verify that the given point is on the curve. find the lines that are (a) tangent and (b) normal to the curve…

Explanation:

Step1: Verify the point on the curve

Substitute $x = 2$ and $y=\frac{3\pi}{2}$ into $2xy+\pi\sin y$. We get $2\times2\times\frac{3\pi}{2}+\pi\sin\frac{3\pi}{2}=6\pi-\pi = 5\pi$, so the point $(2,\frac{3\pi}{2})$ is on the curve.

Step2: Differentiate the equation implicitly

Differentiate $2xy+\pi\sin y = 5\pi$ with respect to $x$. Using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$ for $2xy$ (where $u = 2x$ and $v = y$) and the chain - rule for $\pi\sin y$. We have $2y + 2x\frac{dy}{dx}+\pi\cos y\frac{dy}{dx}=0$.

Step3: Solve for $\frac{dy}{dx}$

Rearrange the terms to isolate $\frac{dy}{dx}$: $2x\frac{dy}{dx}+\pi\cos y\frac{dy}{dx}=- 2y$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2x+\pi\cos y)=-2y$. So, $\frac{dy}{dx}=\frac{-2y}{2x+\pi\cos y}$.

Step4: Find the slope of the tangent line

Substitute $x = 2$ and $y=\frac{3\pi}{2}$ into $\frac{dy}{dx}$: $\frac{dy}{dx}\big|_{x = 2,y=\frac{3\pi}{2}}=\frac{-2\times\frac{3\pi}{2}}{2\times2+\pi\cos\frac{3\pi}{2}}=\frac{-3\pi}{4 + 0}=-\frac{3\pi}{4}$. Using the point - slope form of a line $y - y_1=m(x - x_1)$ (where $(x_1,y_1)=(2,\frac{3\pi}{2})$ and $m =-\frac{3\pi}{4}$), we get $y-\frac{3\pi}{2}=-\frac{3\pi}{4}(x - 2)$. Simplify to $y=-\frac{3\pi}{4}x+3\pi$.

Step5: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. If the slope of the tangent line is $m_t=-\frac{3\pi}{4}$, then the slope of the normal line $m_n=\frac{4}{3\pi}$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(2,\frac{3\pi}{2})$ and $m=\frac{4}{3\pi}$, we have $y-\frac{3\pi}{2}=\frac{4}{3\pi}(x - 2)$. Expand: $y-\frac{3\pi}{2}=\frac{4}{3\pi}x-\frac{8}{3\pi}$. $y=\frac{4}{3\pi}x-\frac{8}{3\pi}+\frac{3\pi}{2}=\frac{4}{3\pi}x+\frac{-16 + 9\pi^{2}}{6\pi}=\frac{4}{3\pi}x+\frac{9\pi^{2}-16}{6\pi}$.

Answer:

a. $y =-\frac{3\pi}{4}x + 3\pi$ b. $y=\frac{4}{3\pi}x+\frac{9\pi^{2}-16}{6\pi}$