Find the angle between vectors r(t) and F(t) Question?

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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. I honestly don't even know where to begin, what's the ii and ij all about? I've never seen that before buy my teacher says it's important... how is this calculus I in high school?

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.

PLEASE help, i'm desperate. Thanks!
 
anyone?? How do i deal with the "ii" and ij" - ive only ever heard of i and j, i dont know what to do when theyre combined like that
 
That double i and double j are wacky. I have never seen that either.

You are correct. This is more like Calc III, not I. Look under differentiation and integration of vector-valued functions.

Here is what I believe the problem is:

\(\displaystyle r(t)=t^{2}i+tj\)

\(\displaystyle r'(t)=2ti+j\)

\(\displaystyle r(t)\cdot r'(t)=2t^{3}+t\)

\(\displaystyle ||r(t)||=\sqrt{t^{4}+t^{2}}, \;\ ||r'(t)||=\sqrt{4t^{2}+1}\)

\(\displaystyle cos{\theta}=\frac{2t^{3}+t}{\sqrt{t^{4}+t^{2}}\sqrt{4t^{2}+1}}\)

\(\displaystyle {\theta}=cos^{-1}\left(\frac{2t^{3}+t}{\sqrt{t^{4}+t^{2}}\sqrt{4t^{2}+1}}\right)\)

\(\displaystyle {\theta}\neq \frac{\pi}{2} \;\ \forall \;\ t\)

Now, I will leave you finish up.
 
chrisduluk said:
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

that surely is a misprint.

Before invention of tensorial notation - dyadic notation such as ii was in use - but nobody uses those anymore (anyway those are not vectors).

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. I honestly don't even know where to begin, what's the ii and ij all about? I've never seen that before buy my teacher says it's important... how is this calculus I in high school?

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.

PLEASE help, i'm desperate. Thanks!
Click to expand...
 
Yeah i figured that ii and ij just meant i and j.

So how do i find a numerical value for theta? what do i plug in. also how do i find the extrema? I remember setting the 1st deriv to 0 and setting the 2nd deriv to 0, but what will each value give me? And what do i do about finding t so that theyre orthogonal? Thanks guys... i've really never seen this before and have no idea how to do it. My teacher said it's important but wont explain it to me as this point...
 
Read Galactus's reply very carefully.

He almost did the whole problem for you (with the assumption that the problem had extra 'i's).
 
I didn't get that for theta(t)

I got: theta(t) = arccos ((t^2+1) / (sqrt (t^2+1) * sqrt(4t^2+1)))

is this correct? I used theta(t) = r(t)*r'(t) / mag r(t) * mag r'(t)

but what are the first and second derivatives? I tried using my TI89 Titanium but it stated too many arguments? And i never learned how to do by hand...

To find the extrema i must find the values of t where the first deriv is 0. Then the values of t where second deriv is 0.

also, when i graphed theta(t), i got the values:

x=-20 y=59.9
x=-2 y=57.2
x=0 y=0
x=2 y=57.2
x=10 y=59.9

I'm assuming there's a horix asymptote at 60, so how can there be a max? Is my graph incorrect for theta(t)? I'm guessing the min is 0?

Thanks for any more help
 
chrisduluk said:
I didn't get that for theta(t)

I got: theta(t) = arccos ((t^2+1) / (sqrt (t^2+1) * sqrt(4t^2+1)))<<< Show work - like galuctus did - starting from definition of r(t) and r'(t)

is this correct? <<< We would not know - unless you show work

I used theta(t) = r(t)*r'(t) / mag r(t) * mag r'(t)

but what are the first and second derivatives? I tried using my TI89 Titanium but it stated too many arguments? And i never learned how to do by hand...

To find the extrema i must find the values of t where the first deriv is 0. Then the values of t where second deriv is 0.

also, when i graphed theta(t), i got the values:

x=-20 y=59.9
x=-2 y=57.2
x=0 y=0
x=2 y=57.2
x=10 y=59.9

I'm assuming there's a horix asymptote at 60, so how can there be a max? Is my graph incorrect for theta(t)? I'm guessing the min is 0?

Thanks for any more help
Click to expand...
 
r(t)=t^2+t
r'(t)=2t+1

theta(t)= r(t)*r'(t) / mag r(t) * mag r'(t)
theta(t)= arccos((t^2+t)*(2t+1))/It^2+1I*I2t+1I
= arccos((t^2*2t) + (t*1))/(sqrt((t^2)^2 + (t)^2) * sqrt((2t)^2 + 1^2)))
= arccos ((t^2+1)/((sqrt(t^2+1) * sqrt(4t^2+1))

Is this wrong?
 
chrisduluk said:
r(t)=t^2+t <<< remember these are vectors - need to have i & j
r'(t)=2t+1<<< remember these are vectors - need to have i & j

cos(theta(t))= r(t)*r'(t) / mag r(t) * mag r'(t)<<< remember these are vectors - in the numerator, you need to use "dot product" - study galactus's solution
theta(t)= arccos((t^2+t)*(2t+1))/It^2+1I*I2t+1I
= arccos((t^2*2t) + (t*1))/(sqrt((t^2)^2 + (t)^2) * sqrt((2t)^2 + 1^2)))<<<How does that come from above?
= arccos ( (t^2+1) /((sqrt(t^2+1) * sqrt(4t^2+1)) <<<How does that come from above?

Is this wrong? - Yes
Click to expand...
 
I just don't see where the answer a few responses up came from. I don't understand what equation was used and why mine isn't right.
 
I think you may be making this tougher than need be, Chris.

They just want you to graph it and then estimate from that.

I gave r(t) and r'(t) in my first post. They gave you r(t), then r'(t) is easily gotten by just differentiating.

\(\displaystyle cos{\theta}=\frac{r(t)\cdot r'(t)}{||r(t)||\cdot ||r'(t)||}=\frac{2t^{3}+t}{\sqrt{t^{4}+t^{2}}\sqrt{4t^{2}+1}}\)

We can see from the graph that \(\displaystyle {\theta}=.34\approx 19.47\) degrees with a max at \(\displaystyle \frac{1}{2}\)

There is no orthogonality because \(\displaystyle {\theta}\neq \frac{\pi}{2}\) fro any t.
 

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i tried finding theta(t) again and i see where i went wrong with my equations. So i tried graphing the correct theta(t) function but did not get that same graph, it just came out as a horiz line. does the mode need to me in radians or degrees? The arccos, or is it cos^-1 or is is cos? is getting me confused when typing it in.

And as far as finding values of t so that r(t) and r'(t) are ortho, are you sure you did this for r(t) and r'(t) and not theta(t) by accident? I don't know why the question would ask for values if there aren't any, i thought there would always be a pot the graphs would be ortho since one is the derivative of the other.
 
The mode must be in radians. Never use degrees in calc unless instructed to do so.

When \(\displaystyle {\theta}=\frac{\pi}{2}\) cosine equals 0.

So, \(\displaystyle 2t^{3}+t=0\)

\(\displaystyle t=0, \;\ \frac{1}{\sqrt{2}}i, \;\ \frac{-1}{\sqrt{2}}i\)
 
so these are the values for t which r(t) and r'(t) are orthogonal?
 
i finally got this thing to graph, but i also have points in the second quadrant. Wouldnt these points count as max?
 
at t= 0

r(t) = 0 --- in that case what does orthogonality mean?
 
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