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/(y-siny) 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/(1-siny)^2 * 1/(1-siny)

= -cosy/(1-siny)^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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