Question
If $\text{x}=\text{a}(\cos2\text{t}+2\text{t}\sin2\text{t})\ \text{and}\ \text{y}=\text{a}(\sin2\text{t}-2\text{t}\cos2\text{t}),$ then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
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Answer
$\text{x}=\text{a}(\cos2\text{t}+2\text{t}\sin2\text{t})$
$\frac{\text{dx}}{\text{dt}}=-2\text{a}\sin2\text{t}+2\text{a}\sin2\text{t}+4\text{at}\cos2\text{t}=4\text{at}\cos2\text{t}$
$\text{y}=\text{a}(\sin2\text{t}-2\text{t}\cos2\text{t})$
$\frac{\text{dy}}{\text{dt}}=2\text{a}\cos2\text{t}-2\text{a}\cos2\text{t}+4\text{at}\sin2\text{t}=4\text{at}\sin2\text{t}$
$\frac{\text{dy}}{\text{dx}}=\tan2\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}(\tan2\text{t})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sec^22\text{t}\frac{\text{d}}{\text{dx}}(\text{t})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sec^22\text{t}\times\frac{1}{4\text{at}\cos2\text{t}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{2\text{a}}\sec^32\text{t}$
$\frac{\text{dx}}{\text{dt}}=-2\text{a}\sin2\text{t}+2\text{a}\sin2\text{t}+4\text{at}\cos2\text{t}=4\text{at}\cos2\text{t}$
$\text{y}=\text{a}(\sin2\text{t}-2\text{t}\cos2\text{t})$
$\frac{\text{dy}}{\text{dt}}=2\text{a}\cos2\text{t}-2\text{a}\cos2\text{t}+4\text{at}\sin2\text{t}=4\text{at}\sin2\text{t}$
$\frac{\text{dy}}{\text{dx}}=\tan2\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}(\tan2\text{t})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sec^22\text{t}\frac{\text{d}}{\text{dx}}(\text{t})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sec^22\text{t}\times\frac{1}{4\text{at}\cos2\text{t}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{2\text{a}}\sec^32\text{t}$
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