Chapter 6 Test
Find the rectangular coordinate of (-3, -29pi/7)
(-2.70, 1.30)
Find all polar coordinates of polar coordinate (1.5, -20 degrees) for -2pi <= x <=2pi
(1.5 , -pi/9) (-1.5, -10pi/9) (-1.5, 8pi/9) (1.5, 17pi/9)
Line segments always have a restriction of...
0<= t <= 1
Find all polar coordinates of polar coordinate (2, pi/6) for -2pi <= x <=2pi
1. (2, pi/6) 2. Draw a picture of scenario 1rst quadrant and postive number 3. (-2, 7pi/6) 4.(2, - 11pi/6) 5.(-2, -5pi/6)
When given a magnitude and degree in bearings of two vectors and asked for the combined magnitude and direction angle ...
1. Convert bearing angle to a trig angle to find the component form for both vectors: <mag*cos(trig angle) , mag*sin(trig angle)> 2. Add the the two component forms to find the component form of the resultant vector 3. Find the magnitude of the resultant 4. Find the angle inside the triangle formed by the componets and the resultant 5. Find the bearing for the direction angle
Find a unit vector in the direction of the given vector w = -i - 2j
1. Convert i's and j's to component form w= -1<1,0> - 2<0,1> W = <-1, 0> - <0, 2> W = <-1, -2> 2. Determine the magnitude of the vector |w| = sqrt( (-1)^2 + (-2)^2 ) |w| = sqrt(5) 3.Using the magnitude determine the unit vector unit vector = 1/sqrt(5) * <-1, -2> unit vector = <-1/sqrt(5), -2/sqrt(5)> unit vector = < - sqrt(5) / 5, -2sqrt(5) / 5>
Determine wheter the vectors u and v are parallel, orthogonal, or neither u = <2, -7> v = <-4, 14>
1. Determine Slope slope of u = (-7/2) slope of v = (14/-4) --> (-7/2) Same slopes = Parallel Vectors
Determine wheter the vectors u and v are parallel, orthogonal, or neither u = <5, -6> v = <-12, -10>
1. Determine Slope slope of u = (-6/5) slope of v = (-10/-12) --> (5/6) 2. Determine if orthognal by finding the dot product u*v = 0 They are Orthogonal Vectors
Let vector U = <-1, 3> and vector V = <2,4> Find the component form of the vector 2u - 4v
1. Determine the component form of vector 2u and vector -4v 2u = 2 * <-1, 3> --> <-2, 6> -4v = -4 * <2, 4> --> <-8, -16> 2. Using vector arithmitic add the first terms and the second terms <-2,6> + <-8, -16> --> <-10, -10> Vector 2u - 4v = <-10, -10>
Find a unit vector in the direction of the given vector u = <-2, 4>
1. Determine the magnitude of the vector |u| = sqrt( (-2)^2 + (4)^2 ) |u| = sqrt(20) |u| = 2sqrt(5) 2. Using the magnitude determine the unit vector unit vector = 1/s2qrt(5) * <-2, 4> unit vector = <-2/2sqrt(5), 4/2sqrt(5)> unit vector = <-1/sqrt(5), 2/sqrt(5)> unit vector = <-sqrt(5)/5, 2sqrt(5)/5>
Plotting points with given polar coordinate (-1 , 2pi/5)
1. Draw picture and convert 2pi/5 to degree 2pi/5 = 72 degrees 1rst quadrant 2. Then draw from magnitude 0 backwards -1 Dot is placed in 3rd quadrant because magnitude is negative
Plotting points with given polar coordinate (3, 4pi/3)
1. Draw picture and find 4pi/3 3rd quadrant 60 angle 2. Then go magnitude from 0 length of 3
Prove that the vectors u and v are orthogonal u = <2, 3> v = < 3/2, -1>
1. Find if the dot product of the two vectors is 0 u*v = 2* 3/2 + 3 * -1 u*v = 3 + -3 u*v = 0
Find the magnitude and direction angle of the vector <-1,2>
1. Find magnitude sqrt( (-1)^2 + (2)^2 ) mag: sqrt(5) 2. Find the direction angle Find the angle inside the triangle using trig angles tan -1 ( 2/1) a = 63.4 degrees Identify the Quadrant 2nd Quadrant = theta = 180 - a Calculate Theta = 180 - 63.4 theta = 116.6 degrees 3. State magnitude and direction angle sqrt(5) @ 116.6 degrees
Find the magnitude and direction angle of the vector <3,4>
1. Find magnitude sqrt( (3)^2 + (4)^2 ) mag: 5 2. Find the direction angle <3, 4> is over 3, up 4 so it would be in the first quadrant and have an x value of 3 and a y value of 4 Using this knowledge find the angle of the triangle formed. tan-1 (4/3) = 53.1 degrees 4. 5 @ 53.1 degrees
Find the parametrization for the curve The line through the points (-2,5) and (4,2)
1. Find slope vector -2 --> 4 = 6 5 --> 2 = -3 <6, -3> 2. Plug into parametrization formula <x,y> = slope vector * t + a starting point <x,y> = <6,-3> * t + (4,2) 3. Assign x and y values x = 6t + 4 y = -3t + 2
Find vectorv with the given magnitude and same direction as u |v| = 2 u = <3, -3>
1. Find the angle of the triangle tan -1 (3/3) = 45 degrees 2. Use triangle angle to determine trig angle Quadrant 4: 360 - 45 = 315 3. Find the component form of vector v < 2 * cos(315) , 2* sin(315) > <1.414, -1.414>
Let Q = (3,4) and S = (2, -8) Find the component form and magnitude of 2QS
1. Find the component form of 2QS by finding the distance in the x and distance in the y 2QS = 2 * <-1, -12> 2QS = <-2, -24> 2. Find the magnitude of 2QS using the component form |2QS| = sqrt( (-2)^2 + (-24)^2 ) |2QS| = sqrt(580) |2QS| = 2sqrt(145) Component form of 2QS = <-2, -24> Magnitude of 2QS = 2sqrt(145)
Let P = (-2,2) amd Q = (3,4) Find the component form and magnitude of vector PQ
1. Find the component form of PQ by finding the distance in the x and distance in the y PQ = <5,2> 2. Find the magnitude of PQ using the component form |PQ| = sqrt( (5)^2 + (2)^2 ) |PQ| = sqrt(29) Component form of PQ = <5,2> Magnitude of PQ = sqrt(29)
Prove that vector RS and vector PQ are equivilant by showing that they represent the same vector R=(-4,7) S = (-1,5) P = (0,0) Q=(3, -2)
1. Find the component form of RS by finding the distance in the x and distance in the y RS = <-4,7> --> <-1,5> RS = <3, -2> 2. Find the component form of PQ by finding the distance in the x and distance in the y PQ = <0,0> --> <3,-2> PQ = <3,-2> 3. Compare RS = PQ <3,-2> = <3,-2>
Prove that vector RS and vector PQ are equivilant by showing that they represent the same vector R=(2,1) S = (0,-1) P = (1,4) Q=(-1, 2)
1. Find the component form of RS by finding the distance in the x and distance in the y RS = <2,1> --> <0,-1> RS = <-2, -2> 2. Find the component form of PQ by finding the distance in the x and distance in the y PQ = <1,4> --> <-1,2> PQ = <-2,-2> 3. Compare RS = PQ <-2,-2> = <-2,-2>
Find the angles between the vectors u = <-4, -3> v = <-1, 5>
1. Find the dot product u*v = -11 2. Find the magnitude of each vector |u| = sqrt( (-4)^2 + (-3)^2 ) |u| = 5 |v| = sqrt( (-1)^2 + (5)^2 ) |v| = sqrt(26) 3. Plug in magnitude and dot product into formula theta = cos -1 (-11 / 5 * sqrt(26) ) theta = 115.6
Find the vector projection of u onto v. Then write u as a sum of two orthogonal vectors one of which is projection of u onto v. u = <-8, 3> v = <-6, -2>
1. Find the dot product u*v = 42 2. Find the magnitude of v |v| = sqrt( (-6)^2 + (-2)^2 ) |v| = sqrt(40) 3. Plug into the formula ProjvU = (42 / sqrt(40)^2) * <-6, -2> ProjvU = <-6.3, -2.1> What will make the dot product 0 and = to u -8 - -6.3 = -1.7 3 - -2.1 = 5.1 u = <-6.3, -2.1> + <-1.7, 5.1>
Find a parametrization for the curve The line segment with endpoints (5,2) and (-2,-4)
1. Find the slope vector <-7, -6> 2. Plug into formula <x,y> = <-7,-6> * t + (5,2) x = -7t + 5 y = -6t + 2 0<= t <=1
Find the x intercept A and the y intercept B and the coordinates of the point P so that AP is perpendicular to the line and |AP| = 1 Given 3x - 4y = 12
1. Find the x and y intercepts x intercept is found when y = 0 3x - 4(0) = 12 3x = 12 x = 4 A(4,0) y intercept if found when x = 0 3(0) - 4y = 12 -4y = 12 y = -3 B(0,-3) AB<-4,-3> ; Slope = -3/-4 = 3/4 perpendicular slope = -4/3 2.From slope determine point P (3, -4) or (-3,4) 3. Find the magnitude of point P sqrt( (3)^2 + (-4)^2 ) = 5 4. Find the unit vector of point P because it asks for the magnitude of AP to be 1 (3, -4) * 1/5 and (-3,4) * 1/5 (.06, -0.8) and (-0.6, 0.8) 5.Add to A (4,0) P = (4.6, -.8) or (3.4, 0.8)
Find the x intercept A and the y intercept B and the coordinates of the point P so that AP is perpendicular to the line and |AP| = 1 Given 3x-7y = 21
1. Find x and y intercepts x intercept is when y = 0 3x - 7(0) = 21 3x = 21 x = 7 A (7,0) y intercept is when x = 0 3(0) - 7y = 21 -7y = 21 y = -3 B(0, -3) 2. Use perpendicular slope to determine possible points AB = <-7, -3> slope = -3/-7 = 3/7 perpendicular slope = -7/3 Possible points: (-3, 7) or (3, -7) 3. Find unit vector of possible points magnitude = sqrt(58) (-3, 7) * 1/sqrt(58) or (3,-7) * 1/sqrt(58) 4. Add on to cooridate A (-0.394, 0.919) + (7,0) --> (6.606, .919) or (.394, -.919) + (7,0) --> (7.394, -0.919)
A 2000 lb car is parked on a street that makes an angle of 12 degrees with the horizontal. Find the the force required to keep the car from sliding down the hill and the force perpendicular to the street.
1. Gravational force mag * sin(x) 2000 * sin(12) F = 415.823 2. Horizontal component mag * cos(12) 1956.295
Plotting points with given polar coordinate
1. If negative magnitude, draw backwards 2. If negative angle start from 0 going backwards
Find the work done by a force of 30 N acting in the direction <2,2> in moving an object 3 m from (0,0) to a point on the line y = (1/2)x in the first quadrant
1. Not in the same direction, so |F| * |AB| * cos(x) 2. Find the angle between the two vectors by drawing a picture 1 line <2,2> and the other <2,1> 3. Use tan-1 to find angle of both triangles, and subtract tan-1(2/2) = 45 tan-1(1/2) = 26.6 final angle = 18.4 4. Plug into formula and solve 30 * 3 *cos(18.4) 85.399
Find the work done by a force of 12 N acting in the direction <1,2> in moving an object 4m from <0,0> to <4,0>
1. Not same direction, but a constant force in any direction = |F|*|AB| *cos(x) 2.Draw a picture based on the direction of the force 1rst quadrant over 1, up 2 3. Find angle of triangle using tan-1 tan-1 (2/1) = 63.4 degrees 4. Plug values into formula 12 * 4 * cos(63.4) 5. Solve 21.492
The angl between a 200lb force and AB = 2i +3j is 30 degrees. Find the work done by F moving an object from A to B.
1. Not same direction, so |F| * |AB| * cos(x) 2. Find magnitude of AB 2<1,0> + 3<0,1) <2,0> + <0,3> <2,3> sqrt( (2)^2 + (3)^2) = sqrt(13) 3. Solve 200 * sqrt(13) * cos 30 624.5
Find u*v given theta is the angle between u and v theta = 150 |u| = 3 |v| = 8
1. Plug numbers into formula cos 150 = (u*v) / 3 * 8 cos 150 = (u*v) /24 cos 150 * 24 = u*v u*v = -20.785
Find the work done by lifting a 2600lb car 5.5ft
1. Same direction = |F| * |AB| 2600 * 5.5ft W = 14300
Use an algebreic method to eliminate the parameter x = 2t - 3, y = 9 -4t; 3<= t <= 5
1. Solve 1 for t x = 2t-3 x + 3 = 2t/2 x/2 + 3/2 = t 2. Substitute y = 9 - 4(x/2 + 3/2) y = 9 -2x - 6 3. Write answer replacing restriction t with solved t y = -2x + 3; <3<= x/2 + 3/2 <= 5
Use an algebreic method to eliminate the parameter x= t^2, y = t + 1
1. Solve for t sqrt(x) = t 2. Substitute y = sqrt(x) +1
x = 2sin t, y = 2 cos t ; 0<= t <= 3pie/2
1. Solve t sin -1 (x/2) = t y = 2 cos ( sin-1(x/2) ); 0 <= sin-1(x/2) <= 3pie/2
Use an algebreic method to eliminate the parameter x = t, y = t^3 -2t +3
1. Substitute y = (x)^3 -2x +3
Use an algebreic method to eliminate the parameter x = 1 + t, y = t
1. Substitution x= 1 + y 2. Write in slope intercept form y = x -1
A person is sitting on a sled on the side of a hill inclined at 60 degrees the person and the sled weigh 160lbs. What is the force required to keep the sled from sliding down the hill.
1. Think of problem as a gravitational force holding the sled. 2. Find the gravitational force using mag * sin(x) 160 * sin(60) F = 138.564
When given an equation in y = mx+b format and asked for parametric form then...
1. Turn y into <x,y>, x into t, and b into 1 of the two starting points 2. Turn m into a slope vector by finding the distance between the x values and the distance between the y values <x1 --> x2, y1 --> y2> <sx, sy> 3. <x,y> = <sx,sy>t + (px, py) 4. Write as a parametric {x = sx*t + px} {y = sy*t + py}
Find the dot product of u and v u = <4,5> v = <-3, -7>
1. Use dot product formula to find dot product 4 * -3 + 5 * -7 u*v = -47
Find the dot product of u and v u = <5,3> v = <12,4>
1. Use dot product formula to find dot product 5* 12 + 3*4 u*v = 72
Find the rectangular coordinate of (3, 2pi/3)
1. Use formula x = r*cos(theta) and y = r*sin(theta) x = 3 * cos(2pi/3), y = 3* sin(2pi/3) x = 3 *-1/2, y = 3*sqrt(3) /2 x = -3/2, y= 3*sqrt(3) / 2
Find the rectangular coordinate of (-2, 60)
1. Use polar to rectangular formula x= r*cos(theta) y = r*sin(theta) x = -2 * cos 60 ; y = -2 * sin 60 x = -2 * 1/2 ; y = -2 * sqrt(3)/2 x = -1 ; y = -sqrt(3)
Let vector U = <-1, 3> and vector W = <2, -5> Find the compontent form of the vector U - W
1. Using vector arithmitic simply subtract the first terms and the last terms <-1, 3> - <2, -5> --> Vector U-W = <-3, 8> Vector U+V = <-3, 8>
Covert polar equation to rectangular equation and identify the graph r= -3 sin(x)
1. When cos or sin is used multiply by r r *r = -3 r sin(x) r^2 = -3rsin(x) 2. r^2 turns into x^2 + y^2 and rsin(x) = y x^2 + y^2 = -3y 3. Add 3y x^2 + y^2 +3y ______ = 0 4. Complete the square x^2 + (y + 3/2)^2 = 9/4 A circle at center ( 0, -3/2) with radius of (3/2)
Covert polar equation to rectangular equation and identify the graph r cscx = 1
1. When cscx or secx is on the left then -divide by csc or sec -Multiply by r -Simplify r = 1/csc x r = sin x r^2 = rsinx x^2 + y^2 = y X^2 + Y^2 + y = 0 x^2 + (y - 1/2)^2 = 1/4
Covert polar equation to rectangular equation and identify the graph r= 3 sec(x)
1. When sec or csc are used multiply by cos or sin respectivly r * cos(x) = 3sec(x) * cos(x) x = 3 Vertical line at x = 3
Find vector v given u = <2,3>, u*v = 10, |v|^2 = 17
1. Write an equation for the dot product, Solve for one of the two variables 2x + 3y = 10 2x = 10 - 3y x = - 3/2y + 5 2. Write the magnitude of v as an equation |v| ^2 = sqrt( x^2 + y^2) ^2 = 17 x^2 + y^2 =17 3.Plug in solved variable (-3/2y + 5)^2 + y^2 = 17 4. Simplify (-3/2y + 5) * (-3/2y + 5) + y^2 9/4y^2 -15/2y -15/2y + 25 + y^2 = 17 13/4y^2 -15y + 8 = 0 4. Remove denominators by multiply all by 4 13y^2 -60y +32 5. Plug into quadratic formula x = -(-60) +- sqrt( (-60)^2 - 4(13)(32) / 2(13) y = 4 or 8/13 6. Plug y into 1 of the equations to find the x value and list as vector v 2x + 3(4) = 10 2x + 12 =10 2x = -2 x= -1 v = <-1, 4> or 2x + 3(8/13) = 10 2x +24/13 = 10 2x = 106/13 x = 53/13 v = <53/13 , 8/13>
Determine wheter the vectors u and v are parallel, orthogonal, or neither u = <15, -12> v = <-4, 5>
1.Determine Slope slope u = (-12/15) --> (-4/5) slope v = (5/-4) 2. Determine if the vectors are orthogonal by finding the dot product U*v = -120 3. The vectors are neither parallel or orthoganal
When given a parametric equation where x = ...t and y = ...t , and asked for slope intercept form then...
1.Find t and subsitute in for either variable to get the equation to equal only x and ys 2. Write in y = mx +b format
Find the component form of vector v |v| = 18 and Given angle is 25. Reference image on pg 464 #29
1.Find the x component using cos: |v| * cos(theta) x component = 18 * cos(25) x component = 16.313 2. Find the y component using sin: |v| * sin(theta) y component = 18 * sin(25) y component = 7.607 3. <16.313, 7.607>
Let vector U = <-1, 3> and vector V = <2,4> Find the component form of the vector U+V
1.Using vector arithmitic simply add the first terms together and the last terms together <-1, 3> + <2,4> --> vector U+V = <1, 7> vector U+V = <1, 7>
Identify max r value r = 3cos3x
3
Given r = 3 sin3x How many pedals are there?
3 Odd "b" in a*cosbx is the number of pedals
Given r = 3 cos2x How many pedals are there?
4 Even number "b" in a*cosbx is 2 times the number of pedals
Identify max r value r = 2 + 3cosx
5
Given 7(cos 135i + sin135j) State the magnitude and direction angle
7 @ 135 degrees
Component form
< distance in the left/right direction, distance in the up/down direction >
What is vector j in component form?
<0, 1>
What is vector i in component form?
<1, 0>
Formula to Find magnitude given a vectors components
<x,y> sqrt( (x)^2 + (y)^2 )
Identify the graph r = 4+4cosx
Cartiod
When given restrictions for t for graphing
Change t max and t min
When given a magnitude and degree in bearings and asked for the component form ...
Convert bearing angle to a trig angle to find the component form: <mag*cos(trig angle) , mag*sin(trig angle)>
An airplane is flying on a bearing of 335 at 530 mph. Find the component form of the velocity of the airplane
Convert bearing angle to a trig angle: Draw the 335 degree angle in bearings Find the angle in triangle - 65 degrees Identify the quadrant - (2nd Qaudrant) Apply appropriate formula - 180 - a Trig angle is 115 Find the component form: <mag*cos(trig angle) , mag*sin(trig angle)> <530 * cos(115) , 530 * sin(115)> <-223.988, 480.343>
limacon: a/b >= 2
Convex limacon
Identify the graph r = 5 + 4sinx
Dimpled limacon
limacon: 1 < a/b < 2
Dimpled limacon
Going from rectangular coordinates to polar coordinates
Given (x,y) find (r, x) r = sqrt( (x)^2 + (y)^2 ) x = tan-1 (y / x) x can be multiple things in polar coordinates, so draw a picture with the angle found. restriction -2pie < x <2pie
Convert the rectangular equation to polar form 2x - 3y = 5
If given both x and y, and not a circle then - convert x and y to r format - factor out r - divide by what is left in parentheses 2x - 3y = 5 2rcos x - 3rsin x = 5 r(2cosx - 3sinx) = 5 r = 5/(2cosx - 3sinx)
Convert the rectangular equation to polar form (x-3)^2 + (y+3)^2 = 18
If given in circular form then -Factor -Simplify and subtract constant to get right side = 0 -Combined x^2 and y^2 to make r^2 -Make other variables their respective r form -Divide by r -Get r alone (x-3)(x-3) + (y+3)(y+3) = 18 x^2 - 6x + 9 + y^2 +6y + 9 = 18 x^2 + y^2 - 6x + 6y = 0 r^2 - 6rcosx + 6rsinx = 0 r - 6cosx + 6sinx = 0 r = 6cosx - 6sinx
Convert the rectangular equation to polar form (x-3)^2 + y^2 = 9
If given in circular form then -Factor -Simplify and subtract constant to get right side = 0 -Combined x^2 and y^2 to make r^2 -Make other variables their respective r form -Divide by r -Get r alone (x-3)(x-3) + y^2 = 9 x^2 -6x + 9 + y^2 = 9 x^2 + y^2 -6x = 0 r^2 -6rcosx = 0 r - 6cosx = 0 r = 6cosx
Convert the rectangular equation to polar form x = 2
If given x or y change to its respective r format rcosx = x rsinx = y x = 2 rcosx = 2 r = 2/cosx r = 2 * 1/cosx r = 2 sec x
Two vectors are parallel ...
If they have the same slope.
Identify the graph r^2 = sin2x; 0<= x <= 2pi
Lemniscate
r^2 = a^2 * sin2x or r^2 = a^2 * cos2x
Lemniscate Type in calulator as r = sqrt(a^2 * cos2x) r= - sqrt(a^2 * cos2x)
Identify the graph theta = pi/3
Line; y = sqrt(3)x
Find the length of each pedal of the polar curve r = 2 + 4sin 2x
Long 6; Short 2 Take the absolute value of the minimum r value and maximum r value
When given a function to input into calculator to look at graph in parametric form...
Make sure in parametric mode and radians
Find the rectangular coordinate of (1.5 , 7pi/3)
Memorized angle denoms are 6,4,3 (3/4, 3sqrt(3)/4)
a * cosbx or a * sinbx
Rose Curve If b is odd, then it is the number of pedals If b is even, then the number of pedals is 2b a makes bigger or smaller
Identify the graph r = 2sin3x
Rose curve with 3 pedals
Identify the graph r = 2x
Spiral
If force is a constant magnitude in any direction then...
W = F*AB (dot product) or W = |F| * |AB| * cos(x)
If the force and direction of motion are in the same direction then ...
W = |F| * |AB| |F| is how much weight , |AB| = how far Work = magnitude of the force * magnitude of the second vector
For trig angles only When finding the angle of direction for a vector, you need find the angle inside the triangle and then use these formulas to determine the angle of direction. Use a as the angle inside the triangle and theta as the direction angle of the magnitude
When in Quad 1: theta = a When in Quad 2: theta = 180 - a When in Quad 3: theta = 180 + a When in Quad 4: theta = 360 - a
Rose Curve
a * cosbx or a * sinbx
limacon curve
a +- bcosx or a+-sinbx
limacon: a = b
cartiod
Identify the graph r = 3
circle, centered at (0,0) with a radius of 3
Slope can be determined for vectors through their
component form
Identify the graph r = 5 + 2 cosx
convex limacon
The angle between two vectors is
cos-1( u*v / |u| * |v| )
Vectors
describe motion, not location Directed line segment Has direction and magnitutde
Two vectors are orthogonal ...
if and only if their dot product is 0
How is magnitude represented
like absolute value but with an arrow over the variable
Identify the graph r = 2 + 5 cos x
limacion with inner loop
a +- bcosx or a+-sinbx
limacon curve
limacon: 0 < a/b < 1
limacon with inner loop
limacon with inner loop
limacon: 0 < a/b < 1
Dimpled limacon
limacon: 1 < a/b < 2
cartiod
limacon: a = b
Convex limacon
limacon: a/b >= 2
A vector is a unit vector when the ...
magnitude of the vector = 1
When asked to match a polar equation with its graph ...
make sure you are in polar mode, and radians are set
spiral of archimedes
r = theta
Lemniscate
r^2 = a^2 * sin2x or r^2 = a^2 * cos2x
r = theta
spiral of archimedes
You find the unit vector of a normal vector by ...
take the vector in component form * 1/magnitude of the vector vector u = 1/|v| * vector v
The dot product of two vectors is
x * x + y * y
Going from polar coordinates to rectangular coordinates...
x = r * cos(x) y = r* sin(x) 1. If an known angle measurement, use an exact value.
Identify if the graph has symmetry r = 4-3cosx
x axis symmetry
Identify if the graph has symmetry r = 5cos2x
x axis symmetry y axis symmetry orgin symmetry of 180 degrees
Identify if the graph has symmetry r = 3+3sinx
y axis symmetry
