1. ## Calculus help needed

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)
since the lines pass through (-2,-9), we have by the point-slope form:

$\displaystyle y + 9 = m(x + 2)$

$\displaystyle \Rightarrow y = mx + (2m - 9)$ ....................(1)

the lines will be of the above form. it is left for us to now find m

since the lines touch the parabola $\displaystyle y = x^2 + 3$, we have:

$\displaystyle mx + (2m - 9) = x^2 + 3$

$\displaystyle \Rightarrow x^2 - mx + (12 - 2m) = 0$

$\displaystyle \Rightarrow x = \frac {m \pm \sqrt{m^2 + 8m - 48}}2$

now, the thing about tangent lines, is that they touch the curve at one point, that is, we need to ensure that the above quadratic has one solution. to ensure this, we need the discriminant to be zero, hence we must solve

$\displaystyle m^2 + 8m - 48 = 0$

$\displaystyle \Rightarrow \boxed{m = -12}$ and $\displaystyle \boxed{m = 4}$

plugging these values for m into (1), we find that the lines are:

$\displaystyle \boxed{y = -12x - 33}$ and $\displaystyle \boxed{y = 4x - 1}$

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
by implicit differentiation, we have
$\displaystyle y' + \cos y ~y' = 1$

$\displaystyle \Rightarrow y'(1 + \cos y) = 1$

$\displaystyle \Rightarrow y' = \frac 1{1 + \cos y}$

b) Write an equation for each vertical tangent to the curve.
we vertical tangent lines, that is, tangent lines with infinite slope, if the derivative is undefined.

thus we must solve

$\displaystyle 1 + \cos y = 0$

$\displaystyle \Rightarrow \cos y = -1$

$\displaystyle \Rightarrow y = (2n + 1) \pi$ for $\displaystyle n$ an integer

plugging this into our original equation, we have:

$\displaystyle (2n + 1) \pi + \sin (2n + 1) \pi = x + 1$

$\displaystyle \Rightarrow x = (2n + 1) \pi - 1$ for $\displaystyle n$ an integer

these are our vertical tangent lines

c) Find d^2y/dx^2 in terms of y
$\displaystyle y' = \frac 1{1 + \cos y}$

by the quotient rule:

$\displaystyle y'' = \frac {\sin y ~y'}{(1 + \cos y)^2}$

and you can plug in the expression for y' if you wish