# Thread: Motion in the xy-plane

1. ## Motion in the xy-plane

I need help with this problem.. it asks me for the accleration and velocity and i already know to find those i need to take the first and second derivatives. but the part i don't get is "Find an equation in x and y whose graph is the path of the particle."

r(t) = (t+1)i + ((t^2)-1)j

2. Originally Posted by aeubz
I need help with this problem.. it asks me for the accleration and velocity and i already know to find those i need to take the first and second derivatives. but the part i don't get is "Find an equation in x and y whose graph is the path of the particle."

r(t) = (t+1)i + ((t^2)-1)j
The x component of r(t) would be x(t), so $x(t) = t + 1$. Similarly $y(t) = t^2 - 1$. This is a parametric set of equations describing the curve. Do you know how to eliminate the parameter?

-Dan

3. Hello, aeubz!

Find an equation in $x$ and $y$ whose graph is the path of the particle:
. . . $r(t) \;= \;(t+1)i + (t^2-1)j$

We have parametric equations: . $\begin{Bmatrix}x &=& t + 1 & {\color{blue}[1]}\\ y &=& t^2-1 & {\color{blue}[2]} \end{Bmatrix}$
Eliminate the parameter.

Solve [1] for $t\!:\;\;t \:=\:x-1$

Substitute into [2]: . $y \:=\x-1)^2 + 1" alt="y \:=\x-1)^2 + 1" />

Therefore: . $y \;=\;x^2 - 2x + 2$

4. thanks! you mean y = x^2 - 2x , but thank you for the help!

5. Originally Posted by aeubz
thanks! you mean y = x^2 - 2x , but thank you for the help!
No he didn't.
$(x - 1)^2 + 1 = (x^2 - 2x + 1) + 1 = x^2 - 2x + 2$

-Dan