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

Explanation:

Step1: Differentiate the given equation implicitly

Differentiate $4xy+\pi\sin y = 11\pi$ with respect to $x$. Using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$ for $4xy$ (where $u = 4x$ and $v = y$) and the chain - rule for $\pi\sin y$. The derivative of $4xy$ is $4y+4x\frac{dy}{dx}$, and the derivative of $\pi\sin y$ is $\pi\cos y\frac{dy}{dx}$, and the derivative of $11\pi$ is $0$. So, $4y + 4x\frac{dy}{dx}+\pi\cos y\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(4x+\pi\cos y)=- 4y$. Then $\frac{dy}{dx}=\frac{-4y}{4x+\pi\cos y}$.

Step3: Find the slope of the tangent line

Substitute $x = 1$ and $y=\frac{5\pi}{2}$ into $\frac{dy}{dx}$. $\cos(\frac{5\pi}{2}) = 0$, so $\frac{dy}{dx}\big|_{x = 1,y=\frac{5\pi}{2}}=\frac{-4\times\frac{5\pi}{2}}{4\times1+\pi\times0}=\frac{-10\pi}{4}=-\frac{5\pi}{2}$.

Step4: Use the point - slope form to find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,\frac{5\pi}{2})$ and $m =-\frac{5\pi}{2}$. $y-\frac{5\pi}{2}=-\frac{5\pi}{2}(x - 1)$. Expand: $y-\frac{5\pi}{2}=-\frac{5\pi}{2}x+\frac{5\pi}{2}$. $y=-\frac{5\pi}{2}x + 5\pi$.

Answer:

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