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2020-10-18 14:27:57 -07:00
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docs/chapters/inflections/content.en-GB.md
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@@ -29,8 +29,8 @@ Of course, before we do our aligned check, let's see what happens if we compute
\[
\begin{aligned}
& Bézier(t) = x_1(1-t)^3 + 3x_2(1-t)^2t + 3x_3(1-t)t^2 + x_4t^3 \\
& Bézier^\prime(t) = a(1-t)^2 + 2b(1-t)t + ct^2\ \left\{ a=3(x_2-x_1),b=3(x_3-x_2),c=3(x_4-x_3) \right\} \\
& Bézier^{\prime\prime}(t) = u(1-t) + vt\ \left\{ u=2(b-a),v=2(c-b) \right\}\
& Bézier^\prime(t) = a(1-t)^2 + 2b(1-t)t + ct^2~ \left\{ a=3(x_2-x_1),b=3(x_3-x_2),c=3(x_4-x_3) \right\} \\
& Bézier^{\prime\prime}(t) = u(1-t) + vt~\left\{ u=2(b-a),v=2(c-b) \right\}\
\end{aligned}
\]
@@ -75,7 +75,7 @@ That's a lot easier to work with: we see a fair number of terms that we can comp
c = x_2 \cdot y_3 \\
d = x_4 \cdot y_3
\end{matrix}\right\}
\ C(t) = (-3a + 2b + 3c - d)t^2 + (3a - b - 3c)t + (c - a)
~C(t) = (-3a + 2b + 3c - d)t^2 + (3a - b - 3c)t + (c - a)
\]
This is a plain quadratic curve, and we know how to solve *C(t) = 0*; we use the quadratic formula:
@@ -86,7 +86,7 @@ This is a plain quadratic curve, and we know how to solve *C(t) = 0*; we use the
y =& 3a - b - 3c \\
z =& c - a
\end{matrix}\right\}
\ C(t) = 0 \ \Rightarrow\ t = \frac{-y \pm \sqrt{y^2 - 4 x z}}{2x}
~C(t) = 0 ~\Rightarrow~t = \frac{-y \pm \sqrt{y^2 - 4 x z}}{2x}
\]
We can easily compute this value *if* the discriminator isn't a negative number (because we only want real roots, not complex roots), and *if* *x* is not zero, because divisions by zero are rather useless.