If x= 1+ t^2 ,y= 3+2 t , find dy dx - Vidyadip

Question

If $\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{y}=\frac{3+2\log\text{t}}{\text{t}},$ find $\frac{\text{dy}}{\text{dx}}$

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Answer

$\text{x}=3\sin\text{t}-\sin3\text{t},\text{y}=3\cos\text{t}-\cos3\text{t}$
$\frac{\text{dx}}{\text{dt}}=3\cos\text{t}-3\cos3\text{t}$
$\frac{\text{dy}}{\text{dt}}=-3\sin\text{t}+3\sin3\text{t}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{3\sin\text{t}+3\sin3\text{t}}{3\cos\text{t}-3\cos3\text{t}}$
When $\text{t}=\frac{\pi}{3}$
$\frac{\text{dy}}{\text{dx}}=\frac{-3\sin\big(\frac{\pi}{3}\big)+3\sin(\pi)}{3\cos\big(\frac{\pi}{3}\big)-3\cos(\pi)}=\frac{3\times\frac{\sqrt{3}}{2}+0}{3\times\frac{1}{2}-3(-1)}=-\frac{1}{\sqrt{3}}$

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