Is there an easier way to find the Principle Unit Normal Vector?
October 8, 2009 5:52 PM Subscribe
Does anyone know of a less tedious method for finding the Principle Unit Normal Vector using Calculus?
That is, N(t) = T'(t) / ||T'(t)|| , which is ridiculous, and I want no part of it. I'm hoping for something that doesn't involve finding the magnitude of T'(t). Thanks for the help!
That is, N(t) = T'(t) / ||T'(t)|| , which is ridiculous, and I want no part of it. I'm hoping for something that doesn't involve finding the magnitude of T'(t). Thanks for the help!
I've never heard of a "principle" unit normal vector, but if you want a generalized unit vector usually, you need to divide by the magnitude. So you are probably not going to be happy (although why normalization is so onerous is also puzzling me).
posted by DU at 7:11 PM on October 8, 2009
posted by DU at 7:11 PM on October 8, 2009
Response by poster: Well, the problem is just tedium. If you have a vector in the form of t i + t^2 j + 2t^3 k, for instance, your first step is to find T(t), which is r'(t) / || r'(t) || . 1 / sqrt(1 + 4t^2 + 36t^4) * <> Trivial, so far. But to find N(t), you now have to find T'(t) / || T'(t) ||.
T'(t) = (8t + 144t^3)/(1+4t^2+36t^4)^(3/2) j + (48t^2 + 864t^4)/(1 + 4t^2 + 36t^4)^(3/2) k
To find N(t), all of that is then normalized. I'm not saying it's impossible, I'm just saying it's tedious and cumbersome, and I'm curious if there are any short-cuts to doing it that doesn't involve these enormous nested ratios.>
posted by hishtafel at 7:38 PM on October 8, 2009
T'(t) = (8t + 144t^3)/(1+4t^2+36t^4)^(3/2) j + (48t^2 + 864t^4)/(1 + 4t^2 + 36t^4)^(3/2) k
To find N(t), all of that is then normalized. I'm not saying it's impossible, I'm just saying it's tedious and cumbersome, and I'm curious if there are any short-cuts to doing it that doesn't involve these enormous nested ratios.>
posted by hishtafel at 7:38 PM on October 8, 2009
Response by poster: Hmm, I lost a bit there. 3rd sentence, inside the <> should be <>>>
posted by hishtafel at 7:41 PM on October 8, 2009
posted by hishtafel at 7:41 PM on October 8, 2009
Response by poster: Hah, it's changing it to a tag. It should be ( i + 2t j + 6t^2 k )
posted by hishtafel at 7:42 PM on October 8, 2009
posted by hishtafel at 7:42 PM on October 8, 2009
Best answer: The answer is, I'm pretty sure, no.
Notice that if you did find an "easier" way, you could compute the norm by simply dividing any nonzero component of N by the corresponding component of T' — so what you're asking for is really equivalent to finding an easier way to compute the norm.
posted by hattifattener at 7:55 PM on October 8, 2009
Notice that if you did find an "easier" way, you could compute the norm by simply dividing any nonzero component of N by the corresponding component of T' — so what you're asking for is really equivalent to finding an easier way to compute the norm.
posted by hattifattener at 7:55 PM on October 8, 2009
Best answer: I don't think there's a way around it. Mathematica or Maple or Matlab or some other algebra-manipulation software will help.
posted by chicago2penn at 8:58 PM on October 8, 2009
posted by chicago2penn at 8:58 PM on October 8, 2009
Here is a link explaining how to do it (is this your book?). Also, don't use Maple or Mathematica until you understand it. Yeah, it's boring, but you're probably going to have to regurgitate it on your exam.
posted by vas deference at 2:25 PM on October 9, 2009
posted by vas deference at 2:25 PM on October 9, 2009
This thread is closed to new comments.
posted by leahwrenn at 7:08 PM on October 8, 2009