Wind chart for vector calculus If a particle starts its motion at the origin of a Cartesian coordinate
system under the influence of a force $F(t) = (1, x(t))$ where $t \ge 0$, find
an equation of the path it follows.
This seems like a wind chart question, my teacher didn't specify. I've never found an equation for a wind chart question I just drew charts so I'm confused. Any suggestions?
 A: If you know the force on a particle, then using $F=ma$ you know the acceleration. From there, you can integrate in order to find the position. The force is given in each direction, so we can solve for each component separately.
$$ \frac{d^2x}{dt^2} = \frac{1}{m}$$
$$ \frac{d^2y}{dt^2} = \frac{x(t)}{m}$$
We have no idea what $x(t)$ is, so we can't directly compute the $y$ components yet. But we can easily integrate the $x$ component to find position. The particle starts it's motion at $t=0$ and begins at the origin which gives enough information to calculate the constants.
$$ \frac{dx}{dt} = \frac{t}{m} + C_1, \quad C_1=0$$ 
$$ x(t) = \frac{t^2}{2m} + C_2 \quad C_2 = 0$$
Now we know $x(t)$, so we can solve the y part of the equations by integrating the same way.
$$ \frac{dy}{dt} = \frac{t^3}{6m^2} + C_3 \quad C_3 = 0$$
$$ y(t) = \frac{t^4}{24m^2} + C_4 \quad C_4 = 0$$
All the constants turn out to be zero because the particle starts at zero velocity at the origin. This made the calculations very straightforward. The more constants you have to carry through, the more complicated finding new constants will be.
A: I'll assume $x(t)$ is the x-coordinate of the particle at time $t$.
Let the particle's position at time $t$ be $p(t) = (x(t),y(t))$.  Then the particle's velocity is $v(t) = (x'(t),y'(t))$, and the particle's acceleration is $a(t) = (x''(t),y''(t))$
Let $m$ be the mass of the particle.  Then, using $F = ma$, we see that
\begin{equation}
(1,x(t)) = m (x''(t),y''(t)).
\end{equation}
In other words,
\begin{align*}
x''(t) &= \frac{1}{m} , \\
y''(t) &= \frac{1}{m} x(t).
\end{align*}
We can solve the first equation to find a formula for $x(t)$.
Then, we can plug this formula into the second equation,
and then find a formula for $y(t)$.
