Differentiate Archimedes's spiral I read that the only problem of differential calculus Archimedes solved was constructing the tangent to his spiral,
$$r = a + b\theta$$
I would like to differentiate it but I don't know much about differentiating polar functions and can't find this particular problem online. Without giving me a full course in differential geometry, how does one calculate the tangent to the curve at $\theta$?
 A: A typical point on the curve is $(x,y)=((a+b\theta)\cos\theta,(a+b\theta)\sin\theta)$.
By the product rule, we have
\begin{align}
dx & = (b\cos\theta-(a+b\theta)\sin\theta)\,d\theta, \\
dy & = (b\sin\theta+(a+b\theta)\cos\theta)\,d\theta.
\end{align}
So the slope at $(x,y)$ is
$$
\frac{dy}{dx} = \frac{b\sin\theta+(a+b\theta)\cos\theta}{b\cos\theta-(a+b\theta)\sin\theta}.
$$
In case $a=0$, this becomes
$$
\frac{\frac y\theta + x}{\frac x \theta - y} = \frac{y+\theta x}{x-\theta y}. \tag{$*$}
$$
I don't actually know whether this last form is useful in drawing the graph.
One can also write $(*)$ as
$$
\frac{\frac y x + \theta}{1 - \theta\frac y x} = \frac{\tan\theta+\theta}{1-\theta\tan\theta} = \frac{\tan\tan\eta+\tan\eta}{1-\tan\eta\tan\tan\eta} = \tan(\eta+\tan\eta)
$$
where $\theta=\tan\eta$.
A: Let $r(\theta)=a+b\theta$ the equation of the Archimedean spiral. 
The cartesian coordinates of a point with polar coordinates $(r,\theta)$ are 
$$\left\{\begin{align}
x(r,\theta)&=r\cos\theta\\
y(r,\theta)&=r\sin\theta
\end{align}\right.
$$
so a point on the spiral has coordinates
$$\left\{\begin{align}
x(\theta)&=r(\theta)\cos\theta=(a+b\theta)\cos\theta\\
y(\theta)&=r(\theta)\sin\theta=(a+b\theta)\sin\theta
\end{align}\right.
$$
Differentiating we have
$$
\left\{\begin{align}
x'(\theta)&=b\cos\theta-(a+b\theta)\sin\theta\\
y'(\theta)&=b\sin\theta+(a+b\theta)\cos\theta
\end{align}\right.
$$
The parametric equation of a line tangent to the spiral at the point $(r(\theta_0),\theta_0)$ is
$$
\left\{\begin{align}
x(\theta)&=x(\theta_0)+x'(\theta_0)(\theta-\theta_0)=x(\theta_0)+[b\cos\theta_0-(a+b\theta_0)\sin\theta_0](\theta-\theta_0)\\
y(\theta)&=y(\theta_0)+y'(\theta_0)(\theta-\theta_0)=x(\theta_0)+[b\sin\theta_0+(a+b\theta_0)\cos\theta_0](\theta-\theta_0)
\end{align}\right.
$$
or in cartesian form
$$
y(\theta)-y(\theta_0)=\frac{y'(\theta_0)}{x'(\theta_0)}[x(\theta)-x(\theta_0)]=\frac{b\sin\theta_0+(a+b\theta_0)\cos\theta_0}{b\cos\theta_0-(a+b\theta_0)\sin\theta_0}[x(\theta)-x(\theta_0)]
$$
where the slope of the line is
$$
\frac{y'(\theta)}{x'(\theta)}=\frac{\operatorname{d}y}{\operatorname{d}x}=\frac{b\sin\theta+(a+b\theta)\cos\theta}{b\cos\theta-(a+b\theta)\sin\theta}=\frac{b\tan\theta+(a+b\theta)}{b-(a+b\theta)\tan\theta}
$$
A: You can parametrize it in the cartesian plane as
$$s(t)=\begin{pmatrix}bt\cos(t-a/b)\\bt\sin(t-a/b)\end{pmatrix}$$
So the direction of the tangent line at a point $s(t)$ is given by
$$s'(t)=\begin{pmatrix}b\cos(t-a/b)-bt\sin(t-a/b)\\b\sin(t-a/b)+bt\cos(t-a/b)\end{pmatrix}$$
A: This is much easier to swallow in complex variables. Let
$$z=(a+b\theta)e^{i\theta}$$
then
$$\dot z=[(a+b\theta)i+b]e^{i\theta}=[(a+b\theta)i+b][\text{cos}\theta + i\text{sin}\theta]$$
from which the previous results for $\dot x$ and $\dot y$ can be readily obtained. However, we don't need that to get the tangent, because it is, in fact, just $\dot z$, with the caveat that you need to express it as an angle, let's call it $\Theta$. Thus,
$$\Theta=\text{tan}^{-1}\frac{\mathfrak{Im}(\dot z)}{\mathfrak{Re}(\dot z)}$$
