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Math Help - Finding where a function = 0

  1. #1
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    Finding where a function = 0

    I need to find where

    20cos(5t) + 4sin(5t) = 0 for the first and second time after the initial t = 0

    I get to

    arctan (-5) = 5t

    I can take arctan of -5 to get -1.3... then divide by 5, but that doesn't help me.

    I need help quickly please.
    Last edited by fourierT; March 27th 2014 at 04:19 AM.
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  2. #2
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    Re: Finding where a function = 0

    Quote Originally Posted by fourierT View Post
    I need to find where

    20cos(5t) + 4sin(5t) = 0 for the first and second time after the initial t = 0

    I get to

    arctan (-5) = 5t

    I can take arctan of -5 to get -1.3... then divide by 5, but that doesn't help me.

    I need help quickly please.
    ok first rewrite this as a single trig function with a phase offset.

    $20\cos(5t)+4\sin(5t)=A\cos(5t+\phi)$

    $A=\sqrt{20^2+4^2}=\sqrt{416}$

    $\phi = \arctan\left(\dfrac{4}{20}\right)\approx 0.197~ rad$

    Now the cosine will be 0 for all odd multiples of $\dfrac{\pi}{2}$ so we look for

    $5t+\phi=\dfrac{2k+1}{2}\pi$

    $t=\dfrac{1}{5}\left(\dfrac{2k+1}{2}\pi-\phi\right)$

    you can probably take it from here.
    Thanks from fourierT
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  3. #3
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    Re: Finding where a function = 0

    Quote Originally Posted by romsek View Post
    ok first rewrite this as a single trig function with a phase offset.

    $20\cos(5t)+4\sin(5t)=A\cos(5t+\phi)$

    $A=\sqrt{20^2+4^2}=\sqrt{416}$

    $\phi = \arctan\left(\dfrac{4}{20}\right)\approx 0.197~ rad$

    Now the cosine will be 0 for all odd multiples of $\dfrac{\pi}{2}$ so we look for

    $5t+\phi=\dfrac{2k+1}{2}\pi$

    $t=\dfrac{1}{5}\left(\dfrac{2k+1}{2}\pi-\phi\right)$

    you can probably take it from here.
    $t=\dfrac{1}{5}\left(\dfrac{3}{2}\pi-\phi\right)$

    and

    $t=\dfrac{1}{5}\left(\dfrac{5}{2}\pi-\phi\right)$


    Are these correct?
    Last edited by fourierT; March 27th 2014 at 04:42 AM.
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  4. #4
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    Re: Finding where a function = 0

    Your first way of attacking it would have worked but you goofed somewhere.

    $20\cos(5t)+4\sin(5t)=0$

    $20\cos(5t)=-4\sin(5t)$

    $-5=\tan(5t)$

    ah I see what happened. You just took the arctan here ignoring the fact that the tangent function is periodic. If you do this

    $-5 = \tan(5t+k\pi)$ and solve for $k=0,1$ you'll arrive at the same answers.
    Thanks from fourierT
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  5. #5
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    Re: Finding where a function = 0

    Quote Originally Posted by romsek View Post
    Your first way of attacking it would have worked but you goofed somewhere.

    $20\cos(5t)+4\sin(5t)=0$

    $20\cos(5t)=-4\sin(5t)$

    $-5=\tan(5t)$

    ah I see what happened. You just took the arctan here ignoring the fact that the tangent function is periodic. If you do this

    $-5 = \tan(5t+k\pi)$ and solve for $k=0,1$ you'll arrive at the same answers.
    Wouldn't I solve for k(1,2) because it asks for where it = 0 after the initial t =0?
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  6. #6
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    Re: Finding where a function = 0

    yes, you're correct.
    Thanks from fourierT
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  7. #7
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    Re: Finding where a function = 0

    Quote Originally Posted by romsek View Post
    yes, you're correct.
    Thanks for your help
    Last edited by fourierT; March 27th 2014 at 05:14 AM.
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