Constructing tangent to a curve in $\mathbb{R}^2$

Question

I've been studying Basic Mathematics for Physics courses.

While teaching about derivatives my prof. said that there are actually two points the tangent at a point passes through (and those points are almost coincident). Symbolically $x_0$ and $x_0+\mathrm dx$. Clearly this sounds absurd because a tangent by definition touches any curve only at a single point. But also, this puts forth the discrepancy that at least two points are required to construct a line.

I thought that if we somehow know the curvature of the curve we would indeed be able to construct a tangent using only a single point by using definition of curvature, $\kappa=1/r$ and follow as we do in the case of a circle.

But how to actually measure the curvature of curve using only its derivative? Feel free to present a model of constructing a tangent this way or any other that you may find relevant and fitting into intermediate-to-advanced calculus courses.

Answer 1

Your intuition is wrong. Knowing the curvature does not help you to construct the tangent. Just think that circles of any radius can be tangent to a given straight line.

And computing the curvature requires both the first and second derivative.

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The intuition behind the two-points construction is that the chord by the target point and an extra point will tend to the tangent when the two points get closer.

Answer 2

Consider the differential quotient $$f'(x_0) = \lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}.$$ Here $f'(x_0)$ is the slope of the tangent line to $f$ at $x_0$ with $x_0$ given.

For this quotient, the tangent line is $f(x)-f(x_0) = f'(x_0)(x-x_0)$ or better $f(x) = f'(x_0)(x-x_0)+f(x_0)$