Question
If $\text{x}=3\cot-2\cos^3\text{t},\text{y}=3\sin\text{t}-2\sin^3\text{t}$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
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Answer
Given,
$\text{x}=3\cot-2\cos^3\text{t},$
$\text{y}=3\sin\text{t}-2\sin^3\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}-6\cos^2\text{t}(-\sin\text{t})$
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}+6\cos^2\text{t}\sin\text{t}$
And $\text{y}=3\sin\text{t}-2\sin^2\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dy}}{\text{dt}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
Now,
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot-2\sin^2\text{t}\cos\text{t}}{-\sin\text{t}+2\cos^2\text{t}\sin\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot[1-2\sin^2\text{t]}}{\sin\text{t}[2\cos^2\text{t}-1]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cot\text{t}$
Differentiating both w.r.t. x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}(\cot\text{x})}{\text{dx}}=-\text{cosec}^2\text{x}$
$\text{x}=3\cot-2\cos^3\text{t},$
$\text{y}=3\sin\text{t}-2\sin^3\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}-6\cos^2\text{t}(-\sin\text{t})$
$\frac{\text{dx}}{\text{dt}}=-3\sin\text{t}+6\cos^2\text{t}\sin\text{t}$
And $\text{y}=3\sin\text{t}-2\sin^2\text{t}$
Differentiating both w.r.t. t,
$\frac{\text{dy}}{\text{dt}}=3\cos\text{t}-6\sin^2\text{t}\cos\text{t}$
Now,
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot-2\sin^2\text{t}\cos\text{t}}{-\sin\text{t}+2\cos^2\text{t}\sin\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cot[1-2\sin^2\text{t]}}{\sin\text{t}[2\cos^2\text{t}-1]}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cot\text{t}$
Differentiating both w.r.t. x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}(\cot\text{x})}{\text{dx}}=-\text{cosec}^2\text{x}$
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