Limits:
If the bottom is 0 don’t assume the limit does not exist right away, first factor and cancel or use your calculator.
Related Rates:
1. Identify all variables and equations.
2. Identify what you are looking for.
3. Make a 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.
For example, The variables x and y are differentiable functions of t and are related by the equation y = 2x^3 - x + 4. When x = 2, dx/dt = -1. Find dy/dt when x = 2.Since everything is given you can skip straight to the derivative.dy/dt=6x^2dx/dt - dx/dtNow plug in all your givens in order to find dy/dt.dy/dt=6(2)^2(-1) - (-1)dy/dt= -23
Volume by disks:
The formula is pi S[r(x)]^2dx
Volume by washers is the same as volume by disks except it has a hole in it. That means it is bounded by two graphs. The formula for volume by washers is Pi S top^2 - bottom^2 dx.
Umm my ap score seems to be dropping every single test, so I need help.
Mainly with ln and e integration and derivatives, related rates, substitution integration, remembering all the area/volume formulas, and stuff I need to figure out on my own. Even when I know what to do and know the way I have to do it I have a lot of trouble applying it and never get the answer. SO HELP.
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Volume is little different because there are two ways to find the volume of a region, depending on the region itself. The two methods are discs and washers.
ReplyDeleteDiscs: (π)int [R(x)]² dx [a,b]
Example:
(π)int √(sinx)² dx [0,π]
(π)int sinx dx [0,π]
(π)(-cosx) [0,π] -cos(π)-(-cos(0))
π(1+1)=2π
Washers: (π)int (Top equation)²-(Bottom equation)² dx [a,b]
Example:
√(x) and (x²)
(π)int (√(x))² - ((x²))² [0,1]
((1/2)x^2) - ((1/5)x^5) 1/2(1)-(1/5)(1) - [1/2(0) -(1/5)(0)]= 3/10
(π)(3/10)= (3π)/10
An easy way to identify ln integration is that the top of the fraction will always be the derivative of the bottom. Your u will be the bottom of the fraction, so you plug u into ln |u|. Hope this helps with ln integration!
ReplyDeletedy/dx e^x = e^x * x'
ReplyDeletewhich often gets confused with:
dy/dx ln(x) = 1/x * x'