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Math Help - Speed Problem

  1. #1
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    Speed Problem

    Question: When a particle moves in a plane so that x=2cos3t and y=sin3t, then the speed on the interval 0 =< t =< \frac{\pi}{2} is a maximum when t equals ...

    A. 0

    B. \frac{\pi}{6}

    C. \frac{\pi}{4}

    D. \frac{\pi}{3}

    E. \frac{\pi}{2}

    I took the derivatives of both x and y equations and put them in the speed formula. But now that I have the derivatives in the speed formula, I'm not sure what the next step is algebraically.

    I have, all in the square root,

    (36sin3t)^2 + (9cos3t)^2

    So how do I expand that out and what are the next steps?
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  2. #2
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    Hi

    You mean 36 \sin^2(3t) + 9 \cos^2 (3t) I suppose ?

    Use \cos^2(3t) = 1 - \sin^2(3t)
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  3. #3
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    Quote Originally Posted by Lord Darkin View Post
    Question: When a particle moves in a plane so that x=2cos3t and y=sin3t, then the speed on the interval 0 =< t =< \frac{\pi}{2} is a maximum when t equals ...

    A. 0

    B. \frac{\pi}{6}

    C. \frac{\pi}{4}

    D. \frac{\pi}{3}

    E. \frac{\pi}{2}

    I took the derivatives of both x and y equations and put them in the speed formula. But now that I have the derivatives in the speed formula, I'm not sure what the next step is algebraically.

    I have, all in the square root,

    (36sin3t)^2 + (9cos3t)^2

    So how do I expand that out and what are the next steps?
    Wouldn't an application of the chain rule be easier?

    \frac{dx}{dt} = -6sin(3t)

    \frac{dy}{dt} = 3cos(3t)

    \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}  = -\frac{3cos(3t)}{6sin(3t)} = -\frac{1}{2}cot(3t)

    IIRC a maximum occurs when \frac{dy}{dx} = 0

    -\frac{cos(3t)}{2sin(3t)} = 0

    cos(3t) = 0

    3t = \frac{\pi}{2}

    t = \frac{\pi}{6}
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  4. #4
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    I'm still a little shaky on this. My teacher, on the answer key he gave me, gave me this all under the square root:

    36 \sin^2(3t) + 9 \cos^2 (3t)

    Then he wrote:

    <br />
72sin(3t)cos(t)*3 + 18cos(3t)(-sin(3t))*3 = 0<br />

    Simplifies to:

    <br />
81sin(6t)<br />

    I'm not sure how to get to the part starting with 72 sin 3 etc. I think I understand the concept of this problem, I just want to know the algebra on the answer key.

    Also, answer is pi/6
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  5. #5
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    \sqrt{36 \sin^2(3t) + 9 \cos^2 (3t)} is maximum when 36 \sin^2(3t) + 9 \cos^2 (3t) is maximum which means when the derivative of 36 \sin^2(3t) + 9 \cos^2 (3t) is equal to 0

    Derivative of
    \sin^2(3t)
    The derivative of u^2(t) is 2 \:u(t)\: u'(t)
    The derivative of u(t) = \sin(3t) is 3 \:\cos(3t)
    Therefore the derivative of
    \sin^2(3t) is 2 \:sin(3t)\: 3 \:\cos(3t) = 6 \:sin(3t) \:cos(3t)

    In the same way the derivative of \cos^2(3t) is 2 \:cos(3t)\: (-3) \:\sin(3t) = -6 \:sin(3t) \:cos(3t)

    Therefore the derivative of 36 \sin^2(3t) + 9 \cos^2 (3t) is 36 \times 6 \:sin(3t) \:cos(3t) - 9 \times 6 \:sin(3t) \:cos(3t) = 162 \:sin(3t) \:cos(3t)

    Using \sin(2x) = 2 \:\sin x \: \cos x the derivative of 36 \sin^2(3t) + 9 \cos^2 (3t) is 81 \: \sin(6t) which is equal to 0 when t = k\:\frac{\pi}{6}

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  6. #6
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    Thanks for the help. I think I get it.

    1 last thing.

    If I have

    sin3(\frac{\pi}{3})

    does that simplify to

    sin(\pi)?
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  7. #7
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Lord Darkin View Post
    Thanks for the help. I think I get it.

    1 last thing.

    If I have

    sin3(\frac{\pi}{3})

    does that simplify to

    sin(\pi)?
    No, it won't.

    \sin \left[3\left(\frac{\pi}{3}\right)\right] is the graph of sin \left(\frac{\pi}{3}\right) but 'squashed' by a factor of 3 such that a period is now \frac{2\pi}{3}
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  8. #8
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    I get it. Thanks.
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