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Math Help - Tangent to the curve from a parametric equation

  1. #1
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    Tangent to the curve from a parametric equation

    Hello,

    I've got this question I've been labouring over.

    The parametric equations of a curve are : x = 1 + 2sin^2Ɵ, y = 4tanƟ.
    From a previous question I found dy/dx to be 1 / (sinƟcos^3Ɵ).

    Now it says

    "Find the equation of the tangent to the curve at the point where Ɵ = π/4, giving your answer in the form y = mx + c.
    So what I did was first try get a value for x and y by plugging in π/4 in the parametric equations.

    x = 1 + (sin/4)^2
    x = 1.00

    y = 4tan(π/4)
    y = 0.05

    Then I put in π/4 into the dy/dx equation.

    1 / (sinπ/4)((cosπ/4)^3)
    = 72.95

    Using the point-slope rule:

    y - y1 = m(x - x1)
    y - 0.05 = 72.95(x - 1)
    y = 72.95x - 72.95 + 0.05
    y = 72.95x + 73

    According to the mark scheme, that is WAY off, as their answer is y = 4x - 4. Where did I go wrong? Sorry if this is in the wrong category.
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  2. #2
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    Re: Tangent to the curve from a parametric equation

    You made several computational mistakes. Were you using a calculator perhaps? This problem should be solved without one.

    When \theta = \pi/4:
    x(\theta) = x(\pi/4) = 1+2\sin^2(\pi/4) = 1 + 2 \left(\frac{\sqrt{2}}{2}\right)^2 = 1 + 2\left(\frac{2}{4}\right) = 1 + 1 = 2.

    y(\theta) = y(\pi/4) = 4\tan(\pi/4) = 4(1) = 4.


    \frac{dy}{dx}(\theta) = \frac{dy}{dx}(\pi/4) = \frac{1}{(\sin(\pi/4)\cos^3(\pi/4)} = \frac{1}{(\sqrt{2}/2)(\sqrt{2}/2)^3}.

    = \frac{1}{(\sqrt{2}/2)^4} = \frac{1}{(\sqrt{16}/16)} = \frac{1}{(4/16)} = \frac{1}{(1/4)} = 4.


    So when \theta = \pi/4, it's at the point (2, 4) and has tangent line of slope 4.
    Last edited by johnsomeone; September 26th 2012 at 07:06 AM.
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  3. #3
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    Re: Tangent to the curve from a parametric equation

    Wow, you lost me.

    How did you get from 2\sin^2(\pi/4) to 2 \left(\frac{\sqrt{2}}{2}\right)^2 ?
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  4. #4
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    Re: Tangent to the curve from a parametric equation

    Another approach to find the slope of the tangent line:

    \frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta  }{dx}=\left(4\sec^2(\theta) \right)\left(\frac{1}{2\sin(2\theta)} \right)=\frac{2\sec^2(\theta)}{\sin(2\theta)}

    Hence:

    \frac{dy}{dx}\left|_{\theta=\frac{\pi}{4}}=\frac{4  }{1}=4

    We could also eliminate the parameter \theta, to find the equivalent Cartesian equation:

    x=1+2\sin^2\left(\tan^{-1}\left(\frac{y}{4} \right) \right)=1+\frac{2y^2}{y^2+16}

    xy^2+16x=3y^2+16

    Implicitly differentiate with respect to x:

    x\cdot2y\cdot\frac{dy}{dx}+y^2+16=6y\frac{dy}{dx}

    \frac{dy}{dx}=\frac{y^2+16}{2y(3-x)}

    When \theta=\frac{\pi}{4} we have (x,y)=(2,4) hence, at this point, we have:

    \frac{dy}{dx}=\frac{4^2+16}{2\cdot4(3-2)}=\frac{32}{8}=4
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  5. #5
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    Re: Tangent to the curve from a parametric equation

    Quote Originally Posted by yorkey View Post
    Wow, you lost me.

    How did you get from 2\sin^2(\pi/4) to 2 \left(\frac{\sqrt{2}}{2}\right)^2 ?
    It seems you don't know that \sin(\pi/4) = \frac{\sqrt{2}}{2}.

    If that's so, then you don't understand some pretty basic trigonometry. If you're taking a calculus class, then that's going to cause you serious problems. I'd suggest you take measures - self-study, a tutor, whatever - to get caught up with trigonometry.
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  6. #6
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    Re: Tangent to the curve from a parametric equation

    The trigonometric rules in my textbook I know and understand. They are:

    y = sinx
    dy/dx = cosx

    y = cosx
    dy/dx = -sinx

    y = tanx
    dy/dx = sec^2x

    But it seems that they've thrown me something they haven't taught me. I'm just asking what is that rule you're talking about. Is there a more general law for it?
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  7. #7
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    Re: Tangent to the curve from a parametric equation

    Quote Originally Posted by yorkey View Post
    The trigonometric rules in my textbook I know and understand. They are:

    y = sinx
    dy/dx = cosx

    y = cosx
    dy/dx = -sinx

    y = tanx
    dy/dx = sec^2x

    But it seems that they've thrown me something they haven't taught me. I'm just asking what is that rule you're talking about. Is there a more general law for it?
    it's called the unit circle ... you're expected to have been exposed to it prior to enrolling in a calculus course.

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  8. #8
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    Re: Tangent to the curve from a parametric equation

    Right, I've got my work cut out for myself. Thanks for the help, everyone. :-)
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