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.
3 sin y + 5x = y²; (π²/5,π)
a. verify that the point is on the given curve. evaluate each side of the equation separat
when x = π²/5 and y = π, 3 sin y + 5x = and y² = .
(type exact answers, using π as needed.)
does the point lie on the curve 3 sin y + 5x = y²?
a. yes, because the values found for each side of the equation are equal.
b. no, because the values found for each side of the equation are unequal.
c. yes, because the value found for y² is greater than the value found for 3 sin y + 5x
d.

Explanation:

Step1: Evaluate left - hand side

Substitute $x = \frac{\pi^{2}}{5}$ and $y=\pi$ into $3\sin y + 5x$. Since $\sin\pi=0$, we have $3\sin\pi+5\times\frac{\pi^{2}}{5}=0 + \pi^{2}=\pi^{2}$.

Step2: Evaluate right - hand side

Substitute $y = \pi$ into $y^{2}$. We get $y^{2}=\pi^{2}$.

Step3: Determine if point lies on curve

Since the left - hand side ($3\sin y + 5x$) and the right - hand side ($y^{2}$) are equal when $x=\frac{\pi^{2}}{5}$ and $y = \pi$, the point lies on the curve.

Step4: Differentiate the equation implicitly

Differentiate $3\sin y+5x = y^{2}$ with respect to $x$.
The derivative of $3\sin y$ with respect to $x$ is $3\cos y\frac{dy}{dx}$ (by the chain rule), the derivative of $5x$ with respect to $x$ is $5$, and the derivative of $y^{2}$ with respect to $x$ is $2y\frac{dy}{dx}$.
So, $3\cos y\frac{dy}{dx}+5 = 2y\frac{dy}{dx}$.

Step5: Solve for $\frac{dy}{dx}$

Rearrange the equation $3\cos y\frac{dy}{dx}+5 = 2y\frac{dy}{dx}$ to get $\frac{dy}{dx}$ terms on one side:
$3\cos y\frac{dy}{dx}-2y\frac{dy}{dx}=-5$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(3\cos y - 2y)=-5$.
Then $\frac{dy}{dx}=\frac{-5}{3\cos y - 2y}$.

Step6: Find the slope at the given point

Substitute $y = \pi$ into $\frac{dy}{dx}=\frac{-5}{3\cos y - 2y}$. Since $\cos\pi=-1$, we have $\frac{dy}{dx}=\frac{-5}{3\times(-1)-2\pi}=\frac{5}{3 + 2\pi}$.

Step7: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(\frac{\pi^{2}}{5},\pi)$ and $m=\frac{5}{3 + 2\pi}$.
$y-\pi=\frac{5}{3 + 2\pi}(x-\frac{\pi^{2}}{5})$.
Multiply through by $3 + 2\pi$: $(3 + 2\pi)(y-\pi)=5(x-\frac{\pi^{2}}{5})$.
Expand: $(3 + 2\pi)y-(3 + 2\pi)\pi=5x-\pi^{2}$.
$(3 + 2\pi)y=5x-\pi^{2}+(3 + 2\pi)\pi$.
$(3 + 2\pi)y=5x-\pi^{2}+3\pi+2\pi^{2}$.
$(3 + 2\pi)y=5x+\pi^{2}+3\pi$.
$y=\frac{5}{3 + 2\pi}x+\frac{\pi^{2}+3\pi}{3 + 2\pi}$.

Answer:

a. When $x=\frac{\pi^{2}}{5}$ and $y = \pi$, $3\sin y+5x=\pi^{2}$ and $y^{2}=\pi^{2}$. The answer is A. Yes, because the values found for each side of the equation are equal.
b. The equation of the tangent line is $y=\frac{5}{3 + 2\pi}x+\frac{\pi^{2}+3\pi}{3 + 2\pi}$