1. Sep 26, 2010

### SOHAWONG

Q:Suppose that r1(t) and r2(t) are vector-valued functions in 2-
space. Explain why solving the equation r1(t) = r2(t) may not produce
all of the points where the graphs of these functions intersect.

In usual function,i always find the intersection by f(x)=g(x)

but how come we can't find all the points when these are vector-valued functions

2. Sep 26, 2010

### HallsofIvy

Consider r1(t)= <cos(t)+ 1/2, sin(t)> and r2(t)= <cos(t)- 1/2, sin(t)> . Do you see that r1 gives vectors that rotate around a circle of radius 1 with center at (-1/2, 0) and that r2 gives vectors that rotate around a circle of radius 1 with center at (1/2, 0)?

Those circles intersect at $\left(0,\sqrt{3}/2\right)$ and $\left(0, -\sqrt{3}/2\right)$ but if you set r1 and r2 equal, you get <cos(t)+ 1/2, sin(t)>= <cos(t)- 1/2, sin(t)> so that cos(t)+ 1/2= cos(t)- 1/2 which is not true for any t!

The problem is that the two circles reach the points of intersection at different t. If you think of these as two object moving around circular paths with t as time, then the two paths cross but the objects are at those points of intersection at different times.

You would do better to write <cos(t)+ 1/2, sin(t)>= <cos(s)- 1/2, sin(s)> so that you have two equations, cos(t)+ 1/2= cos(s)- 1/2 and sin(t)= sin(s) for the two unknown numbers. Obviously, s= t will satisfy the second equation but then the first is unsolvable. Since it is true that $sin(\pi/2- x)= sin(x)$ $s= \pi/2- t$ will also work. But then $cos(s)= cos(\pi/2- t)= cos(t)$ and we have the same problem- the two cosines cancel leaving 1/2= -1/2 which is impossible.

It is also true that $sin(x- \pi)= sin(x)$ so we could have $s= \pi- t$ Then cos(t)+ 1/2= cos(s)- 1/2 becomes $cos(t)+ 1/2= cos(\pi- t)- 1/2= -cos(t)- 1/2$ or $2cos(t)= -1$, $cos(t)= -1/2$ so that $t= 2\pi/3$ and $s= \pi- 2\pi/3= \pi/3$. Now we have $<cos(t)+ 1/2, sin(t)>$$= <cos(2\pi/3)+ 1/2, sin(2\pi/3)>$$= <-1/2+ 1/2, -\sqrt{3}/2>= <0, -\sqrt{3}/2>$ and $<cos(s)- 1/2, sin(s)>= <cos(\pi/3)- 1/2, sin(\pi/3)>$$= <1/2- 1/2, +\sqrt{3}/2>= <0, \sqrt{3}/2>$.

Last edited by a moderator: Sep 26, 2010
3. Sep 26, 2010

### SOHAWONG

thx a lot!
so,would it only happen in vector-valued function?