So this week we continued taking AP practice TESTS. They were kinda hard, but we’ve learned a lot! After the non-calculator portion Monday, we took the calculator portion on Tuesday. We got them back on Wednesday and worked out the entire two test for 3 days and [for me] it helped to know new tricks and step by step instructions. BIG THANKS goes to JOHN!!!
Okay, well let’s go over some stuff shall we:
related rates:
Steps:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation(s) involving your variables (only have 1 unknown)
5. Take the derivative with respect to TIME!
6. Substitute in the Derivative and solve
limits:
Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.
Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative)
Rule #3 - When the degrees are equal, the limit is the coeffecients.
linierazation:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Second Derivative:
first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
once you have everything plugged in and ready to go you then solve for d^2y/d^2x
Intermediate Value Theorem:
1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
* basically you cannot skip any y value
HOW TO FIND THE EQUATION OF A TANGENT LINE: [I always miss this one or do it wrong for some reason…so I went look up the directions of how to solve them.]
1. take f1(x)
2. plug x in to find your slope m
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
Can anyone explain implicit derivatives when there’s like, 4x-2xy+3y^2….the 2xy kinda gets me off…
Oh, and the questions that give me a graph then I have to find the areas which means I have to break it all up and then put all the areas together…can we go over that?
~ElliE~
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2xy is the product rule.
ReplyDeleteThe 2 is a constant so you can pull it out first then do product rule of the xy.
2 ((x) (dy/dx) + y)
The 2 is distributed to the x dy/dx and the y
2x dy/dx + 2y
The implicit derivative of the whole equation would be:
2x dy/dx + 2y + 6y dy/dx
Then solve for dy/dx: -2y/ 2x + 6y
A 2 goes into each: -y/ x+3y
For the questions with the graphs and you are asked to find the area, break the graph into triangles and rectangles and find the area of each shape. The area of a triangle is 1/2 b X h, and the area of a rectangle w X h.
ReplyDeleteA sample problem is number 5 on the non-calculator AP from two weeks ago. It says the graph of a piecewise linear functions f, for 0 <or=x < or =8, is shown above. What is the value of the integral of f(x) dx on [0,8]?
Divide the graph up.
Above:
The first is a triangle (1/2) (2)(2) = 2
Then a rectangle (2)(2) = 4
Triangle (1/2) (1)(2) = 1
Below:
Triangle (1/2)(1)(2) = 1
Triangle (1/2) (2)(2) = 2
Next, add the areas from the graph above the axis together and the ones from below the axis together.
Above: 2+4+1= 7
Below: 1+2 = 3
Then subtract the area below from the area above.
7-3 = 4
THANKSS!!!
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