Vectors/Polars Quiz
other ways to write (6, π/6)
(-6, 7π/6) -you would go to the 7π/6 line and since it is -6 you would trace from that line to the other side of the circle (180°) and that would be π/6 -to go from this to (6, π/6), you can get rid of the negative 6 and do 7π/6- 6π/6 (180°) to get π/6
symmetries about the line θ= π/2 (y-axis)
(-r, -θ) or (r, π-θ)
other ways to write (3, -3π/2)
(3, π/2) -start at 0π and go clockwise (instead of the normal counterclockwise) to where 3π/2 would be and then go three up.
A hiker in the woods travels along the path described by the parametric equations x(t) = 80-t, y(t)=1/2t A bear leaves another area of the woods to the west and travels along the path described by the parametric equations x(t)=4t, y(t)=20 + 4/3(t) (a) do the pathways of the hiker and the bear intersect? (b) do the hiker and the bear collide? Explain.
(a) t=80-x(t) y(t)=1/2(80-x(t)) y(t)=40-1/2x(t) t=x(t)/4 y(t)=20 + 4/3(xt/4) y(t)=20+xt/3 yes they intersect (b) 80-t=4t 80=5t t=16 1/2t=20 + 4/3t 3/6t- 8/6t=20 -5/6t=20 t=-24 No, because at t=16, the x and y values are not the same, which shows they don't collide
A particle moves in the xy plane according to the parametric equations x1 = t^2 + 1, y1=-5t+6 and a second particle moves in the xy plane according to the parametric equation x2=2t, y2= t^2 (a) find all points (x, y) where the paths of the particles intersect (b) determine if the particles collide. That is, are the particles at the same position at the same time.
(b) x1=x2 t²+1=2t t²-2t+1=0 (t-1)(t-1)=0 t=1 y1=y2 -5t+6=t² 0=t²+5t-6 (t+6)(t-1)=0 t=1 or t=-6 (but can't be -6 because it can't be negative) t=1 The particles collide at t=1 at (2,1) (a) (2,1) x1-1=t² t=±√(x-1) y1=-5√(x-1) +6 t=x2/2 y=x²/4
Describing parametric curves (i) a sketch of the parametric curve (including direction of motion) based on the equation you get by eliminating the parameter (ii) limits on x and y (iii) a range of t's for a single trace of the parametric curve (iv) the number of traces of the curve the particle makes if an overall range of t's is provided in the problem x=6cos(3t), y=6sin(3t)
(i) make a data table and draw the circle (ii) -6≤x≤6, -6≤y≤6 (iii) since period of sin(3t), cos(3t) is 2π/3, let 0≤t≤2π/3 (iv) each additional trace is 2π/3+2π/3(n)
symmetries about the polar axis (x-axis)
(r, -θ) or (-r, π-θ) -if they map to the same equation, there is polar axis symmetry
symmetries about the pole (origin)
(r, π+θ) or (-r, θ)
resultant forces
(sum of forces acting on an object)
The cranes are lifting a crate that weighs 18,728 lbs. Determine the tensions T1 and T2 in the cable. T1= 41.5 degrees T2=157.7 degrees
-the vectors are in equilibrium so they form triangles within them -also magnitude is degrees away from 0 degrees on the graph (like in a unit circle) w=<0, -18728> (the box is pulling down) T1=<|T1|cos41.5, |T2|sin41.5> T2=<|T2|cos157.7, |T2|sin157.7> Horizontal (equilibrium) |T1|cos41.5 + |T2|cos157.7 = 0 Vertical (equilibrium) |T2|sin41.5 + |T2|sin157.7 + -18728 =0 using the horizontal component |T1|=(-|T2|cos157.7)/cos41.5 plug this in the vertical equilibrium sin41.5((-|T2|cos157.7)/cos41.5) + |T2|sin157.7 = 18728 |T2|=(((-cos157.7 × sin41.5)/cos41.5) +sin157.7)=18728 |T2|=15,633 lbs
dot product
-this is sometimes called the scalar product and is going to represent "directional growth", that is, a vector's growth in a particular direction, not necessarily its own if u= <x,y> and its magnitude is √(x²+y²) then the dot product of a vector within itself is u × u = (√(x²+y²))²= |u|²
Vectors in the coordinate plane
-to describe our vectors more accurately, we place them in the coordinate plane. -we denote these vectors from their initial point (tail) to terminal point (head)
graphs of polar equations
-we can graph equations in polar coordinates using a variety of strategies, from plotting points to using symmetries
graph r=2
-you can make a table of θ and r values or you can make the circle formula by finding the rectangular formula √(x²+y²) = 2 x²+y²=4
zero vector
0=<0,0>
write down a set of parametric equations for the given equations that meet the given extra conditions (if any) y=3x²-ln(4x+2)
1. let t=4x+2 t-2=4x t-2/4=x so, x=1/4t-1/2 y=3(1/4t-1/2)²-lnt t>0 2.x=t y=3t²-ln(4t+2) t>-1/2 3. x=√t y=3t-ln(4√t + 2) t>0
r=3cos2θ using zeroes of r
1. r=0 when cos2θ=0 2. cosθ=0 at θ=π/2 and 3π/4 3. so cos2θ=0 at π/4 and 3π/8
graph the following polar curves and determine all points of intersection for r=4cosθ, r=4sin2θ
4cosθ=4sin2θ cosθ=sin2θ cosθ=2sinθcosθ 0=2sincosθ-cosθ 0=cosθ(2sinθ-1) cosθ=0 or sinθ=1/2 θ=π/2, 3π/2 θ=π/6, 5π/6 (r, θ) r=4cosθ r=4(0) (because at θ=π/2, 3π/2, r=0) (0, π/2) or (0, 3π/2) r=4cos(π/6)=2√3 (2√3, π/6) same as (√3, 11π/6) r=4cos(5π/6) (-2√3, 5π/6) same as (2√3, 11π/6)
Scalar quantity
A scalar quantity is fully described by its magnitude (a number representing the quantity)
vector quantity
A vector quantity is fully described by its magnitude and direction (represented by an angle measure)
two forces, A and B, act on an object. A has a direction of N20E and B has a direction of S50E. The resultant force has magnitude 500 and direction N36E. Determine the magnitudes of A and B to the nearest thousandth and write the forces A and B in the component form.
A=<|A|cos70, |A|sin70> B=<|B|cos320, |B|sin320> R=<500cos54, 500sin54> A+B=R 1. |A|cos70 + |B|cos320 = 500cos54 2. |A|sin20 - |B|sin320= 500sin54 1. |A|=(500cos54-|B|cos40)/cos70 2. ((500cos54-|B|cos40)/cos70)sin70 - |B|sin40 = 500sin54 (sin70/cos70)=tan70 500cos54tan70 - |B|cos40tan70 - |B|sin40 = 500sin540 |B|(-cos40tan70-sin40)=500sin54-500cos54tan70 |B|=(500sin54-500cos54tan70/(-cos40tan70-sin40) |B|=146.664 |A|=530.793 -use at least 6 decimal points when you have to round to the nearest thousandth
A girl pulls a wagon along a level path for a distance of 44 m. The handle of the wagon makes an angle of 22 degrees above horizontal. (b) How much work is done if she applies the same force from the handle at an angle of 11 degrees?
F: <87cos11, 87sin11> D: <44, 0> F x D= (44 x 87cos11) + 0= 3757.67 J -more energy is required if θ is smaller
A person lifts a package weighing 75 N. If she lifts it 1.2 m off the floor, what work has she done?
F= <0, 75> D= <0, 1.2> w= N x M= 90 J F x d= 0 + (75 x 1.2)= 90
A girl pulls a wagon along a level path for a distance of 44 m. The handle of the wagon makes an angle of 22 degrees above horizontal. (a) If she pulls on the handle with a force of 87 N, how much work is done?
F= <87cos22, 87sin22> D= <44, 0> F x D= (44 x 87cos22) + 0= 3549.26 J
A sailor pulls a boat along a dock using a rope at an angle of 60 degrees with the horizontal. How much work does the sailor do if he exerts a force of 255 N on the rope and pulls the boat 3 m?
F=<255cos60, 255sin60> D=<3, 0> F x D= (3 x 255cos60) + 0= 382.5 J
Thinking about vectors
If we think of vectors as "growth in a direction" (think about the velocity of an object over time), then when we add vectors (u + v), we are combining the "growth" of each vector, which may create a larger or smaller resulting vector. When we multiply by a scalar, (i.e. 3u or 1/2u), we are increasing (or decreasing) the magnitude of a vector in its own direction
An airplane heads due north at 450 miles per hour. The plane experiences a 40 mile per hour crosswind flowing in the direction N 30 degrees E. Express the vectors for the velocity of the plane, P, relative to the air and the velocity of the crosswind, w.
P= <0, 450> (<450cos90, 450sin90>) W= <40cos60, 40sin60>= <20, 20√3>
Below is a table showing the number of sunscreen bottles sold and ice cream cones sold at a local beach on different days.
S(I) I(S) -while these values are correlated, they are not causative -what is the more likely root cause of the increase in ice cream sales and sunscreen sales? Temperature! -Temperature is the parameter (root) that determines ice cream sales and sunscreen sales
Brian is pushing a block up a hill to get stronger. He decides to take a break and his friend Olivia holds the block in place while he rests. If she is applying 100 lbs of force on the block, how much does it weigh? Assume the hill is at 20-degree incline and the block and hill are both smooth with no friction.
There is 3 forces at work: w= weight due to gravity= <0, |w|> a= force experienced by driveway (normal force) b= force that Olivia has to overcome to keep block in place in a right triangle vector w is the hypotenuse (going straight down from the road, it forms the vertical side of the side of the incline triangle) vector a is the vertical leg (going straight down the box (the same tilt)) vector b is the horizontal leg (on the horizontal side of the object) angle a (the angle across from a) is 70 degrees and angle b is 20 degrees |b| (magnitude of vector b) is 100 (because it's the force she applies to overcome the force pushing it down) sin20= |b|/|w| |w|=100/sin200 |w|=292.380 lbs.
limacon with inner loop
a < b
dimpled limacon
a > b
What can we picture a vector as?
a directed line segment -tells which direction the force is operation -distance would be length of the vector
convex limacon
a ≥ 2b or a/b ≥ 2
Dot product formula
a ⊗ b = |a||b|cosθ
polar coordinate system
another way to describe the location can be found by using distance and rotation about the origin, which defines the polar coordinate system. Instead of (x,y) we use (r, θ) to denote the location -origin: "pole" -y-axis: θ=π/2 -x-axis: 0°
things needed to memorize for π+θ or π-θ in symmetries for cos2θ same for cos4θ
cos(4π+4θ)=cos4θ cos(4π-4θ)=cos4θ
things needed to memorize for π+θ or π-θ in symmetries for cos
cos(π-θ)=cosπcosθ+sinπsinθ = (-cosθ) + 0= -cosθ cos(π+θ)=cosπcosθ-sinπsinθ = (-cosθ) - 0= -cosθ
Angle between two vectors
cosθ= A x B/|A||B|
If u = <a₁,b₁> , then what is cu?
cu= <ca₁, cb₁>
vectors
displacement, velocity, acceleration (force, weight)
scalars
distance, speed, work, power, mass, volume, length
lemniscates
figure eight shaped curves
eliminating parameters
first, sketch the graph of the parametric equation -make a data table with three variables (t, x, and y) -plug in values of t into both equations and then find the values of x and y then, eliminate the parameter to find the rectangular equation -isolate t in one equation -plug t in to the other equation in order to find x or y, and that new equation is the parametric equation
quick tips
graphs having r=acos(nθ) will always start at the polar axis graphs having r=asin(nθ) will always start at π/2n
three different forces in a right triangle
in a right triangle vector w is the hypotenuse (going straight down from the road, it forms the vertical side of the side of the incline triangle) vector a is the vertical leg (going straight down the box (the same tilt)) vector b is the horizontal leg (on the horizontal side of the object)
A car drives 500 feet on a road that is inclined 12 degrees to the horizontal, as shown in the figure below. The car weighs 2500 pounds. Gravity acts straight down on the car with a constant force F= -2500j. (a) Find the force experienced by the road due to the weight of the car. (b) Find the work done by the car in overcoming gravity.
in this right triangle: hypotenuse is w which is 2500 vector a is the vertical leg (which we are trying to find) magnitude of vector b is the horizontal leg which is 500 1. draw the vector of the weight so that it forms a right triangle with the given angle (12 degrees) so now, angle b is 12 degrees and angle a is 78 degrees (a) cos12 = |a|/-2500 = |a|=-2500cos12 |a|= 2445.369 lbs (magnitude is not negative) (b) F= <0, -2500> D=<500cos12, 500sin12> w=F x D= 0+ (-2500 x 500sin12) w=-259889.6135 ft x lbs
elimination for x=4-2t, y= 3 + 6t, 0≤t≤5
make a data table for t, x, and y but only from t=0 to t=5 values of x: -6≤ x ≤4 values of y: 3 ≤ y ≤33 (this is because of the restriction of t) from t=0 to t=5, our trace of the curve is that way 1. solve for t in x(t) equation x=4-2t x-4=-2t x-4/-2 = t 2. substitute into y(t) equation y=3+6(x-4/-2) y=3+(-3)(x-4) y=-3x+15
Orthogonal vectors corollary
orthogonal=perpendicular if a⊥b, then a x b = 0
An airplane heads due north at 450 miles per hour. The plane experiences a 40 mile per hour crosswind flowing in the direction N 30 degrees E. Find the true velocity of the plane as a vector.
p + w = <20, 450 + 20√3>
If the two polar graphs r=1/(cosθ-sinθ) and r^2=181 are converted to Cartesian (rectangular) coordinates, find all of the coordinates (x,y) of the intersection
r=1/(x/r - y/r) r=1/(x-y/r) (x-y/r)r=1 x-y=1 y=x-1 plug y into x²+y²=181 so now it's x²+(x²-2x+1)=181 2x²-2x-180=0 2(x²-x-90)=0 2(x-10)(x+9)=0 x=10, x=-9 answer: (10,9) (-9, -10)
express r=2cosθ in rectangular form
r=2(x/r) r²=2x x²+y² = 2x x² - 2x + y²= 0 (x-1)² + y²= 1 (complete the square)
express r=4sinθ + 2 in rectangular form
r=4(y/r)+2 (r=4(y/r)+2)r r²=4y + 2r x² + y² = 4y + 2√(x² + y²) x²+y²-4y= 2√(x² + y²) (x²+y²-4y)²=4(x²+y²)
circle w/center at the origin (pole)
r=a
4-leaved rose
r=a cos 2θ
8-leaved rose
r=a cos 4θ
5-leaved rose
r=a cos 5θ
cardioid
r=a+bcosθ, a=b
roses
r=acos(nθ) r=asin(nθ) n-leaved if n is odd 2n-leaved if n is even
3-leaved rose
r=acos3θ
circle with center on the x-axis
r=acosθ
circle with center on the y-axis
r=asinθ
limacons
r=a±bcosθ r=a±bsinθ (a>0, b>0) increase in a, increase in circular shape increase in b, increase in weird shape -orientation depends on the trigonometric function (sin or cos) and the sign of b
spiral curve
r=aθ -but in the picture, curve goes up to the x-axis
convert (2, 2√3) to polar coordinates
r=√((2)²+(2√3)²) = √4+12 = 4 θ=arctan(√3)= π/3 (r, θ) = (4, π/3)
Express y=x² in polar form
rsinθ=(rcosθ)² rsinθ=r²cos²θ sinθ=rcos²θ r=sinθ/cos²θ= sinθ/cosθ x 1/cosθ r=tanθsecθ
express y=1 in polar form
rsinθ=1 r=1/sinθ = cscθ
lemniscate equation (cos)
r²=a²cos2θ
lemniscate equation (sin)
r²=a²sin2θ
things needed to memorize for π+θ or π-θ in symmetries for sin2θ (same for sin4θ)
sin(2π-2θ)=sin2πcos2θ-sin2θcos2π= 0-(sin2θ)= -sin2θ sin(2π+2θ)=sin2πcos2θ+sin2θcos2π= 0 + (sinθ)= sinθ
things needed to memorize for π+θ or π-θ in symmetries for sin
sin(π-θ)=sinπcosθ-sinθcosπ= 0 - (-sinθ)= sinθ sin(π+θ)=sinπcosθ+sinθcosπ= 0 + (-sinθ)= -sinθ
An airplane heads due north at 450 miles per hour. The plane experiences a 40 mile per hour crosswind flowing in the direction N 30 degrees E. Find the true speed and direction of the plane.
speed=magnitude direction: θ= arctan (450 + 20√3/20) = 87.64° N 2.36° E or E 87.64°N magnitude: |p+w|= √(20)²+(450 + 20√3)²)= 485.054 mph
vector subtraction
the difference of two vectors, u - v, is represented by the other diagonal of the parallelogram formed -it's basically, v is now negative so the arrow is facing the opposite direction so it's u + (-v) and that's the diagonal
Three teams are pulling at a same point. Let |F1|= 500 lbs, |F2|=|F3|=400 lbs. F1=30 degrees F2=135 degrees F3=240 degrees
the magnitude should be degrees away from 0 degrees on the graph (like in a unit circle) F1=<500cos30, 500sin30> F2=<400cos135, 400sin135> F3=<400cos240, 400sin240> resultant force: F1+F2+F3 = <-207.14, 207.14>
Parameters
third variables that is causative not correlative -time is a frequent parameter
dot product alternate version let u= <ax, ay> and v=<bx, by>
u ⊗ v= axbx + ayby
If u = <a₁,b₁> and v = <a₂,b₂>, then what is u+v?
u+v = <a₁ + a₂,b₁ + b₂>
If u = <a₁,b₁> and v = <a₂,b₂>, then what is u-v?
u-v = <a₁ - a₂,b₁ - b₂>
component form of a vector
u= <horizontal distance, vertical distance> or u= <|u|cosθ, |u|sinθ> (applications) or u= i + j
Find the unit vector of v= <3,4>
v/|v| = <3,4>/5 = <3/5, 4/5>
What about when determining u's growth in another direction?
we use the unit vector, sometimes called direction vector the unit vector of v is described by the expression v/|v|, where |v/|v||=1
vector addition
when adding two vectors, u + v, the result is the diagonal of the parallelogram formed by the diagram
rectangular (Cartesian) coordinate system
when we define a function, y=f(x), we are using a rectangular (cartesian) coordinate system. This mapping is described by the set of points (x,y) that lie in the coordinate plane, where the origin is the center.
Work
work= force x displacement W= F x d= |F||d|cosθ -units usually used for force: Newtons (N) or Pounds (lbs) -units usually used for displacement: Meters (m) or Feet (ft) -newton- meters called joules -work: the energy required for the task to happen ex: a horse pulling a cart, a student lifting her backpack from the ground, pushing a shopping cart
elimination x=3sint, y=-3cost, 0≤t≤2π
x/3=sint y/-3=cost 1. for trig, use identities sin²t + cos²t=1 x²/9 + y²/9 =1 x² + y² = 9 circle with the trace going counterclockwise
convert (3, 4π/3) to rectangular coordinates
x= 3cos(4π/3)= -3/2 y= 3sin(4π/3)= -3√3/2 (x, y) = (-3/2, -3√3/2)
write down a set of parametric equations for the given equations that meet the given extra conditions (if any) x² + y² = 36 and the parametric curve resulting from the parametric equation should be at (6,0) when t=0 and the curve should have a counter clockwise rotation
x=6cost y=6sint -make the three variable data table and for t=0 make x=6 and y=0 0≤t≤2π (one trace) or 0≤t≤2π+2πn (n+1 traces)
what about clockwise rotation? x² + y² = 36 and the parametric curve resulting from the parametric equation should be at (6,0) when t=0 and the curve should have a counter clockwise rotation
x=6cost y=-6sint t≥0
elimination x=e^t, y=cos(e^2t), 0≤t≤ln(2π)
x=e^t = y=cos((e^t)^2) y=cos(x^2) 1≤x≤√2π -1≤y≤1
vertical lines
x=k -> rcosθ=k which can be written as r=ksecθ
elimination of x=√(t+1), y = 1/(t+1) , t>-1
x=√(t+1) x²=t+1 x²-1 =t y=1/(x²-1)+1 = 1/x² y=1/x² 0<x<∞ 0<x<∞ trace is towards positive infinity (the right)
Describing parametric curves (i) a sketch of the parametric curve (including direction of motion) based on the equation you get by eliminating the parameter (ii) limits on x and y (iii) a range of t's for a single trace of the parametric curve (iv) the number of traces of the curve the particle makes if an overall range of t's is provided in the problem x=e^t, y=e^-t
y=1/e^t=1/x (i) graph y=1/x (ii) 0<x<∞ (x>0), 0<y<∞ (y>0) (iii) since e^t>0 and e^-t>0 for all t the trace is -∞<t<∞ (iv) only one trace is possible t-> ∞ (to the right)
horizontal lines
y=k -> rsinθ=k which can be written as r=kcscθ
key relationships
y=rsinθ x=rcosθ polar to rectangular tanθ=(y/x) r=√(x² + y²) r²=x² + y² rectangular to polar
zeroes of r
zeroes of r is when r=0
Length of a vector
|cu|=|c||u|
magnitude
|u|= √(x²+y²) -speed (mi/hr) -this is the hypotenuse (diagonal)
direction of a vector
θ check quadrants with angle tanθ= y/x -> θ=arctan(y/x)
an example with parameters on the unit circle
θ is the parameter upon which x and y depend x=cosθ y=sinθ parametric equations which becomes: x² + y² =1 (cartesian equation) through elimination often we need to switch between parametric equations and cartesian equations
lines with non-zero slopes
θ=k