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Math Help - Calculating derivitive from trig equation

  1. #1
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    Calculating derivitive from trig equation

    Hi, i need to calculate the derivitive from the following formular:

    y = 3sin (PIE * t / 12) : t is a variable and PIE is 3.14 etc

    I think its: y = PIEcos (PIE * t / 12)

    Or would it be: y = 3PIEcos (PIE * t / 12) ??


    All help is appreciated.
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  2. #2
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    Hello,

    Let a be pi/12 (no e at the end of pi...).

    Your function is now 3 sin(a*t)
    Its derivative if 3*a*cos(a*t)

    That is to say 3*pi/12 *cos(pi*t/12)=pi/4*cos(pi*t/12)
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Ok here you go

    if f(x)=3sin\bigg(\frac{\pi{t}}{12}\bigg)...then f'(x)=3cos\bigg(\frac{\pi{t}}{12}\bigg)*\frac{\pi}  {12}=\frac{\pi}{4}cos\bigg(\frac{\pi{t}}{12}\bigg)
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  4. #4
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    Thanks for your replies. The next step asks, "calculate the rate of change of the depth of the water when t = 19."

    Would the answer to this be 0.203?? (to 3dp)
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    The answer

    would be f'(19)=\frac{\pi{cos\bigg(\frac{19\pi}{12}\bigg)}}  {4}=\frac{(\sqrt{3}-1)\pi{\sqrt{2}}}{16}=.0677
    Last edited by Mathstud28; April 5th 2008 at 10:48 AM.
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  6. #6
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    Awesome, thanks. But if you follow your equation through, i get the answer as 0.575?? Also, where did the square root of 3 come from?
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  7. #7
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    I think that you can have the exact value of cos(19pi/12) by saying that 19pi/12=3pi/2+pi/12

    After that, i don't really know, you can use some methods you study in high school, but it's kinda too far for me
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Wow

    I am sorry I am the typo king today it should be \frac{(\sqrt{3}-1)\pi{\sqrt{2}}}{16}...and this comes from using the basic unit circle
    Last edited by Mathstud28; April 5th 2008 at 10:43 AM.
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  9. #9
    Moo
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    I'd like to see your proof that cos(19pi/12) equals to what you said
    Not that i think it's false, but it's a method i've forgotten...
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  10. #10
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    Sweet, thanks for all your help. Are you able to just quickly explain where you get the square root bits from? It is confusing me a bit.
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Its kinda complicated

    ...do you really want me to go through it?
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  12. #12
    Moo
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    Oh yeah ^^

    If you could give the formula so quickly, it may not be so difficult to tell us how to do ?
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    hah

    ok so persistent cos\bigg(\frac{19\pi}{12}\bigg)=cos\bigg(\frac{3\p  i}{2}+\frac{\pi}{12}\bigg)=cos\bigg(\frac{3\pi}{2}  \bigg)cos\bigg(\frac{\pi}{12}\bigg)-sin\bigg(\frac{3\pi}{2}\bigg)sin\bigg(\frac{\pi}{1  2}\bigg) =0-\bigg(\frac{(\sqrt{3}-1)\sqrt{2})}{4}\bigg)(-1)..thes being commonly known ones...and now multiplying that by \frac{\pi}{4} gives you what you want..now before you ask sin\bigg(\frac{\pi}{12}\bigg) is just one you have to know its just one of the exact values on the unit circle
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  14. #14
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    is just one you have to know
    Really ?
    The commonly learnt values are pi/6, pi/4, pi/3, pi/2 (and 0, pi etc...), but it's not so easy (let's say 3 lines of calculus) to retrieve it with pi/12=1/2*pi/6
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  15. #15
    MHF Contributor Mathstud28's Avatar
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    if you are trying to be sardonic

    if you did rewrite it as sin\bigg(\frac{1}{2}\frac{\pi}{6}\bigg) you would nowhere to go except use another trig identity that sin(2x)=(-1)^{\frac{x}{2\pi}}\sqrt{\frac{1-cos(x)}{2}}..then you'd have to go that cos\bigg(\frac{\pi}{6}\bigg)=\frac{\sqrt{3}}{2}...then go that the sin\bigg(\frac{1}{2}\frac{\pi}{6}\bigg)= is equal to \frac{\pi}{6} imputed into that forumla...for me its easier to remember what the sin\bigg(\frac{\pi}{12}\bigg) is
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