# Thread: Calc test tomorrow.. need help studying

1. ## Calc test tomorrow.. need help studying

I have a calc test tomorrow, and as I was studying, there were some problems I couldn't figure out. Can anyone help me with these, please..?

1) Find the equations of the two tangent lines to the graph of $y=(x^2)+3$ that pass through the point (-2,-9)

2) Consider the curve defined by the equation $y+siny=x+1$ for 0 less than or equal to y less than or equal to 2pi

a) find dy/dx in terms of y

b) Write an equation for each vertical tangent to the curve.

c) Find d^2y/dx^2 in terms of y

thanks so much for the help! i really need it, hehe:]

I have a calc test tomorrow, and as I was studying, there were some problems I couldn't figure out. Can anyone help me with these, please..?

1) Find the equations of the two tangent lines to the graph of $y=(x^2)+3$ that pass through the point (-2,-9)

2) Consider the curve defined by the equation $y+siny=x+1$ for 0 less than or equal to y less than or equal to 2pi

a) find dy/dx in terms of y

b) Write an equation for each vertical tangent to the curve.

c) Find d^2y/dx^2 in terms of y

thanks so much for the help! i really need it, hehe:]
i already responded to this here. do not double posts. it wastes people's time. in fact, i think i see Soroban working on this right now when he could be helping someone else

1) Find the equations of the two tangent lines to the graph of $y\:=\:x^2+3$
that pass through the point $P(-2,\,-9)$

Let the point of tangency be $Q(p,\,p^2+3)$

The slope of the tangent is: . $y \:=\:2x$
At $Q(p,\,p^2+3)$, the slope is: . $m \:=\:2p$

We want the slope of $PQ$ to equal $2p$.

The slope of $PQ$ is: . $\frac{(p^2+3) - (-9)}{p - (-2)} \;=\;\frac{p^2+12}{p+2}$

Then we have: . $\frac{p^2+12}{p+2} \:=\:2p\quad\Rightarrow\quad p^2 + 4p - 12\:=\:0$

. . which factors: . $(p-2)(p+6) \:=\:0$

. . and has roots: . $p \:=\:2,\,-6$

The corresponding y-values are: . $7,\:39$

We have two points of tangency: . $Q(2,\:7),\;\;R(-6,\:39)$

. . At $Q(2,\:7)$, the slope is: . $m = 4$

. . At $R(-6,\:39)$, the slope is: . $m = -12$

You can write the equations of the tangent lines now . . .

2) Consider the curve defined by the equation $y+\sin y=\:x\:+1$ for $0 \leq y \leq 2\pi$

a) Find $\frac{dy}{dx}$ in terms of $y$

b) Write an equation for each vertical tangent to the curve.

c) Find $\frac{d^2\!y}{dx^2}$ in terms of $y$

a) Differentiate implicitly: . $\frac{dy}{dx} + \cos y\cdot\frac{dy}{dx} \:=\:1$

Then: . $\frac{dy}{dx}(1 + \cos y) \:=\:1\quad\Rightarrow\quad\boxed{\frac{dy}{dx} \:=\:\frac{1}{1 + \cos y}}$

b) For a vertical tangent, the slope is undefined; its denominator is zero.

So we have: . $1 + \cos y \:=\:0\quad\Rightarrow\quad\cos y \:=\:-1\quad\Rightarrow\quad y \:=\:\pi$

Substitute into the equation: . $\pi + \sin\pi \:=\:x + 1\quad\Rightarrow\quad\boxed{ x \:=\:\pi-1}$

c) We have: . $\frac{dy}{dx}\;=\;(1 + \cos y)^{-1}$

Then: . $\frac{d^2\!y}{dx^2} \;\;=\;\;-(1+\cos y)^{-2}(-\sin y)\cdot\frac{dy}{dx} \;\;=\;\;\frac{\sin y}{(1+\cos y)^2}\cdot\frac{dy}{dx}
$

Since $\frac{dy}{dx}\,=\,\frac{1}{1+\cos y}$, we have: . $\frac{d^2\!y}{dx^2} \;\;=\;\;\frac{\sin y}{(1+\cos y)^2}\cdot\frac{1}{1+\cos y}$

Therefore: . $\boxed{\frac{d^2\!y}{dx^2} \;\;=\;\;\frac{\sin y}{(1+\cos y)^3}}$