How to sketch a curve given its intrinsic equation I'm wondering what the best method of sketching a curve is, if you know the coordinates of a point on the line and  its intrinsic equation in the form:
$$
s=f(\psi)
$$
where s is arc length from the origin to a point on the curve and where $$tan(\psi )=\frac{dy}{dx}$$
When attempting a problem I tried to convert the equation to cartesian following a method online but I ended up getting an integral which is unsolvable with my level of maths
I'm wondering if there is another method of sketching this curve.
Thank you for any replies
 A: Graphically, you can start from the solution of $f(\psi)=0$ and draw a line segment in the direction of $\psi$. Then take a small increment, compute $s'=f(\psi+\delta)$ and mark the point at distance $s'$ on the line segment. Draw the line segment in the direction of $\phi+\delta$, compute $s''=f(\phi+2\delta)$ and mark the point at distance $s''-s'$...
Iterate at will. This is in fact a graphical resolution of the differential equation.
If you find that too tedious ($\delta$ should be tiny), use a numerical solver such as RK4.

Improved graphical method:
Rather than working with line segments, you can use circular arcs. When you give $\phi$ an increment, the corresponding increment of $s$ allows you to compute the radius of the arc, $$r=\dfrac{\Delta s}{\Delta\psi}.$$
A: Maybe you can invert the given equation $s=f(\psi)$ to an explicit equation of the form
$$\psi=\psi(s)\ .$$
The curve has an arc length parametrization of the form
$$s\mapsto\bigl(x(s),y(s)\bigr)\ ,$$
and one has $$\dot x(s)=\cos\psi(s),\qquad\dot y(s)=\sin\psi(s)\ .$$
This allows to find $s\mapsto\bigl(x(s),y(s)\bigr)$ explicitly or numerically.
A: From the differential equation given, we can tell that the curve (provided we assume it is continuous) has a slope that varies from $-\infty$ to $+\infty$ as $x$ increases. Thus it is something $\cup$-shaped, like, at least at the endpoints, that is for $\psi=\pmπ/2.$
If you gave the exact form of $f,$ perhaps we might be able to say more. But from the differential equation alone, assuming $y$ is a single-valued function of $x,$ then the above is the picture that appears.
