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Math Help - Tangent Curve Questions

  1. #1
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    Tangent Curve Questions

    Couple of curve tangent questions. I have no idea what the 1st is asking.

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Couple of curve tangent questions. I have no idea what the 1st is asking.

    First Question:

    Let curve 1 be: 2x^2 + y^2 = 3 \Rightarrow y = \sqrt {3 - 2x^2}
    Let curve 2 be: x = y^2 \Rightarrow y = \sqrt {x}


    Thus, the curves intersect where:
    \sqrt {3 - 2x^2} = \sqrt {x}

    \Rightarrow 3 - 2x^2 = x

    \Rightarrow 2x^2 + x - 3 = 0

    \Rightarrow (2x + 3)(x - 1) = 0

    \Rightarrow x = - \frac {3}{2} \mbox { or } x = 1

    x = - \frac {3}{2} is extraneous, since y has no solution for this x-value

    when x = 1:

    y = \sqrt {1} = 1

    So (1,1) is the ONLY point of intersection. Now let's find the slopes of the tangent lines of each curve at this point. If the slopes are perpendicular, that is, one is the negative inverse of the other, then we have that the curves are orthogonal.


    For curve 1:

    2x^2 + y^2 = 3

    \Rightarrow 4x + 2y \frac {dy}{dx} = 0

    \Rightarrow \frac {dy}{dx} = - \frac {2x}{y}

    at (1,1), \frac {dy}{dx} = -2


    For curve 2:
    x = y^2

    \Rightarrow 1 = 2y \frac {dy}{dx}

    \Rightarrow \frac {dy}{dx} = \frac {1}{2y}

    at (1,1), \frac {dy}{dx} = \frac {1}{2}

    Thus we see that the curves are orthogonal, since -2 is the negative reciprocal of \frac {1}{2}






    Question 14:

    y = (x - 1)^2 (4 - x)

    By the product rule:

    \frac {dy}{dx} = 2(x - 1)(4 - x) - (x - 1)^2

    \Rightarrow \frac {dy}{dx} = -3(x - 2)^2 + 3 I skipped a lot of steps, i don't think you'll have a problem getting here.

    This is a downward opening parabola, the derivative is maximum at it's vertex. That is, y' is max at (2,3)

    So we have that the tangent of greatest slope occurs at x = 2 and it's value is 3. I think you can take it from here and find the equation of the tangent line
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    Quote Originally Posted by Jhevon View Post
    First Question:

    x = - \frac {3}{2} is extraneous, since y has no solution for this x-value

    when x = 1:

    y = \sqrt {1} = 1

    So (1,1) is the ONLY point of intersection. Now let's find the slopes of the tangent lines of each curve at this point. If the slopes are perpendicular, that is, one is the negative inverse of the other, then we have that the curves are orthogonal.
    Shouldn't (1,-1) also be a point of intersection since 1=y^2 and y= +/- 1 or am I plugging it into the wrong equation here or something?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Shouldn't (1,-1) also be a point of intersection since 1=y^2 and y= +/- 1 or am I plugging it into the wrong equation here or something?
    you're absolutely right, my boy! (1,-1) is also a an intersecting point, my apologies. good looking out
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  5. #5
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    the two curves

    The two curves are perpendicular at the points of intersection.
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  6. #6
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    Quote Originally Posted by curvature View Post
    The two curves are perpendicular at the points of intersection.
    The diagram adds nothing to the answers and is in fact misleading as the
    scale on the x and y axes are different, which alters the apparent angle
    of intersection

    RonL
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  7. #7
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    the largest slope

    the largest slope occurs at the piont of inflection
    Attached Thumbnails Attached Thumbnails Tangent Curve Questions-05jun2007.jpg  
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  8. #8
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    visualize the problem

    Quote Originally Posted by CaptainBlack View Post
    The diagram adds nothing to the answers and is in fact misleading as the
    scale on the x and y axes are different, which alters the apparent angle
    of intersection

    RonL
    I just want to visualize the problem.
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  9. #9
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    wrong

    Quote Originally Posted by Jhevon View Post
    First Question:

    Let curve 1 be: 2x^2 + y^2 = 3 \Rightarrow y = \sqrt {3 - 2x^2}
    Let curve 2 be: x = y^2 \Rightarrow y = \sqrt {x}


    Thus, the curves intersect where:
    \sqrt {3 - 2x^2} = \sqrt {x}

    \Rightarrow 3 - 2x^2 = x

    \Rightarrow 2x^2 + x - 3 = 0

    \Rightarrow (2x + 3)(x - 1) = 0

    \Rightarrow x = - \frac {3}{2} \mbox { or } x = 1

    x = - \frac {3}{2} is extraneous, since y has no solution for this x-value

    when x = 1:

    y = \sqrt {1} = 1

    So (1,1) is the ONLY point of intersection. Now let's find the slopes of the tangent lines of each curve at this point. If the slopes are perpendicular, that is, one is the negative inverse of the other, then we have that the curves are orthogonal.


    For curve 1:

    2x^2 + y^2 = 3

    \Rightarrow 4x + 2y \frac {dy}{dx} = 0

    \Rightarrow \frac {dy}{dx} = - \frac {2x}{y}

    at (1,1), \frac {dy}{dx} = -2


    For curve 2:
    x = y^2

    \Rightarrow 1 = 2y \frac {dy}{dx}

    \Rightarrow \frac {dy}{dx} = \frac {1}{2y}

    at (1,1), \frac {dy}{dx} = \frac {1}{2}

    Thus we see that the curves are orthogonal, since -2 is the negative reciprocal of \frac {1}{2}






    Question 14:

    y = (x - 1)^2 (4 - x)

    By the product rule:

    \frac {dy}{dx} = 2(x - 1)(4 - x) - (x - 1)^2

    \Rightarrow \frac {dy}{dx} = -3(x - 2)^2 + 3 I skipped a lot of steps, i don't think you'll have a problem getting here.

    This is a downward opening parabola, the derivative is maximum at it's vertex. That is, y' is max at (2,3)

    So we have that the tangent of greatest slope occurs at x = 2 and it's value is 3. I think you can take it from here and find the equation of the tangent line
    No, y' is max at (2,2).
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