# 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 $\displaystyle y=(x^2)+3$ that pass through the point (-2,-9)

2) Consider the curve defined by the equation $\displaystyle y+siny=x+1$ 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 $\displaystyle y=(x^2)+3$ that pass through the point (-2,-9)

2) Consider the curve defined by the equation $\displaystyle y+siny=x+1$ 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 $\displaystyle y\:=\:x^2+3$
that pass through the point $\displaystyle P(-2,\,-9)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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