Lesson 6.1 - 6.5 exluding 6.4

Polar plane

A system with point O, the pole, and a ray from O.

Ex write parameter equation of the line through (1,2) with slope (-5/4)

Answers are x = 1 + 4t and y = 2 - 5t / x = 1 - 4t and y = 2 + 5t / etc more options

Cardioid

If the value of a = the value of b then its a limacon

Limacon

If the value of a = the value of b then its a limacon

Ex: eliminate parameter, stretch and indicate orientation.

In the example x = 2cost and y = 2sint then solve for sin/cos then use the formula sin squared plus cos squared = 1, then the equation becomes x^2 + y^2 = 4

What is rectangular form?

basically just normal form

Graphing Parametric equation using table.

Create a table with rows t, x , y. If you have a range you just put in those range numbers, but if no range just put in a bunch of random numbers for t, then just plug those numbers into x and y then plug into the table. Then whatever the the x and y coordinates are for 1 t value is where you put the point. To find the orientation just look at the t value and corresponding points to see which way it is going.

Writing polar complex numbers in normal complex numbers.

Ex 8(cis 2pi/3) You know the triangle has to be in the second quadrant because the angle is 2pi/3. Based on that you know its a 30 60 triangle and can find out the side lengths based on r which we know is 8.

Convert polar equation to rectangular equation.

Ex: r= 6cos0, x=rcos0, x/r = cos0, so we can do 6 times x/r = r, times both sides by r to get r^2 = 6x. r^2 = . x^2 + y^2 = 6x. Then you would move the 6x over and solve for the circle.

Graphs of polar equations all other using tables

Examples: r= 1-sin0 or r^2 =25cos20, so to solve for something like r = 1-sin0 we would make a table with angles 0, pi/2, pi, 3pi/2 then simply plug in those values and then find whatever they are then plot them. After this there will be images showing different names of different shapes. For something like r^2 = 25cos20 you would use values 0, pi/4, pi/2, 3pi/2, pi then plug in and solve and it usually becomes something like an infinity sign. When there is the penis dot structure you have to draw a loop.

Converting rectangular coordinates to polar coordinates.

For example (-1,1) to polar coordinates, we know tan = y/x so it becomes -1, we know tan is -1 at 3pi/4 so that is our first value. To find r we use r = Sqrt(1^2 + -1^2) which means r = sqrt(2). So it would be (sqrt(2), 3pi/3) and u can check using the same as above.

Example) x = 1/ sqrt(t + 1) has restricted values of t > -1

Pictures

Solving for roots of complex numbers

Use DeMoivre's Theorum / look at phone pictures. Ex 4th roots of 1, first you put both into polar forms and find out r^4=1, and 40 = 0+2pi k, then divide by 4 to get 0 =0 +pi/2k. You add up until just before 2 pi because that is 1 rotation so it would become 1(cis0), 1cis(pi/2), etc

Multiplying and dividing complex numbers

When multiplying you just multiply the r values and add the values inside the cis. When dividing you divide the first r by the second r and on the inside of the cis subtract the 2nd one from the first one.

DeMoivre's Theorem

[r(cosx+isinx)]^n=r^n(cos(nx)+isin(nx))

Writing complex numbers in polar form.

ex: 6-6i, first draw out the point 6-6i, 6^2 + 6^2 = r, r =6sqrt(2). Based on where the point is drawn you can find out the angle for example this one would be 7pi/4 because its in the 4th quadrant and 6/-6 tan is 1 so it works. So therefore the final would be 6sqrt(2)(cis 7pi/4)

Converting polar coordinates to rectangular coordinates

x= rcos0, y = rsin0, tan0 = y/x, r = sqrt(x^2 + y^2). So for example converting (3, 7pi/6) we know x=rcos0 so it becomes 3cos (7pi/6) which = -3sqrt(3)/2, then substitute in y for 3sin (7pi/6) which equals -3/2. And u can check to see if they end up in the same place when u graph them uk.

Convert rectangular equation to polar equation.

y=4, rsin0 = 4, r = 4/sin0, r =4csc0... For example we have (x-3)^2 + (y-2)^2=13. First we multiply out to get x^2-6x + y^2 - 4y = 13. Whole numbers factor out then we turn x^2 + y^2 into r^2 and subtituion x and y with rcos0 and rsin0. Then we are left like r^2 - 6rcos0 - 4rsin0) then factor out r to get r(r-6cos0-4sin0) we have r=0 and r-6cos0-4sin0=0. r-0 not possible so this is answer. I have pictures on phone too. r-6cos0-4sin0=0

Graphs of polar equations: rose

Example: r = 3cos20, the 3 before the cos is the length of the petals, the number before the angle 2 represents the number of petals for odd numbers that is the exact number of petals and even numbers you times by two. First step is to solve by asymptotes, set equation to 0 = 3cos20 and divide by 3, 0 =cos20 then sort of swap it so it's 20 = cos(0), so it becomes 20= pi/2 but you add pi because cosine can equal 0 also at 3pi/2 just be aware of that because it can be plus 2pi sometimes. So we are left with 20 = pi/2 + pi n, then divide by 2 to get 0 = pi/4 + pi/2n, then go up to just below 2pi so our asymptotes are pi/4, 3pi/4, 5pi/4, 7pi/4. You find the asymptotes the same for odd and even number of petals but for even this is all you need because you just find the midpoint between the petals and then go 3 out etc. For odd you have another step to find where the petals are because they are in between every other asymptote. So we have 4sin30, set equal to 4 and if it was something like -sin30 you would set equal to -1, so we have 1=sin30, then we sort of swap again to get 30=sin1 and sin of one is pi/2 and takes a full rotation to get back so we have 30= pi/2 + 2pi k, then divide by three to get 0 = pi/6 + 2pi/3k which would be where our petals would be on an odd rose so I would just use pi/6 for the first petal and it just alternates

Finding polar coordinates given rectangular coordinates.

For example (-1,1), First you would graph it which would lie in the second quadrant, use pythag theorem to solve for r which would be sqrt(2). Then find the angle created by the point which would be 3pi/4 bc its in the second quadrant. So it would be (Sqrt(2), 3pi/4) and the negative would be (-3, 7pi/4) and adding 2pi to each if u wanted all solutions.

How to plot points in polar coordinate system.

For example P(2,60 degrees) you would find wherever 60 degrees is then draw a ray in that line, then you would go two rings out because of the 2. If there is a negative like (-3,3Pi/4) you would find where the 3pi/4 is but because the 3 is negative you would go 3 in the opposite direction.

Finding all possible coordinates for a point.

For example P(3, pi/3) you would do (3, pi/3 +/- 2pi n), for negative remember it goes backwards so u have to do -3 as well but add pi to the angle so it would be (-3, 4pi/3 +/- 2pi n) To generalize (r, 0 +/- 2pi n) and (-r, (0+pi) +/- 2pi n)

Testing for symmetry of these equations using algebra

For polar symmetry you just substitute 0 for -0 then see if it's the same, usually cosine is the same because cos(-0) = cos(0) , sin and tan are not, for pole you substitute 0 for 0 + pi then use the sin/cos formulas from previous units like sin(0)cos(pi) + cos(0)sin(pi), for sin it's the same sign as of equation it goes like above, for cos it's the oppposite sign as og and goes coscos sinsin, for line/ pi/2 you substitute 0 for 0-pi and do the same as above,

Polar form of a complex numbers

a+bi, a=rcos0 and b=rsin0 same as x and y, so a+bi = rcos0 +isin0 also known as cis. r(cos0 + isin0)

Parametric equation

another way of defining a function (application: projectile motion) ex: x = f(t), y = 2t+3, -2 < t < 4, where t is the parameter.

Graphing parametric equation by eliminating the parameter.

ex: y = 2t + 3, x = t^2, y is easier to solve for t / just solve using normal stuff so t = y-3 / 2 , then plug into x equation to get x = 1/4 (y-3)^2 + 1 (since t is restricted we need to find end points so plug t values in og equations) (these can result in lines, parabolas etc) (remember to put orientation)

note if its just r=4 it would become a circle

note if its just r=4 it would become a circle