## 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
$${da \over db} \quad \text{(change in } a \text{ with respect to } b)$$
-#### Gradient at a point on parametric curve
+### Gradient at a point on parametric curve
$${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
### 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)$$
+