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 + π sin y = 61π, (4, 5π/2). the point is on the curve because when 4 is substituted for x and 5π/2 is substituted for y, the resulting statement is 61π = 61π, which is a true statement (type an exact answer, using π as needed.) a. write the equation of the tangent line to the curve at (4, 5π/2). y = - 5π/8 x + 5π (type an equation. type an exact answer, using π as needed.) b. write the equation of the normal line to the curve at (4, 5π/2). (type an equation. type an exact answer, using π as needed.)

Explanation:

Step1: Differentiate the given equation 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 $\frac{dy}{dx}$

Rearrange the terms to isolate $\frac{dy}{dx}$:
$6x\frac{dy}{dx}+\pi\cos y\frac{dy}{dx}=- 6y$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(6x+\pi\cos y)=-6y$.
So, $\frac{dy}{dx}=\frac{-6y}{6x + \pi\cos y}$.

Step3: Find the slope of the tangent line at the point $(4,\frac{5\pi}{2})$

Substitute $x = 4$ and $y=\frac{5\pi}{2}$ into $\frac{dy}{dx}$:
$\cos(\frac{5\pi}{2}) = 0$, so $\frac{dy}{dx}\big|_{x = 4,y=\frac{5\pi}{2}}=\frac{-6\times\frac{5\pi}{2}}{6\times4+\pi\times0}=-\frac{5\pi}{8}$.
The equation of a line in point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,\frac{5\pi}{2})$ and $m =-\frac{5\pi}{8}$.
$y-\frac{5\pi}{2}=-\frac{5\pi}{8}(x - 4)$.
$y-\frac{5\pi}{2}=-\frac{5\pi}{8}x+\frac{5\pi}{2}$.
$y=-\frac{5\pi}{8}x + 5\pi$.

Step4: Find the slope of the normal line

The slope of the normal line $m_n$ is the negative reciprocal of the slope of the tangent line.
Since the slope of the tangent line $m_t=-\frac{5\pi}{8}$, then $m_n=\frac{8}{5\pi}$.
Using the point - slope form $y - y_1=m_n(x - x_1)$ with $(x_1,y_1)=(4,\frac{5\pi}{2})$ and $m_n=\frac{8}{5\pi}$.
$y-\frac{5\pi}{2}=\frac{8}{5\pi}(x - 4)$.
$y=\frac{8}{5\pi}x-\frac{32}{5\pi}+\frac{5\pi}{2}$.

Answer:

$y=\frac{8}{5\pi}x-\frac{32}{5\pi}+\frac{5\pi}{2}$