How to Find the Vector Equation of a Tangent Line to a Curve at a Given Point

Click For Summary

Homework Help Overview

The problem involves finding the vector equation of a tangent line to a given curve in space at a specific point. The curve is defined by a vector equation involving a parameter \( u \), and the point of interest is (0, 0, 1).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to show work for assistance and explore different methods for finding the tangent line. One participant suggests using the gradient, while another questions the validity of directly substituting the point into the equation without first determining the appropriate parameter \( u \).

Discussion Status

The discussion is ongoing, with participants exploring various approaches and questioning assumptions about the method to find the tangent line. There is no explicit consensus yet, but the dialogue indicates a productive examination of the problem.

Contextual Notes

Participants note the importance of determining the correct value of \( u \) that corresponds to the point (0, 0, 1) before proceeding with further calculations.

ElDavidas
Messages
78
Reaction score
0
Can anybody help me out with this Q?

"A curve R in space has vector equation:

[tex]x = (sin(\pi u), u^2 - 1, u^2 + 3u + 3)[/tex]

u is a real number. Find a vector equation of the tangent line to R at the point (0, 0, 1)"
 
Physics news on Phys.org
What are your ideas or thoughts on how to solve this problem? You need to show some work to get help.
 
Well, I originally thought of taking the gradient of x and then plugging in the values of (0,0,1). Not sure if this is the right way to go about it though.
 
The derivative of a curve at a point is indeed parallel to the tangent line (if the curve has nonzero speed at the point and a tangent line). But you can't just plug in (0, 0, 1) because you only have 1 variable-u. How do you find u such that x(u) = (0, 0, 1)?
 

Similar threads

Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K