Question

solve this question ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{x=\dfrac{1+t}{t^3}\,\,\,\,\,\,and\,\,\,\,\,\,y=\dfrac{3}{2t^2}+\dfrac{2}{t}}$

$\tt{\implies\,x=\dfrac{1}{t^3}+\dfrac{1}{t^2}\,\,\,\,\,\,and\,\,\,\,\,\,y=\dfrac{3}{2t^2}+\dfrac{2}{t}}$

Differentiating each expression w.r.t. t,

$\tt{\implies\,\dfrac{dx}{dt}=-\dfrac{3}{t^4}-\dfrac{2}{t^3}\,\,\,\,\,\,and\,\,\,\,\,\,\dfrac{dy}{dt}=-\dfrac{3}{t^3}-\dfrac{2}{t^2}}$

$\tt{\implies\,\dfrac{dx}{dt}=\dfrac{1}{t}\bigg(-\dfrac{3}{t^3}-\dfrac{2}{t^2}\bigg)\,\,\,\,\,\,and\,\,\,\,\,\,\dfrac{dy}{dt}=-\dfrac{3}{t^3}-\dfrac{2}{t^2}}$

Now,

$\sf{\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{-\dfrac{3}{t^3}-\dfrac{2}{t^2}}{\dfrac{1}{t}\bigg(-\dfrac{3}{t^3}-\dfrac{2}{t^2}\bigg)}}$

$\sf{\implies\dfrac{dy}{dx}=\dfrac{1}{\dfrac{1}{t}}}$

$\sf{\implies\dfrac{dy}{dx}=t}$

Now,

$\sf{\left|\dfrac{dy}{dx}-x\bigg(\dfrac{dy}{dx}\bigg)^3\right|}$

$\sf{=\left|t-\bigg(\dfrac{1+t}{t^3}\bigg)\cdot\,t^3\right|}$

$\sf{=\left|t-(1+t)\right|}$

$\sf{=\left|t-1-t\right|}$

$\sf{=\left|-1\right|}$

$\sf{=1}$