Angle of Tangent to Radius in Polar Coordinates, proof.

Question

While studying differential equations by George F. Simmons I have encountered this theorem:

Let C be a curve embedded in a plane defined by polar coordinates.Let P be the point at ⟨r,θ⟩. Then the angle ψ made by the tangent to C at P with the radial coordinate is given by: $tanψ=r\frac{dθ}{dr}$

I don't understand why $dr$ is shown as such despite the fact that that $|OQ|-dr$ is not equal to $|OP|$. Isn't $dr$ supposed to be the difference of those two radii?

Answer 1

The diagram shows $\mathrm d\theta$ almost equal to $\theta$ and is therefore a little confusing. In the limit where $r\mathrm d\theta\to 0,$ $|OQ|-\mathrm d r$ is equal to $|OP|.$ That is, the two values approach each other in the limit of infinitesimal $\mathrm d\theta.$