1. ## Parametric representation/Lagrange

Let p0, p1, and p2 be points in the plane.

a) Specify a parametric representation of a planar curve C(t)
that satisfies C(t0) = p0, C(t1) = p1, and C(t2) = p2.
The components of C should be quadratic polynomials in t.

having alot of trouble with this. Im not quite sure where to begin.

2. If you set the x coordinate to $at^2+bt+c$, and substitute the x-coordinate part of each of your three conditions, you get three equations of the form $at_i^2+bt_i+c=\text{x coordinate of }p_i$. Since the $t_i$ and $p_i$ are known values, you have 3 equations in the three unknowns a, b, and c. Then you do the same thing for the y coordinate.

You might find this simpler than solving 3 equations in 3 unknowns (twice):
Lagrange polynomial - Wikipedia, the free encyclopedia

Post again in this thread if you're still having trouble.

- Hollywood

3. Originally Posted by hollywood
If you set the x coordinate to $at^2+bt+c$, and substitute the x-coordinate part of each of your three conditions, you get three equations of the form $at_i^2+bt_i+c=\text{x coordinate of }p_i$. Since the $t_i$ and $p_i$ are known values, you have 3 equations in the three unknowns a, b, and c. Then you do the same thing for the y coordinate.

You might find this simpler than solving 3 equations in 3 unknowns (twice):
Lagrange polynomial - Wikipedia, the free encyclopedia

Post again in this thread if you're still having trouble.

- Hollywood
yes im having trouble setting this up.
i got help for another question in this thread.

http://www.mathhelpforum.com/math-he...rpolation.html

is it similar?

is it:

x(t) = C0(t)x0 + C1(t)x1 + C2(t)x2
y(t) = C0(t)y0 + C1(t)y1 + C2(t)y2

or am i way off?

4. It is the same as the other thread, except you have both x and y to solve, and you have variables instead of numbers.

Let $p_0=(x_0,y_0)\text{, }p_1=(x_1,y_1)\text{, and }p_2=(x_2,y_2)$.

Also, let $C(t)=(x(t),y(t))$.

So we are looking for:

$x(t)\text{ satisfying }x(t_0)=x_0\text{, }x(t_1)=x_1\text{, and }x(t_2)=x_2$, and

$y(t)\text{ satisfying }y(t_0)=y_0\text{, }y(t_1)=y_1\text{, and }y(t_2)=y_2$.

Each of these is solved by the same process as the other thread. For the x-coordinate:

$x(t)=x_0\ \frac{t-t_1}{t_0-t_1}\ \frac{t-t_2}{t_0-t_2}+x_1\ \frac{t-t_0}{t_1-t_0}\ \frac{t-t_2}{t_1-t_2}+x_2\ \frac{t-t_0}{t_2-t_0}\ \frac{t-t_1}{t_2-t_1}$

and of course the y-coordinate is similar.

- Hollywood

5. Originally Posted by hollywood
It is the same as the other thread, except you have both x and y to solve, and you have variables instead of numbers.

Let $p_0=(x_0,y_0)\text{, }p_1=(x_1,y_1)\text{, and }p_2=(x_2,y_2)$.

Also, let $C(t)=(x(t),y(t))$.

So we are looking for:

$x(t)\text{ satisfying }x(t_0)=x_0\text{, }x(t_1)=x_1\text{, and }x(t_2)=x_2$, and

$y(t)\text{ satisfying }y(t_0)=y_0\text{, }y(t_1)=y_1\text{, and }y(t_2)=y_2$.

Each of these is solved by the same process as the other thread. For the x-coordinate:

$x(t)=x_0\ \frac{t-t_1}{t_0-t_1}\ \frac{t-t_2}{t_0-t_2}+x_1\ \frac{t-t_0}{t_1-t_0}\ \frac{t-t_2}{t_1-t_2}+x_2\ \frac{t-t_0}{t_2-t_0}\ \frac{t-t_1}{t_2-t_1}$

and of course the y-coordinate is similar.

- Hollywood
oh ok so by doing it for the x and for the y ill get $C(t)=(x(t),y(t))$?

where x(t) and y(t) will be a function?

6. Yes, that's correct.

- Hollywood