Parameterized tangent line to a parameterized curve

Cauchy1789
Messages
43
Reaction score
0

Homework Statement



I seem to remember that a parameterized a(t) curve in \mathbb{R}^3 that one can construct the tangent from the slope of a'(t) and the curve itself.

such that the tangent line L = a(t) + s * a'(t) to a. This is supposedly a straight line in \mathbb{R}^3.
To make a long question. What theorem allows me to construct the tangent in such a way? confused:
 
Physics news on Phys.org
Cauchy1789 said:
I seem to remember that a parameterized a(t) curve in \mathbb{R}^3 that one can construct the tangent from the slope of a'(t) and the curve itself.

such that the tangent line L = a(t) + s * a'(t) to a. This is supposedly a straight line in \mathbb{R}^3.
To make a long question. What theorem allows me to construct the tangent in such a way? confused:

Hi Cauchy1789! :smile:

It's the theorem that says that the slope of the tangent euqals the derivative …

so, for fixed t, L(s) = s * a'(t) + constant …

and the constant has to be a(t) because L(0) = a(t). :wink:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top