Calculus
posted by Penny on .
Consider the curve defined by y + cosy = x +1 for 0 =< y =< 2pi....
a. Find dy/dx in terms of y. *I got 1/(ysiny) but I feel like that's wrong.
b. Write an equation for each vertical tangent to the curve.
c. find d^2y/dx^2 in terms of y.
d. Sketch a graph of the curve using a table of values for 0 =< y =< 2pi.
Table of values? Vertical tangents? Finding the second derivative? Way out of my league. I'd REALLY appreciate the help!!!

Way out of your league? Are you taking calculus? These topics should be familiar to you.
a.
y + cosy = x+1
y'  siny y' = 1
y'(1  siny) = 1
y' = 1/(1  siny)
b. vertical tangents occur where y' is infinite. That means where the denominator is zero.
1  siny = 0 where siny = 1, or y = pi/2
when y=pi/2, 5pi/2, ... (4k+1)pi/2
y + cosy = x+1
(4k+1)pi/2 + 0 = x+1
x = (4k+1)pi/2  1
so, the vertical tangents are the lines x = (4k+1)pi/2  1
c.
y' = 1/(1  siny)
y'' = 1/(1  siny)^2 * (cosy) y'
= cosy/(1siny)^2 * 1/(1siny)
= cosy/(1siny)^3
d. go to wolframalpha and type
plot y + cosy = x + 1
and it will show the serpentine curve, making it clear that there are many vertical tangents.