].
    .
     
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