[spec] start vector calculus

diff --git a/spec/calculus.md b/spec/calculus.md
index 9d9ed36d9d26ef611b9fc4f9d8598ab528949ff8..3ed01c3fa748c39df9f0f503d883f99f11d24975 100644 (file)
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -347,13 +347,11 @@ where $f(x) > g(x)$
 
 ## Length of a curve
 
-$$L = \int^b_a \sqrt{1 + ({dy \over dx})^2} \> dx$$
+$$L = \int^b_a \sqrt{1 + ({dy \over dx})^2} \> dx \quad \text{(Cartesian)}$$
 
-Evaluate on CAS. Or use Interactive $\rightarrow$ Calculation $\rightarrow$ Line $\rightarrow$ `arcLen`.
-
-### Parametric curve
+$$L = \int^b_a \sqrt{{dx \over dt} + ({dy \over dt})^2} \> dt \quad \text{(parametric)}$$
 
-$$l = \int^b_a \sqrt{({dx \over dt})^2 + ({dy \over dt})^2} \> dt$$
+Evaluate on CAS. Or use Interactive $\rightarrow$ Calculation $\rightarrow$ Line $\rightarrow$ `arcLen`.
 
 ## Rates
 
@@ -361,7 +359,7 @@ $$l = \int^b_a \sqrt{({dx \over dt})^2 + ({dy \over dt})^2} \> dt$$
 
 $${da \over db} \quad \text{(change in } a \text{ with respect to } b)$$
 
-#### Gradient at a point on parametric curve
+### Gradient at a point on parametric curve
 
 $${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
 
@@ -384,3 +382,44 @@ If $f$ is continuous on $[a, b]$, then
 $$\int^b_a f(x) \> dx = F(b) - F(a)$$
 
 where $F$ is any antiderivative of $f$
+
+## Differential equations
+
+One or more derivatives
+
+**Order** - highest power inside derivative  
+**Degree** - highest power of highest derivative  
+e.g. ${\left(dy^2 \over d^2 x\right)}^3$: order 2, degree 3
+
+### Verifying solutions
+
+Start with $y=\dots$, and differentiate. Substitute into original equation.
+
+### Function of the dependent variable
+
+If ${dy \over dx}=g(y)$, then ${dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}$. Integrate both sides to solve equation. Only add $c$ on one side. Express $e^c$ as $A$.
+
+### Mixing problems
+
+$$\left({dm \over dt}\right)_\Sigma = \left({dm \over dt}\right)_{\text{in}} - \left({dm \over dt}\)_{\text{out}}$$
+
+### Separation of variables
+
+If ${dy \over dx}=f(x)g(y)$, then:
+
+$$\int f(x) \> dx = \int {1 \over g(y)} \> dy$$
+
+### Using definite integrals to solve DEs
+
+Used for situations where solutions to ${dy \over dx} = f(x)$ is not required.
+
+In some cases, it may not be possible to obtain an exact solution.
+
+Approximate solutions can be found by numerically evaluating a definite integral.
+
+### Using Euler's method to solve a differential equation
+
+$${{f(x+h) - f(x)} \over h } \approx f^\prime (x) \quad \text{for small } h$$
+
+$$\implies f(x+h) \approx f(x) + hf^\prime(x)$$
+