Question

a. verify that the given point lies on the curve.
b. determine an equation of the line tangent to the curve at the given point.
(sin y + 6x=y^{2};(\frac{pi^{2}}{6},pi))
a. verify that the point is on the given curve. evaluate each side of the equation separately.
when (x = \frac{pi^{2}}{6}) and (y=pi,sin y + 6x=square) and (y^{2}=square).
(type exact answers, using (pi) as needed.)

Explanation:

Step1: Evaluate left - hand side

Substitute $x = \frac{\pi^{2}}{6}$ and $y=\pi$ into $\sin y + 6x$.
$\sin(\pi)+6\times\frac{\pi^{2}}{6}=0 + \pi^{2}=\pi^{2}$

Step2: Evaluate right - hand side

Substitute $y = \pi$ into $y^{2}$.
$y^{2}=\pi^{2}$
Since the left - hand side equals the right - hand side when $x=\frac{\pi^{2}}{6}$ and $y = \pi$, the point lies on the curve.

For part b:
First, differentiate the equation $\sin y+6x = y^{2}$ implicitly with respect to $x$.
The derivative of $\sin y$ with respect to $x$ is $\cos y\cdot y'$ (by the chain rule), the derivative of $6x$ with respect to $x$ is $6$, and the derivative of $y^{2}$ with respect to $x$ is $2y\cdot y'$.
So we have $\cos y\cdot y'+6 = 2y\cdot y'$.
Solve for $y'$:
$\cos y\cdot y'-2y\cdot y'=-6$
$y'(\cos y - 2y)=-6$
$y'=\frac{6}{2y-\cos y}$
Substitute $x = \frac{\pi^{2}}{6}$ and $y=\pi$ into $y'$:
$y'=\frac{6}{2\pi-\cos(\pi)}=\frac{6}{2\pi + 1}$
The equation of the tangent line using the point - slope form $y - y_{1}=m(x - x_{1})$ where $(x_{1},y_{1})=(\frac{\pi^{2}}{6},\pi)$ and $m=\frac{6}{2\pi + 1}$ is:
$y-\pi=\frac{6}{2\pi + 1}(x-\frac{\pi^{2}}{6})$
$y=\frac{6}{2\pi + 1}x-\frac{\pi^{2}}{2\pi + 1}+\pi$

Answer:

a. The point $(\frac{\pi^{2}}{6},\pi)$ lies on the curve since when $x = \frac{\pi^{2}}{6}$ and $y=\pi$, $\sin y+6x=\pi^{2}$ and $y^{2}=\pi^{2}$.
b. The equation of the tangent line is $y=\frac{6}{2\pi + 1}x-\frac{\pi^{2}}{2\pi + 1}+\pi$