Question

AP CALC 98 POINTS!!!!!!

Answer 1

Looking at this question again, I don't understand why you're told "for y=11". That doesn't seem relevant at all... So you can disregard the answer I posted a few minutes ago on your other question.

a) With $x^2=-2+y+5\cos y$, differentiate both sides with respect to $x$ to get

$2x=\dfrac{\mathrm dy}{\mathrm dx}-5\sin y\dfrac{\mathrm dy}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x}{1-5\sin y}$

b) The point P occurs at $x=2$, which corresponds to a $y$-coordinate of

$4=-2+y+5\cos y\implies y\approx4.928$

The slope of the line tangent to this point is approximately

$\dfrac{\mathrm dy}{\mathrm dx}\approx\dfrac{2(2)}{1-5\sin(4.928)}\approx0.68$

so the equation of the tangent line is approximately

$y-4.928=0.68(x-2)\implies y=0.68x+3.57$

c) The tangent line to the graphed curve is vertical when $\dfrac{\mathrm dy}{\mathrm dx}$ is undefined. This happens when $1-5\sin y=0$, or $y=\pi-\sin^{-1}\dfrac15+2n\pi$ and $y=\sin^{-1}\dfrac15+2n\pi$ where $n$ is any integer.

In case you're not sure where the general solution came from: We have

$\sin y=\dfrac15$

which has an infinite number of solutions. $\sin^{-1}\dfrac15$ is one of them, which we obtain by taking the inverse sine of both sides of this equation. Since $\sin(\pi-x)=\sin x$, we also know that $\pi-\sin^{-1}\dfrac15$ is a solution. And since $\sin(x+2n\pi)=\sin x$ for integers $n$, we also know that we can add any multiple of $2\pi$ to these two solutions to get infinitely more solutions.