Question

verify that the given point is on the curve. find the lines that are (a) tangent and (b) normal to the curve at the given point.
6xy + \pi\sin y=61\pi, (4,\frac{5\pi}{2})
the point is on the curve because when 4 is substituted for x and \frac{5\pi}{2} is substituted for y, the resulting statement is 61\pi =61\pi, which is a true statement. (type an exact answer, using \pi as needed.)
a. write the equation of the tangent line to the curve at (4,\frac{5\pi}{2})
(type an equation. type an exact answer, using \pi as needed.)

Explanation:

Step1: Differentiate implicitly

Differentiate $6xy+\pi\sin y = 61\pi$ with respect to $x$.
Using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$ for $6xy$ (where $u = 6x$ and $v = y$) and the chain - rule for $\pi\sin y$.
We get $6y+6x\frac{dy}{dx}+\pi\cos y\frac{dy}{dx}=0$.

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

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(6x + \pi\cos y)=-6y$.
Then $\frac{dy}{dx}=\frac{-6y}{6x+\pi\cos y}$.

Step3: Find the slope of the tangent line

Substitute $x = 4$ and $y=\frac{5\pi}{2}$ into $\frac{dy}{dx}$.
$\cos(\frac{5\pi}{2}) = 0$, so $\frac{dy}{dx}\big|_{x = 4,y=\frac{5\pi}{2}}=\frac{-6\times\frac{5\pi}{2}}{6\times4+\pi\times0}=\frac{-15\pi}{24}=-\frac{5\pi}{8}$.

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

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,\frac{5\pi}{2})$ and $m =-\frac{5\pi}{8}$.
$y-\frac{5\pi}{2}=-\frac{5\pi}{8}(x - 4)$.
Expand: $y-\frac{5\pi}{2}=-\frac{5\pi}{8}x+\frac{5\pi}{2}$.
The equation of the tangent line is $y=-\frac{5\pi}{8}x + 5\pi$.

Answer:

$y=-\frac{5\pi}{8}x + 5\pi$