Tangent lines were on there too and the steps for finding the tangent line are:
1. Take the derivative of the equation like normal
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
absolute maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Example:
Integrate (x^2+1) (2x) dx
Since you can't integrate product rule, you know you have to use substitution.
u=x^2 + 1 du= 2x dx
Integrate: u du
1/2 u ^2
Plug u back in
1/2 ( x^2+1)^2 + c
Example 2:
Integrate x (x^2+1)^2 dx
u= x^2 +1 du= 2x
Since you are missing a 2 in the problem, you have to add a 1/2 to get rid of the 2
1/2 S u^2 du
1/2 (1/3) u^3
1/6 (x^2 +1) ^3 + c
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
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
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation 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 derivative and solve.
i am still having problems with the particle equations on our AP's so if someone could work one of those for me that would be suuppper helpful!!!!!!
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the way to look at particle problems is like this in a list
ReplyDeletepostion
velocity
acceleration
When moving from the top to the bottom take the derivative, and when moving from the down to up integrate it.
If the problem asks for velocity when starting with postion take the derivative, or when going from postion to acceleration take the second derivative.
Or if its the other way around it works the same with integration.