[chem] optical & geometric isomers

diff --git a/spec/calculus.md b/spec/calculus.md
index c6818d4a70fc710c8ac1f5a7ef5e08173d0b718e..97bc926a7cb6b59e66ec2eba6c634ac742cb6028 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
 
 $${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
 
@@ -396,3 +394,17 @@ 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$$