After finishing the page on String Art Math, I decided to reverse the problem and consider families of lines that are tangent to a given curve. Consider a function $$y = f(x)$$ that is continuous and differentiable over a range of $$x$$. Its derivative is $$y' = f'(x)$$ and describes the slope of the curve at every point on the curve. In particular, for some point $$(k,f(k))$$, a line passing through $$(k,f(k))$$ with slope $$f'(k)$$ will be tangent to the curve described by the function. We can use the slope and point to find the equation of the tangent line. It is $$y = f'(k)x + f(k) - kf'(k)$$. We can then use this generic solution to find families of lines tangent to curves described by various common functions.

### Example 1: $$y = x^2$$

In this case, $$f(x) = x^2$$ and $$f'(x) = 2x$$. So for some point $$(k, k^2)$$, the tangent line is $$y = 2kx + k^2 - k(2k)$$ or $$y = 2kx - k^2$$. Several lines tangent to $$y = x^2$$.

### Example 2: $$y = {1 \over x}$$

In this case, $$f(x) = {1 \over x}$$ and $$f'(x) = -{1 \over {x^2}}$$. So for some point $$(k, {1 \over k})$$, the tangent line is $$y = -{1 \over k^2}x + {2 \over k}$$. Several lines tangent to $$y = {1 \over x}$$.

### Example 3: $$y = sin(x)$$.

In this case, $$f(x) = sin(x)$$ and $$f'(x) = cos(x)$$. So for some point $$(k, sin(k))$$, the tangent line is $$y = xcos(k) + sin(k) - kcos(k)$$. Several lines tangent to $$y = sin(x)$$.

A tool at desmos.com allows you to animate the tangent lines. Here, for example, is an animated graph of the tangent to the sine.