Finding the line tangent to a curve

[randomjibberi]

Homework Statement

Find an equation for the line tangent to the curve at the point defined by the given value of t.
x = 2 cos t
y = 2 sin t
t = pi/4

Homework Equations

sin (pi/4) = sqrt(2)/2
cos (pi/4) = sqrt(2)/2

The Attempt at a Solution

Determining the slope:
[dy/dt]/[dx/dt] = [2 cos t]/[-2 sin t]
= [2 cos (pi/4)]/[-2 sin (pi/4)]
= [2 (sqrt(2)/2)]/[-2 (sqrt(2)/2)]
= [sqrt(2)]/[-sqrt(2)]
= -1
slope = -1

Finding the line:
y = mx+b
m = -1
2 sin t = -1(2 cos t)+ b
2 sin t = -2 cos t + b
2 sin (pi/4) = -2 cos(pi/4) + b
2 [sqrt(2)/2] = -2 [sqrt(2)/2] + b
sqrt(2) = -sqrt(2) + b
b = 2 sqrt(2)

y = -1x + 2 sqrt(2)

The answer the book has says [y = -x + 2]. I'm not sure what I did wrong with the problem.

 

[Dick]

The book is certainly wrong. y=(-x)+2 doesn't work at all.

 

[HallsofIvy]

For one thing, \sqrt{2}\ne -\sqrt{2}+ 2 so that line does not even pass through the point (\sqrt{2},\sqrt{2}).