Component functions and coordinates of linear transformation

raghad
Messages
5
Reaction score
0
Let A(a, b, c) and A'(a′,b′,c′) be two distinct points in R3. Let f from [0 , 1] to R3 be defined by f(t) = (1 -t) A + t A'. Instead of calling the component functions of f ,(f1, f2, f3) let us simply write f = (x, y, z). Express x; y; z in terms of the coordinates of A and A, and t. I thought that a is the partial derivative of f1 with respect to x, b is the partial derivative of f2 with respect to y, and c is the partial derivative of f3 with respect to z. I am right? Any hint how to find the relation and to find the derivative of f?
 
Last edited:
Physics news on Phys.org
raghad said:
Let A(a, b, c) and A'(a′,b′,c′) be two distinct points in R3. Let f from [0 , 1] to R3 be defined by f(t) = (1 -t) A + t A'. Instead of calling the component functions of f ,(f1, f2, f3) let us simply write f = (x, y, z). Express x; y; z in terms of the coordinates of A and A, and t. I thought that a is the partial derivative of f1 with respect to x, b is the partial derivative of f2 with respect to y, and c is the partial derivative of f3 with respect to z. I am right? Any hint how to find the relation and to find the derivative of f?

I don't see why you think partial derivatives come into this. ##(1-t)A=(1-t)(a,b,c)=((1-t)a,(1-t)b,(1-t)c)##. Now do something similar for ##tA'## and just add them.
 
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