post 27

here we go again.

implicit derivatives.


1. Take the derivative of both sides (implicit derivatives usually have an equals sign)

2. When you take a derivative of a y term, state that. Do so by putting y prime or dy/dx. Like ( taking the derivative of y^2 would be 2y(dy/dx) )

3. Move all of the terms that don't have a dy/dx in them to one side. Factor out a dy/dx out of all the terms that do have it, then divide to finish solving for dy/dx.

First Derivative Test:

1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)

Second Derivative Test:

1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph)

RELATED RATES:
1. identify all variables and equations
2. identify what you are looking for
3. sketch and label
4. write an equations involving your variables (you can only have one unknown so a secondary equation may be given)
5. take the derivative (with respect to time)
6. substitute in derivative and solve

Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers

i dont really understand the particle and bacteria problems. yeerd.