okayyy so everytime we have a monday off all my stuff gets messed uppp and an think that monday is sunday, etc. sorrrrrry!
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
Finding the Area and volume under a curve
Int (Top Equation)-(Bottom Equation) on the interval [a,b]
You can find a and b by setting the equations equal to each other and solving because a and b are their intersections.
If the area is on the y then a and b need to be y values and solved for x.
If the area is on the x then a and b need to be x values and solved for y.
Example: Find the area of the rigion bounded by f(x)=2-x^2 and g(x)=x.
2-x^2=x 2-x^2-x=0 (-x-1)(x+2) x=-1 x=2
int (Top)-(Bottom) dx [-1,2]
To find the top and bottom equation just graph them on your graphing calculator. You'll see that 2-x^2 is on top with x on the bottom.
int (2-x^2)-(x) dx [-1,2]
int (2-x^2-x) = 2x-(1/3)x^3-(1/2)x^2
Solve like an ordinary definite integral.
2(2)-(1/3)((2)^3)-(1/2)((2)^2)-[2(-1)-(1/3)((-1)^3)-(1/2)((-1)^2)]=(3/2)
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.
Discs: (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
i need help with substitution and a refresher on optimization would also be nice.
ohhh and i have no idea how to work my calculator if anyone would like to go over some thingggggggssssss =)
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
ReplyDelete1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
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
For your calculator.
ReplyDeleteSay you want to do this problem:
9(integral)5 x^7 - sin(X)sec(X) + 8x^78.
You would plug this into your calculator by:
1. math > fnInt( > enter
2. plug in the equation
fnInt(x^7-sinxsecx+8x^78
3. put your limitation points:
fnInt(x^7-sinxsecx+8x^78,x,5,9)
4. Press enter.
I got 2.45 x 10^74
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
ReplyDelete1. 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
for your calculator:
ReplyDeleteif you need to take an integral. go to math
then fnint(...)
then type in the actual equation. then after that type ,x,bottom#,top#)<--close paranthesis.
if you need to take a derivative, all i know how to do is trig function derivatives. like if you need the derivative of tan(pi/6)
you would go to math, then fnder(...)
type in the equation. BUTTTTT instead of typing in tan(pi/6) you would type in tan(x)
then ,x,pi/6)<--close paranthesis again.
hope i helped a little.
ALSO, if you are not in RADIANS MODE, this won't work. to make sure you are in radians, click mode, then go down and highlight radians, then click enter.
if you need to take an integral .. plug in the problem into the y= and then hit calc...go to the fnint...put in your lower bounds first and then your higher..hit enter and it shows you the answer
ReplyDeleteif you need to take a derivative .. go to math and number 8 .. put in the problem.. then
, X , lower bound , higher bound ]
HOPE THIS HELPS!