Calculus 3 Questions: Tangent Vectors/Normal Vectors, Arc Length and Curvature?

Original Post

Link

I am looking for some clarification about these two situations.

1. Find the principal unit normal vector to the curve r(t) = 5ti + 4j when t = 3.

I found this answer to be undefined, as T'(t) = 0 and the magnitude of T'(t) = 0, but I'm not sure.

Theorem Needed: N(t) = T'(t) / ||T'(t)||

2. Find the curvature of the curve r(t) = 2cos(π t)i + 3sin(π t)j (π and t are spaced to avoid πt)

I found the magnitude of r'(t) x r''(t) to be 6π^3, but I'm not sure if the magnitude of r'(t) is 3π. My final answer would come out to be 6/27 if these are correct.

Theorem Needed: K = ||T'(t)|| / ||r'(t)|| = ||r'(t) x r''(t)|| / ( ||r'(t)|| )^3

Any input is appreciated. #1 would be about simplifying sqrt( 4π^2sin^2(π t) + 9π^2cos^2(π t) ) and #2 would be about what T'(t) is and if it end up being undefined.

I am mainly wanting to know if I have the correct answers.

Re: Calculus 3 Questions: Tangent Vectors/Normal Vectors, Arc Length and Curvature?

Quote:

Originally Posted by

**drumcorpsguy** Original Post

Link
I am looking for some clarification about these two situations.

1. Find the principal unit normal vector to the curve r(t) = 5ti + 4j when t = 3.

I found this answer to be undefined, as T'(t) = 0 and the magnitude of T'(t) = 0, but I'm not sure.

Theorem Needed: N(t) = T'(t) / ||T'(t)||

The theorem you have is for a curve. This problem gives you a line. The "unit normal" is a vector perpendicular to the line, of length 1. By convention, it is the normal in the "positive" rotation. For this line, which is a horizontal line, that would be j.

Quote:

2. Find the curvature of the curve r(t) = 2cos(π t)i + 3sin(π t)j (π and t are spaced to avoid πt)

I found the magnitude of r'(t) x r''(t) to be 6π^3, but I'm not sure if the magnitude of r'(t) is 3π. My final answer would come out to be 6/27 if these are correct.

Theorem Needed: K = ||T'(t)|| / ||r'(t)|| = ||r'(t) x r''(t)|| / ( ||r'(t)|| )^3

$\displaystyle r'(t)= -2\pi sin(\pi t)i+ 2\pi cos(\pi t)j$. Its length is $\displaystyle \sqrt{4\pi^2 sin^2(\pi t)+ 4\pi^2 cos^2(\pi t)}= 2\pi\sqrt{sin^2(\pi t)+ cos^2(\pi t)}= 2\pi$, not $\displaystyle 3\pi$.

Quote:

Any input is appreciated. #1 would be about simplifying sqrt( 4π^2sin^2(π t) + 9π^2cos^2(π t) ) and #2 would be about what T'(t) is and if it end up being undefined.

I am mainly wanting to know if I have the correct answers.

Re: Calculus 3 Questions: Tangent Vectors/Normal Vectors, Arc Length and Curvature?

The derivative of the unit tangent vector is still 0 though, which would make it undefined.

$\displaystyle r'(t)= -2\pi sin(\pi t)i+ 2\pi cos(\pi t)j$. Its length is $\displaystyle \sqrt{4\pi^2 sin^2(\pi t)+ 4\pi^2 cos^2(\pi t)}= 2\pi\sqrt{sin^2(\pi t)+ cos^2(\pi t)}= 2\pi$, not $\displaystyle 3\pi$.[/QUOTE]

r(t) = 2cos(π t)i + 3sin(π t)j

sin has a 3, not a 2, so the identity does not work. I'm thinking that only the pi^2 can come out from under the radical.

Re: Calculus 3 Questions: Tangent Vectors/Normal Vectors, Arc Length and Curvature?

yep.

1. undefined

2. 3 / [4(1 + (5/4)cos^2(π t))^(3/2)]