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Simple differentiation.

  1. Nov 7, 2009 #1
    The tangent to the curve y=ax^3 at the point (-1,b) is perpendicular to the line y = 2x+3. Find the values of a and b.

    Could someone show me how to answer that?
     
  2. jcsd
  3. Nov 7, 2009 #2

    HallsofIvy

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    First, if (-1, b) is on the curve [itex]y= ax^3[/itex], what is b?

    The two lines given by y= mx+ u and y= nx+ v are perependicular if and only if mn= -1. With y= 2x+ 3, m= 2 so what is n?

    The slope, n, of a tangent line is equal to the derivative of the function at that point. What is the derivative there. What is the derivative of [itex]y= ax^3[/itex] at x= -1?
     
  4. Nov 7, 2009 #3
    y = 2x+3
    y' = 2
    m = 2
    so, a perpendicular gradient to 2 is -1/2.
    y = ax^3
    y' = 3ax^2
    -1/2 = 3ax^2
    -1/2 = 3a(-1)^2
    a = -1/6

    y = -1/6(-1)^3
    b = 1/6.
     
  5. Nov 7, 2009 #4

    HallsofIvy

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    Exactly right! Congratulations.
     
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