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Math Help - Find the length of tangent

  1. #1
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    Find the length of tangent

    Find the length of the portion of the tangent to the curve x = acos^3 \theta , y=a sin^3 \theta intercepted between coordinate axes.
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  2. #2
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    Length of tangent

    Hello zorro
    Quote Originally Posted by zorro View Post
    Find the length of the portion of the tangent to the curve x = acos^3 \theta , y=a sin^3 \theta intercepted between coordinate axes.
    I'll assume the question means 'Find, in terms of \theta, the length ...'. In which case, it's fairly straightforward. Find, in terms of \theta:

    • \frac{dx}{d\theta}
    • \frac{dy}{d\theta}
    • Hence \frac{dy}{dx}
    • The equation of the tangent at the point with parameter \theta

    Then find the values of x and y where this tangent crosses the axes, by putting y = 0 and x = 0 in turn, in this equation.

    Then use Pythagoras Theorem to find the required length.

    Grandad
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  3. #3
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    is this right

    Quote Originally Posted by Grandad View Post
    Hello zorro

    I'll assume the question means 'Find, in terms of \theta, the length ...'. In which case, it's fairly straightforward. Find, in terms of \theta:

    • \frac{dx}{d\theta}
    • \frac{dy}{d\theta}
    • Hence \frac{dy}{dx}
    • The equation of the tangent at the point with parameter \theta
    Then find the values of x and y where this tangent crosses the axes, by putting y = 0 and x = 0 in turn, in this equation.

    Then use Pythagoras Theorem to find the required length.

    Grandad
    --------------------------- ---------------------------

    i have got half of it can u tell me what to do next ..........

    <br />
\frac{dx}{dy} = \frac{sin^2 \theta cos \theta}{- cos^2 \theta sin \theta}<br />

    <br />
= -tan \theta<br />

    f'(0) = - tan (0) = 0

    there the slope of the tangent is 0

    Now the eq of the tangent at pt (x_0,y_0) is (0,0)

    therefore eq is  (y - y_0) = m (x-x_0)

    ie (y-0)=0(x-0)
    ie y=0
    Last edited by zorro; January 4th 2009 at 08:33 PM.
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  4. #4
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    Length of tangent

    Hello zorro
    Quote Originally Posted by zorro View Post
    --------------------------- ---------------------------

    i have got half of it can u tell me what to do next ..........

    <br />
\frac{dx}{dy} = \frac{sin^2 \theta cos \theta}{- cos^2 \theta sin \theta}<br />

    <br />
= -tan \theta<br />

    f'(0) = - tan (0) = 0

    there the slope of the tangent is 0

    Now the eq of the tangent at pt (x_0,y_0) is (0,0)

    therefore eq is  (y - y_0) = m (x-x_0)

    ie (y-0)=0(x-0)
    ie y=0
    Note that you have written \frac{dx}{dy}, instead of \frac{dy}{dx}. Also, perhaps you mis-read my post: I said to find the equation of the tangent at the point whose parameter is \theta, not 0.

    So, using \frac{dy}{dx} = -\tan \theta, the equation of the tangent whose paramter is \theta is:

    y- a\sin^3\theta = -\tan\theta(x-a\cos^3\theta)

    Meets the x-axis where y = 0, and

    - a\sin^3\theta = -\tan\theta(x-a\cos^3\theta)

    \Rightarrow a \sin^2\theta\cos\theta=x-a\cos^3\theta

    \Rightarrow x=a\cos\theta(\sin^2\theta+\cos^2\theta)= ?

    Can you take it from there?

    Grandad

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  5. #5
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    Problem encountered

    Quote Originally Posted by Grandad View Post
    Hello zorro

    Note that you have written \frac{dx}{dy}, instead of \frac{dy}{dx}. Also, perhaps you mis-read my post: I said to find the equation of the tangent at the point whose parameter is \theta, not 0.

    So, using \frac{dy}{dx} = -\tan \theta, the equation of the tangent whose paramter is \theta is:

    y- a\sin^3\theta = -\tan\theta(x-a\cos^3\theta)

    Meets the x-axis where y = 0, and

    - a\sin^3\theta = -\tan\theta(x-a\cos^3\theta)

    \Rightarrow a \sin^2\theta\cos\theta=x-a\cos^3\theta

    \Rightarrow x=a\cos\theta(\sin^2\theta+\cos^2\theta)= ?

    Can you take it from there?

    Grandad

    After this i get this

    \Rightarrow x=a\cos\theta

    But dont we have to find the length of the tangent .........Please advice
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  6. #6
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    Hello zorro

    We have found that the tangent meets the x-axis where x = a\cos\theta.

    In a similar way, it meets the y-axis where x = 0, and therefore:
    y-a\sin^3\theta =-\tan\theta(0-a\cos^3\theta)
    =a\sin\theta\cos^2\theta
    \Rightarrow y=a\sin\theta(\sin^2\theta +\cos^2\theta)
    =a\sin\theta
    So we want the distance between the points (0,a\sin\theta) and (a\cos\theta, 0); which is:
    \sqrt{a^2\cos^2\theta +a^2\sin^2\theta}
    =a
    Grandad
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  7. #7
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    Thanks mite

    Quote Originally Posted by Grandad View Post
    Hello zorro

    We have found that the tangent meets the x-axis where x = a\cos\theta.

    In a similar way, it meets the y-axis where x = 0, and therefore:
    y-a\sin^3\theta =-\tan\theta(0-a\cos^3\theta)
    =a\sin\theta\cos^2\theta
    \Rightarrow y=a\sin\theta(\sin^2\theta +\cos^2\theta)
    =a\sin\theta
    So we want the distance between the points (0,a\sin\theta) and (a\cos\theta, 0); which is:
    \sqrt{a^2\cos^2\theta +a^2\sin^2\theta}
    =a
    Grandad

    cheers
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