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 $\theta$?

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
$$
\frac{y'(\theta)}{x'(\theta)}=\frac{\operatorname{d}y}{\operatorname{d}x}=\frac{b\sin\theta+(a+b\theta)\cos\theta}{b\cos\theta-(a+b\theta)\sin\theta}=\frac{b\tan\theta+(a+b\theta)}{b-(a+b\theta)\tan\theta}
$$
