# Find the angle between vectors r(t) and F(t)

• Jul 28th 2008, 10:51 AM
chrisduluk
Find the angle between vectors r(t) and F(t)
Hi everyone, i have a midterm exam coming up and i'm looking at problems in the section of the book to help myself prepare. Can anyone answer this question for me? I have attempted to answer this problem but without the answer i don't know if i'm right.

Find the angle theta between vectors r(t) and r'(t) as a function of t. Sketch the graph of theta(t) and find any extrema of theta(t). Find any values of t which the vectors r(t) and r'(t) are orthogonal.

and they give vector r(t) = t^2 ii + t ij

And there are no typos, that's really what the question asks.

I tried using the techniques given in the book but i have NO idea where to go with this problem.

Can someone do this one out for me? It's not a homework, i just need to see it done out step by step so i can follow it, along with the answer.

• Jul 28th 2008, 01:11 PM
Mathstud28
Quote:

Originally Posted by chrisduluk
Hi everyone, i have a midterm exam coming up and i'm looking at problems in the section of the book to help myself prepare. Can anyone answer this question for me? I have attempted to answer this problem but without the answer i don't know if i'm right.

Find the angle theta between vectors r(t) and r'(t) as a function of t. Sketch the graph of theta(t) and find any extrema of theta(t). Find any values of t which the vectors r(t) and r'(t) are orthogonal.

and they give vector r(t) = t^2 ii + t ij

And there are no typos, that's really what the question asks.

I tried using the techniques given in the book but i have NO idea where to go with this problem.

Can someone do this one out for me? It's not a homework, i just need to see it done out step by step so i can follow it, along with the answer.

Just use the fact that

$\displaystyle \theta=\arccos\left(\frac{\vec{r}\cdot\vec{r'}}{|| \vec{r}||||\vec{r'}||}\right)$
• Jul 28th 2008, 01:13 PM
TwistedOne151
Dot product
First, remember that for any two non-zero vectors $\displaystyle \mathbf{a}$ and $\displaystyle \mathbf{b}$,
$\displaystyle \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos\theta$, where θ is the angle between the two vectors. Thus the angle between $\displaystyle \mathbf{r}(t)$ and $\displaystyle \mathbf{r}'(t)$ is
$\displaystyle \theta=\arccos\left(\frac{\mathbf{r}\cdot\mathbf{r }'}{|\mathbf{r}||\mathbf{r}'|}\right)$.
As $\displaystyle \mathbf{r}(t)=t^2\mathbf{i}+t\mathbf{j}$,
$\displaystyle \mathbf{r}'(t)=\frac{d}{dt}\left(t^2\mathbf{i}+t\m athbf{j}\right)$
$\displaystyle =2t\mathbf{i}+\mathbf{j}$
Plugging in,
$\displaystyle \theta(t)=\arccos\left(\frac{(t^2\mathbf{i}+t\math bf{j})\cdot(2t\mathbf{i}+\mathbf{j})}{|t^2\mathbf{ i}+t\mathbf{j}||2t\mathbf{i}+\mathbf{j}|}\right)$
$\displaystyle =\arccos\left(\frac{(t^2\cdot2t)+(t\cdot1)}{\sqrt{ (t^2)^2+(t)^2}\sqrt{(2t)^2+(1)^2}}\right)$
$\displaystyle =\arccos\left(\frac{t^2+1}{\sqrt{t^2+1}\sqrt{4t^2+ 1}}\right)$

You should be able to proceed from there.

--Kevin C.
• Jul 28th 2008, 02:12 PM
chrisduluk
what's arcos? How do i simplify that to get an answer?
What's arccos? And is there a way to simplify that answer?
• Jul 28th 2008, 02:39 PM
Plato
Quote:

Originally Posted by chrisduluk
What's arccos? And is there a way to simplify that answer?

You are kidding! Aren’t you?
If you are not, then you do not have the background necessary to understand the answer to this question.
If that is true, then why are you being asked to do this problem?