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Math Help - eqn of normal to the curve

  1. #1
    Member maybeline9216's Avatar
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    Question eqn of normal to the curve

    Find the eqn of the normal to the curve y=e^(3x+2) at the point where x=-1, leaving your answer in terms of e.

    For this qn, im not sure of the simplified version of the answer since you have to leave the answer in terms of e.
    Last edited by maybeline9216; October 7th 2008 at 05:18 AM. Reason: forgot to add brackets
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  2. #2
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    Quote Originally Posted by maybeline9216 View Post
    Find the eqn of the normal to the curve y=e^3x+2 at the point where x=-1, leaving your answer in terms of e.

    For this qn, im not sure of the simplified version of the answer since you have to leave the answer in terms of e.
    1. The point P(-1,e^{-3}+2) belongs to the graph of the function and the normal.

    2. Slope of the tangent m = y':

    y'=3 \cdot e^{3x} and therefore m = \dfrac{3}{e^{3}}

    3. The direction of the normal is r = -\dfrac{e^3}3

    4. Use point-slope-formula of a straight line:

    y-(e^{-3}+2)=-\dfrac{e^3}3 \cdot (x-(-1)) . Solve for y:

    5. Equation of the normal:

    y = -\dfrac{e^3}3 \cdot x -\dfrac{e^3}3+\dfrac1{e^3} + 2
    Attached Thumbnails Attached Thumbnails eqn of normal to the curve-norm_aufegraph.png  
    Last edited by earboth; October 7th 2008 at 05:20 AM. Reason: last line was wrong
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  3. #3
    Member maybeline9216's Avatar
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    umm sorry shld be y= e^(3x+2)
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  4. #4
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    Quote Originally Posted by maybeline9216 View Post
    umm sorry shld be y= e^(3x+2)
    Ähemm!

    The way to solve your question is the same, so I'll give you the results only:

    P(-1, e^{-1})

    y'=3 \cdot e^{3x+2}

    m = \dfrac3{e}

    r = -\dfrac e3

    Equation of the normal:

    y-e^{-1}=-\dfrac e3 \cdot (x+1)

    \boxed{y=-\dfrac e3 \cdot x  -\dfrac e3 +e^{-1}}
    Attached Thumbnails Attached Thumbnails eqn of normal to the curve-norm_zuegrph.png  
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  5. #5
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    Quote Originally Posted by maybeline9216 View Post
    Find the eqn of the normal to the curve y=e^3x+2 at the point where x=-1, leaving your answer in terms of e.

    For this qn, im not sure of the simplified version of the answer since you have to leave the answer in terms of e.
    Let me guess, "qn" is question.

    So what if you have to leave the answer in terms of e? That's okay.
    That's what the qn is asking for.

    I bet you know that the y' is the slope of the tangent line to anywhere on the y-curve.
    And you know too that -1/(y') is the slope of the normal line to anywhere on the y-curve.

    So, you find the y'.
    It should be
    y' = 3e^(3x +2)

    The slope of the normal line?
    Why, m = -1 / 3e^(3x +2), of course.

    Nww you find the point on the y-curve where the normal line is supposed to inject the curve.
    It is given as the point where x = -1.
    So, the y?
    y = e^(3(-1) +2) = e^(-1) = 1/e

    Thus, the point is (-1,1/e)

    To find the equation of the normal line, you can use the point-slope form of the line
    (y -y1) = m(x -x1)

    You aleady have the (x1,y1) ....the (-1,1/e)

    What about the m where x = -1?
    m = -1 / 3e^(3(-1) +2)
    m = -1 / 3e^(-1)
    m = -e/3

    Hence, the normal line is
    (y -1/e) = (-e/3)(x -(-1))
    y -1/e = (-e/3)x -e/3
    y = (-e/3)x -e/3 +1/e ------answer.
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  6. #6
    Member maybeline9216's Avatar
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    Can this be simplified any further???
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  7. #7
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    Quote Originally Posted by maybeline9216 View Post
    Can this be simplified any further???
    You can collect the constant summands:

    <br />
\boxed{y=-\dfrac e3 \cdot x -\dfrac e3 +e^{-1} = -\dfrac e3 \cdot x +\dfrac {3-e^2}{3e}}<br />
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