Question
If $\text{x}=2\cos\text{t}-\cos2\text{t},\text{y}=2\sin\text{t}-\sin2\text{t},$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{2}.$
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Answer
Given,
$\text{x}=2\cos\text{t}-\cos\text{2}\text{t}$
$\text{y}=2\sin\text{t}-\sin\text{2t}$
Differentiating w.r.t. t,
$\frac{\text{dy}}{\text{dx}}=2(-\sin\text{t})-2(-\sin\text{2t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=2\cot-2\cos\text{2t}$
Dividing both:
$\frac{\text{dy}}{\text{dx}}=\frac{2(\cos\text{t}-\cos\text{2t})}{2(\sin\text{2t}-\sin\text{t})}$
Differentiating w.r.t. t,
$\Rightarrow\frac{\text{d}\frac{\text{dy}}{\text{dx}}}{\text{dt}}=\frac{(\sin\text{2t}-\sin\text{t})(-\sin\text{t}+2\sin\text{2t})-(\cos\text{t}-\cos\text{2t})(2\cos\text{2t}-\cos\text{t})}{(\sin\text{2t}-\sin\text{t})^2}$
Dividing:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{(\sin\text{2t}-\sin\text{t})(2\sin\text{t}-\sin\text{t})-(\cos\text{t}-\cos\text{2t})(2\cos\text{2t}-\cos\text{t})}{2(\sin\text{2t}-\sin)^3}$
Putting: $\text{t}=\frac{\pi}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1+2}{-2}=-\frac{3}{2}$
$\text{x}=2\cos\text{t}-\cos\text{2}\text{t}$
$\text{y}=2\sin\text{t}-\sin\text{2t}$
Differentiating w.r.t. t,
$\frac{\text{dy}}{\text{dx}}=2(-\sin\text{t})-2(-\sin\text{2t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=2\cot-2\cos\text{2t}$
Dividing both:
$\frac{\text{dy}}{\text{dx}}=\frac{2(\cos\text{t}-\cos\text{2t})}{2(\sin\text{2t}-\sin\text{t})}$
Differentiating w.r.t. t,
$\Rightarrow\frac{\text{d}\frac{\text{dy}}{\text{dx}}}{\text{dt}}=\frac{(\sin\text{2t}-\sin\text{t})(-\sin\text{t}+2\sin\text{2t})-(\cos\text{t}-\cos\text{2t})(2\cos\text{2t}-\cos\text{t})}{(\sin\text{2t}-\sin\text{t})^2}$
Dividing:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{(\sin\text{2t}-\sin\text{t})(2\sin\text{t}-\sin\text{t})-(\cos\text{t}-\cos\text{2t})(2\cos\text{2t}-\cos\text{t})}{2(\sin\text{2t}-\sin)^3}$
Putting: $\text{t}=\frac{\pi}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1+2}{-2}=-\frac{3}{2}$
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