Post a New Question

Calculus

posted by 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/(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 - ,

    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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question