find an equation of the line tangent to the curve at the point corresponding to the given value of t.\nx =…

Explanation:

Step1: Find $\frac{dx}{dt}$

Using product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = t$ and $v=\sin t$ or $\cos t$. $\frac{dx}{dt}=-\sin t+\sin t + t\cos t=t\cos t$

Step2: Find $\frac{dy}{dt}$

$\frac{dy}{dt}=\cos t-\cos t + t\sin t=t\sin t$

Step3: Find $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, so $\frac{dy}{dx}=\tan t$

Step4: Evaluate the slope at $t = \frac{5\pi}{4}$

Substitute $t=\frac{5\pi}{4}$ into $\frac{dy}{dx}$, we get $m = \tan(\frac{5\pi}{4}) = 1$

Step5: Find the coordinates of the point

When $t=\frac{5\pi}{4}$, $x=\cos(\frac{5\pi}{4})+\frac{5\pi}{4}\sin(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}-\frac{5\pi\sqrt{2}}{8}=-\frac{4\sqrt{2} + 5\pi\sqrt{2}}{8}=-\frac{(4 + 5\pi)\sqrt{2}}{8}$ $y=\sin(\frac{5\pi}{4})-\frac{5\pi}{4}\cos(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}+\frac{5\pi\sqrt{2}}{8}=\frac{-4\sqrt{2}+5\pi\sqrt{2}}{8}=\frac{(5\pi - 4)\sqrt{2}}{8}$

Step6: Use the point - slope form $y - y_0=m(x - x_0)$

$y-\frac{(5\pi - 4)\sqrt{2}}{8}=1\times(x+\frac{(4 + 5\pi)\sqrt{2}}{8})$ $y=x+\frac{(4 + 5\pi)\sqrt{2}}{8}+\frac{(5\pi - 4)\sqrt{2}}{8}$ $y=x+\frac{4\sqrt{2}+5\pi\sqrt{2}+5\pi\sqrt{2}-4\sqrt{2}}{8}$ $y=x+\frac{10\pi\sqrt{2}}{8}=x+\frac{5\pi\sqrt{2}}{4}$

Answer:

$y=x+\frac{5\pi\sqrt{2}}{4}$