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vanhoeak2

  • 3 years ago

dealing with parametric curves if dy/dx = dy/dt/dx/dt = -sin(t)/-2sin(2t) now I need to find d^2y/dx^2 and i'm a bit stuck...

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  1. TuringTest
    • 3 years ago
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    \[\frac{d^2y}{dx^2}=\Large{\frac{d}{dt}\left(\frac{dy}{dx}\right)\over\frac{dx}{dt}}\]

  2. vanhoeak2
    • 3 years ago
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    \[\frac{ \frac{ d }{ dt }(\frac{ -\sin(t) }{ -2\sin(2t) }) }{ -2\sin(2t) }\] then I would have\[\frac{ \frac{ \cos (t) }{ 4\cos(2t)} }{ -2\sin(2t) }\] is that correct?

  3. TuringTest
    • 3 years ago
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    \[\frac d{dt}(\frac{-\sin t}{-2\sin 2t})\neq\frac{\cos t}{4\cos t}\]remember the quotient rule...

  4. TuringTest
    • 3 years ago
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    you may do well to simplify first\[\frac{-\sin t}{-2\sin(2t)}=\frac{\sin t}{4\sin t\cos t}=\frac1{4\cos t}=\frac14\sec t\]

  5. vanhoeak2
    • 3 years ago
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    that makes more sense

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