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Math Help - Parametric equations

  1. #1
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    Parametric equations

    A curve is given parametrically by
    x=a(5\cos\theta+\cos 5\theta)
    y=a(5\sin\theta-\sin 5\theta)
    Find the equation of the normal to the curve at the point with parameter \theta.
    Find those points on the curve at which the normal is also normal to the curve at another point.

    my attempt:
    Gradient
    \frac{dx}{d\theta}=-5a(\sin\theta+\sin 5\theta)
    \frac{dy}{d\theta}=5a(\cos\theta-\cos 5\theta)
    \frac{dy}{dx}=-\frac{\cos\theta-\cos 5\theta}{\sin\theta+\sin 5\theta}
    normal
    \frac{\sin\theta+\sin 5\theta}{cos\theta-\cos 5\theta}=-\frac{2\sin 3\theta\cos 2\theta}{2\sin 3\theta\sin 2\theta}=-\frac{\cos 2\theta}{\sin 2\theta}
    equation
    y-a(5\sin\theta-\sin 5\theta)=-\frac{\cos 2\theta}{\sin 2\theta}(x-a(5\cos\theta+\cos 5\theta))
    y\sin 2\theta-a\sin 2\theta(5\sin\theta-\sin 5\theta)=-x\cos 2\theta+a\cos 2\theta(5\cos\theta+\cos 5\theta)
    y\sin 2\theta+x\cos 2\theta=a[5\cos\theta\cos 2\theta+\cos 2\theta\cos 5\theta+5\sin 2\theta\sin\theta-\sin 2\theta\sin 5\theta]
    y\sin 2\theta+x\cos 2\theta= a[\frac{5}{2}(\cos 3\theta+\cos\theta)+\frac{1}{2}(\cos 7\theta+\cos 3\theta)-\frac{5}{2}(\cos 3\theta-\cos\theta)+\frac{1}{2}(\cos 7\theta-\cos 3\theta)]
    y\sin 2\theta+x\cos 2\theta=a(5\cos\theta+\cos 7\theta)
    The answer is: y\sin 2\theta+x\cos 2\theta=a(5\cos\theta-\cos 3\theta)
    So I got the RHS correct, but where on the LHS is wrong?
    And for finding the points where the normal there is normal at another point, I don't know where to start.
    Thanks for any help
    Last edited by arze; October 22nd 2009 at 05:00 AM. Reason: typo
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  2. #2
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    Hello arze
    Quote Originally Posted by arze View Post
    A curve is given parametrically by
    x=a(5\cos\theta+\cos 5\theta)
    y=a(5\sin\theta-\sin 5\theta)
    Find the equation of the normal to the curve at the point with parameter \theta.
    Find those points on the curve at which the normal is also normal to the curve at another point.

    my attempt:
    Gradient
    \frac{dx}{d\theta}=-5a(\sin\theta+\sin 5\theta)
    \frac{dy}{d\theta}=5a(\cos\theta-\cos 5\theta)
    \frac{dy}{dx}=-\frac{\cos\theta-\cos 5\theta}{\sin\theta+\sin 5\theta}
    normal
    \frac{\sin\theta+\sin 5\theta}{cos\theta-\cos 5\theta}=-\frac{2\sin 3\theta\cos 2\theta}{2\sin 3\theta\sin 2\theta}=-\frac{\cos 2\theta}{\sin 2\theta}
    equation
    y-a(5\sin\theta-\sin 5\theta)=-\frac{\cos 2\theta}{\sin 2\theta}(x-a(5\cos\theta+\cos 5\theta))
    y\sin 2\theta-a\sin 2\theta(5\sin\theta-\sin 5\theta)=-x\cos 2\theta+a\cos 2\theta(5\cos\theta+\cos 5\theta)
    y\sin 2\theta+x\cos 2\theta=a[5\cos\theta\cos 2\theta+\cos 2\theta\cos 5\theta+5\sin 2\theta\sin\theta-\sin 2\theta\sin 5\theta]
    y\sin 2\theta+x\cos 2\theta= a[\frac{5}{2}(\cos 3\theta+\cos\theta)+\frac{1}{2}(\cos 7\theta+\cos 3\theta)-\frac{5}{2}(\cos 3\theta-\cos\theta)+\frac{1}{2}(\cos 7\theta-\cos 3\theta)]
    y\sin 2\theta+x\cos 2\theta=a(5\cos\theta+\cos 7\theta)
    The answer is: y\sin 2\theta+x\cos 2\theta=a(5\cos\theta-\cos 3\theta)
    So I got the RHS correct, but where on the LHS is wrong?
    And for finding the points where the normal there is normal at another point, I don't know where to start.
    Thanks for any help
    Thanks for showing us all your detailed working. I think you are right, and the stated answer is wrong. Here's why:

    The normal must (obviously) pass through the point with parameter \theta. So when x = a(5\cos\theta +\cos5\theta) and y= a(5\sin\theta-\sin5\theta) the value of y\sin 2\theta+x\cos 2\theta is

    a(5\sin\theta-\sin5\theta)\sin2\theta + a(5\cos\theta +\cos5\theta)\cos2\theta

    =5a(\sin\theta\sin2\theta+\cos\theta\cos2\theta) +a(\cos5\theta\cos2\theta-\sin5\theta\sin2\theta)

    =5a\cos\theta+a\cos7\theta

    So your equation contains the original point, has the required gradient and is therefore the correct equation of the normal.

    As far as the second part is concerned, you may find the following idea useful.

    If this line is also normal at a point with parameter \phi, then the gradients at \theta and \phi are equal. So, simplifying your expression for \frac{dy}{dx}, \tan2\theta = \tan2\phi

    \Rightarrow 2\phi = n\pi +2\theta, n \in \mathbb{Z}

    \Rightarrow \phi = \frac{n\pi}{2} +\theta, n \in \mathbb{Z}

    Grandad
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  3. #3
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    Should it be -\cot 2\theta=-\cot 2\phi? Then how to go about finding the value of \theta? I can see that values of \theta in the first quadrant will have values for \phi in the third quadrant, but other than that I don't know how to continue.

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  4. #4
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    Hello arze
    Quote Originally Posted by arze View Post
    Should it be -\cot 2\theta=-\cot 2\phi? Then how to go about finding the value of \theta? I can see that values of \theta in the first quadrant will have values for \phi in the third quadrant, but other than that I don't know how to continue.

    Yes, -\cot 2\theta=-\cot 2\phi, but I was using the rather simpler gradient of the tangent, which gives the same thing in a different format: \tan 2\theta=\tan 2\phi.

    To solve the problem, I should try the following:

    • find the gradient of the chord joining the points with parameters \theta and \phi, simplifying the result using the trig identities for the sums and differences of sines and cosines;


    • use the relation \phi = \theta + \frac{n\pi}{2} for different values of n (I think it might just be for n even and n odd);


    • equate the result to the gradient of the normal, -\cot\theta;


    • and then solve the resulting equation for \theta.


    I don't have time to try this right now, but that's the only method I can think of.

    Grandad
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  5. #5
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    I've been trying the method you suggested, tried it many times over but still can't solve for \theta, although I think it will work, just haven't hit on the right thing.
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  6. #6
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    Hello arze
    Quote Originally Posted by arze View Post
    I've been trying the method you suggested, tried it many times over but still can't solve for \theta, although I think it will work, just haven't hit on the right thing.
    The attached image shows what the graph looks like for a = 1 (I put the values into an Excel spreadsheet). The dots represent the values of \theta every 10^o, or \frac{\pi}{18} radians.

    Among other things, this confirms that the answer given for the normal is incorrect. If you look, for example, where \theta = 0, the point on the graph is (6,0), the gradient is zero, and the normal is therefore x= 6. This is consistent with our answer of y\sin 2\theta+x\cos 2\theta=a(5\cos\theta+\cos 7\theta), but not with the given answer of y\sin 2\theta+x\cos 2\theta=a(5\cos\theta-\cos 3\theta), which gives x = 4 when \theta = 0.

    I thought I had a solution to the second part of the question, based on the method I outlined, but I have found an error which I cannot currently resolve. I think that the solutions are \theta = \frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}, \frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}, but cannot prove this as yet!

    Grandad

    P.S. I have just realised that we have made a mistake in the sign of the gradient of the normal:

    \cos\theta - \cos5\theta=2\sin3\theta\sin2\theta not -2\sin3\theta\sin2\theta

    So \frac{\sin\theta+\sin 5\theta}{\cos\theta-\cos 5\theta}=\frac{2\sin 3\theta\cos 2\theta}{2\sin 3\theta\sin 2\theta}=\frac{\cos 2\theta}{\sin 2\theta}

    Sorry, I don't have time now to follow this up - but perhaps you can correct the working that comes after this.
    Attached Thumbnails Attached Thumbnails Parametric equations-untitled.jpg  
    Last edited by Grandad; October 23rd 2009 at 01:44 PM. Reason: Add PS
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  7. #7
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    Quote Originally Posted by Grandad View Post

    P.S. I have just realised that we have made a mistake in the sign of the gradient of the normal:

    \cos\theta - \cos5\theta=2\sin3\theta\sin2\theta not -2\sin3\theta\sin2\theta

    So \frac{\sin\theta+\sin 5\theta}{\cos\theta-\cos 5\theta}=\frac{2\sin 3\theta\cos 2\theta}{2\sin 3\theta\sin 2\theta}=\frac{\cos 2\theta}{\sin 2\theta}

    Its actually 5\cos\theta+\cos 5\theta and 5\sin\theta-\sin 5\theta
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