3 years ago

Please help in order to pass the grade I’m in now I have to finish my aleaks

Mathematics

1 answer:

xz_007 [3.2K]3 years ago

8 0

U=12 Hopes this helps

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Find dy/dx and d2y/dx2. x = cos(2t), y = cos(t), 0 < t < π

pogonyaev

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$

$y=\cos t\implies\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$x=\cos2t\implies\dfrac{\mathrm dx}{\mathrm dt}=-2\sin2t=-4\sin t\cos t$
$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-\sin t}{-4\sin t\cos t}=\dfrac14\sec t$

Let $z=\dfrac{\mathrm dy}{\mathrm dx}$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm dz}{\mathrm dx}=\dfrac{\mathrm dz}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm dz}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$
$\implies\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac{\mathrm d}{\mathrm dt}\left[\frac14\sec t\right]}{-4\sin t\cos t}=\dfrac{\frac14\sec t\tan t}{-4\sin t\cos t}=-\dfrac1{16}\sec^3t$