Posted by Penny on Tuesday, January 24, 2012 at 6:37pm.
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!!!

Calculus  Steve, Wednesday, January 25, 2012 at 11:58am
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.
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