From 4c492006f24c2d8629f148886b9a83db70dec3e2 Mon Sep 17 00:00:00 2001
From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Wed, 27 Mar 2019 09:44:24 +1100
Subject: [PATCH] [spec] rates

---
 spec/calculus.md | 12 ++++++++++++
 1 file changed, 12 insertions(+)

diff --git a/spec/calculus.md b/spec/calculus.md
index ee88068..c59c4b7 100644
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -245,3 +245,15 @@ $\int k f(x) dx = k \int f(x) dx$
 
 To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function.
 
+## Rates
+
+### Related rates
+
+$${da \over db} \quad \text{change in } a \text{ with respect to } b$$
+
+#### Gradient at a point on parametric curve
+
+$${dy \over dx} = {{dy \over dt} \over {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
+
+$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \over {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$
+
-- 
2.43.2