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Thread: 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 $\displaystyle x = acos^3 \theta $, $\displaystyle y=a sin^3 \theta $ intercepted between coordinate axes.
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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 $\displaystyle x = acos^3 \theta $, $\displaystyle y=a sin^3 \theta $ intercepted between coordinate axes.
    I'll assume the question means 'Find, in terms of $\displaystyle \theta$, the length ...'. In which case, it's fairly straightforward. Find, in terms of $\displaystyle \theta$:

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

    Then find the values of $\displaystyle x$ and $\displaystyle y$ where this tangent crosses the axes, by putting $\displaystyle y = 0$ and $\displaystyle x = 0$ 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 $\displaystyle \theta$, the length ...'. In which case, it's fairly straightforward. Find, in terms of $\displaystyle \theta$:

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

    $\displaystyle
    \frac{dx}{dy} = \frac{sin^2 \theta cos \theta}{- cos^2 \theta sin \theta}
    $

    $\displaystyle
    = -tan \theta
    $

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

    there the slope of the tangent is 0

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

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

    ie (y-0)=0(x-0)
    ie y=0
    Last edited by zorro; Jan 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 ..........

    $\displaystyle
    \frac{dx}{dy} = \frac{sin^2 \theta cos \theta}{- cos^2 \theta sin \theta}
    $

    $\displaystyle
    = -tan \theta
    $

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

    there the slope of the tangent is 0

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

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

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

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

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

    Meets the $\displaystyle x$-axis where $\displaystyle y = 0$, and

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

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

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

    Can you take it from there?

    Grandad

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    Problem encountered

    Quote Originally Posted by Grandad View Post
    Hello zorro

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

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

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

    Meets the $\displaystyle x$-axis where $\displaystyle y = 0$, and

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

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

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

    Can you take it from there?

    Grandad

    After this i get this

    $\displaystyle \Rightarrow x=a\cos\theta$

    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 $\displaystyle x$-axis where $\displaystyle x = a\cos\theta$.

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

    Quote Originally Posted by Grandad View Post
    Hello zorro

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

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

    cheers
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