Finding the curvature of a curve at specific points

[burt]

Attached is a document showing my steps in finding the curvature of a curve for a vector valued function. I'm sorry I didn't type it out here - it is simply too much for me to do. The problem was:
Find the curvature of the curve [MATH]r(t)=<4\cos(2t),3\sin(2t)>[/MATH] at the points [MATH]t=0 \text{and} t=\frac{\pi}{4}[/MATH]. My work is on the attached document. I feel confident about my answer for t=0 but not for my answer with [MATH]t=\frac{\pi}{4}[/MATH] (that one did not match up with an answer key. Any insight?

 

[Steven G]

Regarding t=pi/4 : Your 2nd equal sign is not valid.

 

[burt]

Your 2nd equal sign is not valid

Why not? Isn't sin(pi/4) equal to one?

 

[burt]

Regarding t=pi/4 : Your 2nd equal sign is not valid.

I went through and checked it again - it 9216 divided by 16 is 9/4. squareroot of that is 3/2. divide that by 8 and get 3/16.
I still don't understand what is wrong.

 

[pka]

I will not give my complete course in vector analysis here.
If \(r(t)\) is a vector curve then curvature is
\(\kappa(t)=\dfrac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3}\).

 

[burt]

I will not give my complete course in vector analysis here.
If \(r(t)\) is a vector curve then curvature is
\(\kappa(t)=\dfrac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3}\).

I thought it was [MATH]\frac{|T'(t)|}{|r'(t)|}[/MATH] - with T(t) being the unit tangent vector
Here is the quote from my book:

 

[pka]

I thought it was [MATH]\frac{|T'(t)|}{|r'(t)|}[/MATH] - with T(t) being the unit tangent vector
Here is the quote from my book: View attachment 17209

Well I said that I would debate the point! Mine is a widely used text. I can only suggest that you look at what \(\|T'(t)\|\) is.

 

[Steven G]

Why not? Isn't sin(pi/4) equal to one?

2nd equal sign: What is to the left of the 2nd equal sign has no sin functions and what is to the right of the 2nd equal sign has no sin functions so I do not understand your question.

 

[Steven G]

BTW, sin (pi/4) in fact is not 1. However, sin(pi/2) does equal 1.

 

[burt]

2nd equal sign. What is to the left of the 2nd equal sign has no sin functions and what is to the right of the 2nd equal sign has no sin functions so I do not understand your question.

Is this the part you are referring to?

 

[burt]

BTW, sin (pi/4) in fact is not 1. However, sin(pi/2) does equal 1

Whoops, meant that - All the fours become twos because of the 2t inside the function.

 

[Steven G]

Just count to the 2nd equal sign. What you have to the left of the equal sign is correct (else I would have said that the 1st equal sign is wrong) and what you have to the right of the equal sign is wrong.

For example: 7+2 = 6+3 = 4 +4. The 2nd equal sign is not valid.

 

[burt]

Just count to the 2nd equal sign. What you have to the left of the equal sign is correct (else I would have said that the 1st equal sign is wrong) and what you have to the right of the equal sign is wrong.

For example: 7+2 = 6+3 = 4 +4. The 2nd equal sign is not valid.

I'm sorry, I'm feeling very slow here. Is the part of the equation I attached last time the part you are referencing? Because it all checks out when I look at it.

 

[Steven G]

sqrt(576)/128 I think is not correct.

EDIT: Looking at it now after getting some sleep I realize that I was wrong and sqrt(576)/128 is correct. The computational work you did for t= pi/4 is correct.

 

[burt]

sqrt(576)/128 I think is not correct.

EDIT: Looking at it now after getting some sleep I realize that I was wrong and sqrt(576)/128 is correct. The computational work you did for t= pi/4 is correct.

Thank you for all your help!