# Mechanics: Displacement Vector

1. May 18, 2014

### teme92

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Equations of motion

3. 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.

2. May 18, 2014

### Ray Vickson

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. May 18, 2014

### teme92

So if t is the x-axis and r(t) the y-axis when t=0 r(t) is at a minimum?

4. May 18, 2014

### teme92

Actually is it when the x-compenent is zero. So in this case 1-t=0 so t=1?

5. May 18, 2014

### vela

Staff Emeritus
$\vec{r}(t)$ 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. May 18, 2014

### teme92

Differentiate and let it equal to zero. Then sub t values back into r(t)?

7. May 18, 2014

### vela

Staff Emeritus
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. May 18, 2014

### teme92

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. May 18, 2014

### vela

Staff Emeritus
You're off by a factor of 2.

10. May 18, 2014

### Ray Vickson

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