Mechanics: Displacement Vector

  • Thread starter teme92
  • Start date
  • #1
185
2

Homework Statement



Consider the vector r(t)= (1-t)i + (3+2t)j + (t-4)k which depends on scalar parameter t. For which t0 is the length of r(t0) minimal?

Homework Equations



Equations of motion

The Attempt at a Solution



So I approached this question by saying the length would be minimal when the magnitude is minimal ie. |r|=0.

|r|=√[(1-t)2+(3+2t)2+(t-4)2]

Simplified down I got 3t2+t+13=0

I tried using the quadratic formula to solve this then but couldn't as the value under the square root was negative.

Am I approaching this question wrong then? Any help would be much appreciated.
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement



Consider the vector r(t)= (1-t)i + (3+2t)j + (t-4)k which depends on scalar parameter t. For which t0 is the length of r(t0) minimal?

Homework Equations



Equations of motion

The Attempt at a Solution



So I approached this question by saying the length would be minimal when the magnitude is minimal ie. |r|=0.

|r|=√[(1-t)2+(3+2t)2+(t-4)2]

Simplified down I got 3t2+t+13=0

I tried using the quadratic formula to solve this then but couldn't as the value under the square root was negative.

Am I approaching this question wrong then? Any help would be much appreciated.

Why would you assume that the smallest value of |r| is zero? That would be the case only if the line passes through the origin. Draw a sketch of a similar situation in two dimensions, and look carefully at the picture.
 
  • #3
185
2
So if t is the x-axis and r(t) the y-axis when t=0 r(t) is at a minimum?
 
  • #4
185
2
Actually is it when the x-compenent is zero. So in this case 1-t=0 so t=1?
 
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,093
1,672
##\vec{r}(t)## describes a curve in three dimensions, and ##t## is a parameter telling you where you are on that curve. To sketch the curve, you need to do so in three dimensions. Ray is suggesting you try a simpler example where the curve lies in two dimensions so you can get an idea of how to analyze the problem correctly. For instance, you could try a curve like ##\vec{r}(t) = (t-1)\hat{i} + (t+2)\hat{j}##.

Back to the original problem… You found the length ##\| \vec{r}(t) \| = \sqrt{2(3t^2+t+13)}## as a function of ##t##. You want to find where this function attains a minimum. How do you do that?
 
  • #6
185
2
Differentiate and let it equal to zero. Then sub t values back into r(t)?
 
  • #7
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,093
1,672
Right. The problem is only asking for the value of ##t##, so you don't need to plug the value back into ##\vec{r}(t)##, though you might try it to see if the answer is reasonable.
 
  • #8
185
2
Ok vela thanks for the help. I got t=1/3 in the end. There was chain rule involved in differentiating so it would take a while to type it out. Thanks for the help again.
 
  • #9
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,093
1,672
You're off by a factor of 2.
 
  • #10
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Differentiate and let it equal to zero. Then sub t values back into r(t)?

Even easier: ##|r|## is a minimum if and only if ##r^2## is a minimum (why?), so you might as well just minimize ##r^2 = (1-t)^2 + (3+2t)^2 + (t-4)^2 = 6 t^2 + 2t + 26.##
 

Related Threads on Mechanics: Displacement Vector

Replies
3
Views
2K
  • Last Post
Replies
18
Views
7K
  • Last Post
Replies
1
Views
4K
Replies
1
Views
1K
Replies
1
Views
3K
Replies
2
Views
11K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
15
Views
5K
Top