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lovesit2x

  • 2 years ago

r=sin(4-2theta)

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  1. lovesit2x
    • 2 years ago
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    \[Find dr/d \theta. r=\sin (4-2\theta)\]

  2. lovesit2x
    • 2 years ago
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    @zepdrix

  3. ravina
    • 2 years ago
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    ambra te paundi hai udariyan

  4. zepdrix
    • 2 years ago
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    Hmm so what would the derivative of \(\large \sin\theta\) be? This is another one of those problems where we'll have to work with the chain rule. :)

  5. ravina
    • 2 years ago
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    phul koi vilayti ala lae gaya udd main reh gaya kyarian

  6. lovesit2x
    • 2 years ago
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    cos theta

  7. mathsmind
    • 2 years ago
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    ur previous question was solved @lovesit2x

  8. ravina
    • 2 years ago
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    ho attiye kapurthaliye

  9. lovesit2x
    • 2 years ago
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    yes it was, thanks for your help @mathsmind

  10. zepdrix
    • 2 years ago
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    Yes good! Since the inside of the sine is more than just \(\large \theta\), we have to apply the chain rule, multiplying by the derivative of the inside. \[\huge r'=\cos(4-2\theta)\color{royalblue}{\left(4-2\theta\right)'}\]

  11. lovesit2x
    • 2 years ago
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    cos(4-2theta) (-2) ?

  12. zepdrix
    • 2 years ago
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    Understand the process? We took the derivative of sine, which gave us cosine (with the same contents inside of it). And then we make a copy of the inside, and multiply by it's derivative.

  13. zepdrix
    • 2 years ago
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    Yah sounds good!

  14. lovesit2x
    • 2 years ago
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    :D yay!!

  15. zepdrix
    • 2 years ago
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    \[\huge r'=-2\cos(4-2\theta)\]Yay!

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