• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the line tangent to a curve

  • #1

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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
The book is certainly wrong. y=(-x)+2 doesn't work at all.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
955
For one thing, [itex]\sqrt{2}\ne -\sqrt{2}+ 2[/itex] so that line does not even pass through the point [itex](\sqrt{2},\sqrt{2})[/itex].
 

Related Threads on Finding the line tangent to a curve

  • Last Post
Replies
9
Views
2K
Replies
5
Views
3K
Replies
1
Views
4K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
2K
Replies
4
Views
6K
  • Last Post
Replies
2
Views
6K
Replies
3
Views
10K
  • Last Post
Replies
3
Views
907
Top