# Homework Help: Find the angle between vectors r(t) and r'(t) Question?

1. Jul 31, 2008

### chrisduluk

Hi everyone, i have a midterm exam tomorrow night (Thursday night) 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.

1. Find the angle theta between vectors r(t) and r'(t) as a function of t.
2. Sketch the graph of theta(t)
3. Find any extrema of theta(t).
4. 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.

First off, i have NO idea what ii and ij mean, i'm assuming they just mean i and j?

So far i have r(t)= t^2 + t and r'(t)=2t+1 and r(t)*r'(t)=2t^3 +t
and
theta(t) = arccos( (2t^3 +t) / (sqrt(t^4+t^2) * sqrt(4t^2 +1) ))

Can anyone confirm this is right or wrong? And basically from here i have no idea what to do to answer 2-4 of the problem. My teacher said there would be a problem just like this on my exam, so i REALLY need to see this problem done out so i know what to do.

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. I have until Thursday night, so PLEASE help asap, thanks!

Last edited: Jul 31, 2008
2. Jul 31, 2008

### Defennder

You forgot the arc cos, when using the dot product to find angle in between vectors. Parts 2,3 are standard questions regarding theta(t) once you have found it.

For part 4, you just have to use the dot product and solve t for the conditions where the 2 vectors are perpendicular.

3. Jul 31, 2008

### chrisduluk

OHH!!! i'm sorry i did get the arccos i just didnt put it in. even still i am confused how to graph it and find the values i need... Is the function for theta(t) i have even right? (I fixed it with the arccos in the original post)

4. Jul 31, 2008

### chrisduluk

anyone have any idea where to proceed if this is even correct so far? =/

5. Jul 31, 2008

### Defennder

Yes I believe your expression for theta(t) is right. Now the tough part is graphing it.

6. Jul 31, 2008

### chrisduluk

That's what i'm kind of asking, how do i graph this thing to find the extrema?

And for part 4 of the problem, do i need theta(t) at all? I have no idea how to find the values for t when r(t) and r'(t) are ortho. i know i need to do the dot product, but wont i just get theta(t)? I need help here...

7. Jul 31, 2008

### Defennder

There might be a better way of sketching the graph quickly, but if you are presented with a completely alien function and you're told to sketch it, what's the first thing you should do?

And as for part 4, theta(t) is not the same thing as the dot product of r(t) and r'(t).

8. Jul 31, 2008

### chrisduluk

Ordinarily i would graph it in my calculator. To find extrema of theta(t) i would i set the derivative of theta(t) to 0 and solve for t values to find critical points, then i would set the second derivative to 0 and do the same. Is this how i should find the extrema of theta(t)?

9. Jul 31, 2008

### Ben Niehoff

What is the value of $\vec A \cdot \vec B$ when A and B are orthogonal?

10. Jul 31, 2008

### chrisduluk

HUH?? Please someone help me do the rest of this problem, i have an hour and a half before class this is important =/

11. Jul 31, 2008

### Ben Niehoff

I did help. You should think about the answer to my question and how it applies to your problem.

If you won't take the time to think about how something might apply to your question, then maybe you deserve to get your question wrong.

12. Jul 31, 2008

### chrisduluk

My question is what's A and B. Couldn't i just find t when r(t)xr'(t)=0

13. Aug 1, 2008

### Defennder

If that 'x' is supposed to read as the dot product then you're on the right track. Just solve that equation for the values of t you'll need. As for sketching a graph, you just have to use the 1st and 2nd derivative tests for turning points and the intervals between them to see how the graph looks like.