91Ó°ÊÓ

The parametric curve is $x={\cos }^{3}t,y={\sin }^{3}t,t=\frac{\mathrm{Ï€}}{4}$.

Consider the parametric curves $x=t^2,y=(2-t{)}^2$at $t=\frac{1}{2}$.

The objective is to find the equation of a tangent line for the given parametric equations.

The formula to find the tangent line equation is $y-y_1=m\left(x-x_1\right)$.

First find the slope of the parametric curves by finding the derivative of the parametric curves.

For that we use the formula$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

Now take the parametric equation $x=t^2$.

Differentiate the curve with respect to t.

Then

$\frac{dx}{dt}=\frac{d}{dt}{t}^2$since $x=t^2$

$\frac{dx}{dt}=2t·\frac{dt}{dt}\left[since\frac{d{t}^2}{dt}=2t\right]$

$\frac{dx}{dt}=2t$

Now take the parametric equation $y=(2-t{)}^2$.

Differentiate the curve with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}(2-t{)}^2 \\ \frac{dy}{dt}=2×(2-t\left)\frac{d}{dt}\right(2-t) \\ \frac{dy}{dt}=2(2-t\left)\right(-1) \\ \frac{dy}{dt}=-2(2-t) \end{gathered}$$

Now substitute the values of $\frac{dx}{dt},\frac{dy}{dt}$in the slope formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Then

$\frac{dy}{dx}=\frac{-2(2-t)}{2t}\left[Since\frac{dx}{dt}=2t,\frac{dy}{dt}=-2(2-t)\right]$

$\frac{dy}{dx}=\frac{-(2-t)}{t}$

The slope when $t=\frac{1}{2}$is as follows,

$$\begin{gathered} \frac{dy}{d{x}_{1=1}^2}=\frac{-\left(2-\frac{1}{2}\right)}{\frac{1}{2}} \\ \frac{dy}{d{x}_{1=1}^2}=\frac{-\left(\frac{4-1}{2}\right)}{\frac{1}{2}} \end{gathered}$$

On further simplification,

$$\begin{gathered} \frac{dy}{d{x}_1-\frac{1}{2}}=\frac{-3}{\frac{1}{2}} \\ \frac{dy}{d{x}_1,\frac{1}{2}}=-3 \end{gathered}$$

Thus, the slope of the parametric equation is $\frac{dy}{dx}=m=-3$.