Question

In Exercises 41-44 find an equation for the line tangent to the parametric curve at the given value t f.

$x={\cos }^{3}t,y={\sin }^{3}t,t=\frac{\pi }{4}$.

Short answer

The tangent line at $t=\frac{\pi }{4}$ is $y-\frac{1}{2\sqrt{2}}=-1\left(x-\frac{1}{2\sqrt{2}}\right)$.

Step-by-step solution

Given information

The parametric curve $x={\cos }^{3}t,y={\sin }^{3}t,t=\frac{\pi }{4}$.

Calculation

Consider the parametric curves $x={\cos }^{3}t,y={\sin }^{3}t$ at $t=\frac{\pi }{4}$.

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$x={\cos }^{3}t$.

Differentiate the curve with respect to t.

Then,

$$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}\left({\cos }^{3}t\right)\left[sincex={\cos }^{3}t\right] \\ \frac{dx}{dt}=3{\cos }^{2}t\frac{d}{dt}\left(\cos t\right) \\ \frac{dx}{dt}=3{\cos }^{2}t(-\sin t)\left[since\frac{d}{dt}\cos t=-\sin t\right] \end{gathered}$$

Now take the parametric equation $y={\sin }^{3}t$.

Differentiate the curve with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}\left({\sin }^{3}t\right) \\ \frac{dy}{dt}=3{\sin }^{2}t·\frac{d}{dt}\left(\sin t\right) \\ \frac{dy}{dt}=3{\sin }^{2}t·\cos t\left[since\frac{d}{dt}\sin t=\cos t\right] \end{gathered}$$

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

$$\begin{gathered} \frac{dy}{dx}=\frac{3{\sin }^{2}t\cos t}{-3{\cos }^{2}t\sin t}\left[\sin ce\frac{dx}{dt}=-3{\cos }^{2}t\sin t,\frac{dy}{dt}=3{\sin }^{2}t·\cos t\right] \\ \frac{dy}{dx}=-\frac{\sin t}{\cos t} \end{gathered}$$

The slope when $t=\frac{\pi }{4}$ is as follows,

$\frac{dy}{d{x}_r=\frac{\pi }{4}}=-\frac{\sin \frac{\pi }{4}}{\cos t\frac{\pi }{4}}$

On further simplification,

$$\begin{gathered} \frac{dy}{d{x}_{t-\frac{\pi }{4}}}=-\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} \\ \frac{dy}{d{x}_{t-\frac{\pi }{4}}}=-1 \end{gathered}$$

Thus, the slope of the parametric equation is$\frac{dy}{dx}=m=-1$.

Further calculation

The point $(x,y)$When $t=\frac{\pi }{4}$is,

$(x,y)=\left({\cos }^{3}t,{\sin }^{3}t\right)$

$(x,y)=\left({\cos }^3\frac{\pi }{4},{\sin }^3\frac{\pi }{4}\right)\left[since\right.$by

substituting $t=\frac{\pi }{4}$

$(x,y)=\left(\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}\right)$

Now the slope point formula is$y-y_1=m\left(x-x_1\right)$

The point is $\left(\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}\right)$and the slope $\frac{dy}{dx}=m=-1$.

$y-\frac{1}{2\sqrt{2}}=-1\left(x-\frac{1}{2\sqrt{2}}\right)\left[\text{since}y-y_1=m\left(x-x_1\right)\right]$

Therefore, the tangent line at $t=\frac{\pi }{4}$is $y-\frac{1}{2\sqrt{2}}=-1\left(x-\frac{1}{2\sqrt{2}}\right)$