Determine the curves along the vector field Consider the vector field  $\vec{v} = x \vec\imath + y \vec\jmath$ for $0\le{x}\le{\infty}$ and $0\le{y}\le{\infty}$.
Determine the parametric equations for the curves along which $\vec{v}$ has constant magnitude and for the curves along which $\vec{v}$ has constant direction.
I know how to draw the vector field, and I have already done so for a few points. I am confused on how you would use this vector field to get the parametric equations for curves along the constant magnitude and constant direction. What does it mean to say "along the constant magnitude" and "along the constant direction"? I appreciate any answer I can get. 
 A: Hints: 1) What is the magnitude of $\vec{v}$ at any point $(x,y)$? Once you compute that, it should be clearer what it means for $\vec{v}$ to have constant magnitude. 2) What is the slope of $\vec{v}$ at some point $(x,y)$? Can you use that to attack the second question?
A: a) The magnitude of the field is $\sqrt{x^2+y^2}$. When it is constant, we have 
$$
\sqrt{x^2+y^2} = k
$$
for some constant $k$. In other words, the curves for which the magnitude is constant are circles. Can you find parametric equations of such circles?
b) The curves of the vector field are such that the vectors are tangent to the curves, therefore, they satisfy
\begin{cases}
dx = k\cdot  x \\
dy = k \cdot y
\end{cases}
where $k$ is constant. It follows that for $x,y >0$:
$$k = \frac{dx}{x} = \frac{dy}{y} \quad \Rightarrow \quad \ln x = \ln y+ C \quad \Rightarrow \quad \ln \frac{x}{y} = C \quad \Rightarrow \quad \frac{x}{y} = e^C$$
We have thus shown that the curves are of the form
$$
y= A x
$$
where $A=e^{-C}$ is a positive constant. This is a linear equation, which shows that all curves have constant slope (or direction). 
