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Thread: Tangent to the curve

  1. #1
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    Tangent to the curve

    for what values of a and b will the parabola y =x^2 + ax + b be tangent to the curve y = x^3 at point (1,1)?
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  2. #2
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    Quote Originally Posted by mathamatics112 View Post
    for what values of a and b will the parabola y =x^2 + ax + b be tangent to the curve y = x^3 at point (1,1)?

    So we are going to need to solve some equations....

    I am going to call $\displaystyle f(x)=y=x^3 $

    Starting with $\displaystyle y=x^2+ax+b$ we get

    $\displaystyle 1=1^2+a(1)+b \implies a=-b$

    $\displaystyle f'(x)=3x^2 \implies f'(1)=3$

    Now $\displaystyle y'=2x+a$ using the slope above and the x-coordinate we can solve for a

    $\displaystyle 3=2(1)+a \implies a =1$ so $\displaystyle b=-1$

    $\displaystyle y=x^2+x-1$
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  3. #3
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    First you will want to find the derivative of: $\displaystyle f(x) = x^3$. Using the power rule, $\displaystyle f'(x) = 3x^2$ Now plug in the 1 to find the slope of the curve at x = 1. $\displaystyle 3(1)^2 = 3$. Now you will want to find the derivative of the equation $\displaystyle g(x) = x^2 + ax + b$ and set that equal to the derivative of $\displaystyle f(x)$. So $\displaystyle g'(x) = 2x + a$ and $\displaystyle f'(x) = 3x^2.$
    $\displaystyle 3x^2 = 2x + a$
    Substituting in the 1:
    $\displaystyle
    \\3 = 2 + a
    $
    $\displaystyle
    \\a = 1
    $

    Now that you have the a-value, you will want to solve for b. Set $\displaystyle f(1) = g(1)$ to get the b.

    $\displaystyle
    \\ x^3 = x^2 + x + b
    $
    $\displaystyle
    \\ 1 = 1 + 1 + b
    $
    $\displaystyle
    \\ b = -2
    $

    Finally, the equation of the parabola tangent to the cubic is:
    $\displaystyle
    g(x) = x^2 + x - 1
    $
    Last edited by JoshHJ; Feb 23rd 2009 at 03:57 PM.
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