Question

(a) find y(x).
y(x) = 6\sin(x)+6x\cos(x)\\
find an equation of the tangent line to the curve y at the point (\frac{\pi}{2},3\pi).
y = 6x + 9\frac{\pi}{2}
(b) illustrate part (a) by graphing the curve y and the tangent line on the same screen.

Explanation:

Step1: Recall tangent - line formula

The equation of a tangent line to the curve $y = f(x)$ at the point $(x_0,y_0)$ is given by $y - y_0=f^{\prime}(x_0)(x - x_0)$. We know that $x_0=\frac{\pi}{2}$, $y_0 = 3\pi$, and $y^{\prime}(x)=6\sin(x)+6x\cos(x)$.

Step2: Evaluate the derivative at $x_0$

Substitute $x = \frac{\pi}{2}$ into $y^{\prime}(x)$:
$y^{\prime}(\frac{\pi}{2})=6\sin(\frac{\pi}{2})+6\times\frac{\pi}{2}\cos(\frac{\pi}{2})$.
Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2})=0$, we have $y^{\prime}(\frac{\pi}{2})=6\times1+6\times\frac{\pi}{2}\times0 = 6$.

Step3: Find the equation of the tangent line

Use the point - slope form $y - y_0=y^{\prime}(x_0)(x - x_0)$. Substitute $x_0=\frac{\pi}{2}$, $y_0 = 3\pi$, and $y^{\prime}(x_0)=6$ into it:
$y-3\pi=6(x - \frac{\pi}{2})$.
Expand the right - hand side: $y-3\pi=6x-3\pi$.
Then $y = 6x$.

Answer:

$y = 6x$