Readings this week 1 Parametric Equations Supplement 2 Section 10.1 3 Sections 2.1-2.2
Precalculus Review Quiz session Thursday equations of lines and circles worksheet available at http://www.math.washington.edu/ m124/ (under week 1) print it out before coming to class
Homework All homework assignments on webassign.net First assignment due Tuesday October 5 Help session for signing up with webassign.net Thursday September 30 3:30-5:00 Smith 102
Circle Review The distance formula tells us that a circle with center (h, k) and radius r has the equation (x h) 2 + (y k) 2 = r 2.
Equations of Lines A line is defined by two points. Call them (x 1, y 1 ) and (x 2, y 2 ). Different forms of the equation of the line include 1 y = mx + b where b is the y intercept. 2 point slope m = (y 2 y 1 ) (x 2 x 1 ) 3 parametric equations y 2 y = m(x 2 x) y = y 1 + at, x = x 1 + bt, m = a/b
Perpendicular Lines It is easy to tell if two lines are perpendicular. Suppose the lines have equations y = m 1 x + b 1 and y = m 2 x + b 2. They are perpendicular if m 1 m 2 = 1
Tangent lines to a circle If we know the equation a circle and a point on the circle then we can find the equation of the tangent line. The tangent line is perpendicular to the radial line.
Suppose we have the circle (x h) 2 + (y k) 2 = r 2 and we know that (p, q) is a point on the circle. How can we find the tangent line to the circle at (p, q)? We know one point on the tangent line so we only need to find the slope.
The tangent line is perpendicular to the radial line and m radial = q k p h. Thus the slope of the tangent line is m tangent = 1/m radial = p h q k. Using the point slope formula we get that the equation of the tangent line is y q = m tangent (x p) = p h (x p). q k
Tangent lines to a circle For example the point (1, 7) lies on the circle x 2 + 4x + y 2 6y = 12. Find the equation of the tangent line to the circle at (1, 7) We put the equation in standard form by completing the square. x 2 + 4x + y 2 6y = x 2 + 4x + 4 4 + y 2 6y + 9 9 = (x + 2) 2 4 + (y 3) 2 9 = 12 or (x ( 2)) 2 + (y 3) 2 = 25 and the circle has radius 5 and center ( 2, 3).
Tangent lines to a circle To find the equation of the tangent line to the circle (x ( 2)) 2 + (y 3) 2 = 25 at (1, 7) we use the formula we just derived y q = m 2 (x p) = p h (x p). q k Using the formula we just derived we get the equation of the tangent line is y 7 = 1 ( 2) 7 3 (x 1) = 3 (x 1). 4
It is always good to check your results graphically.
Two Tangents Find all the tangent lines to the circle that go through the point (5, 0). x 2 + (y 1) 2 = 1
All the lines through (5,0) are of the form y = m(x 5). Which of them are tangent to the circle? If the line is tangent to the circle then it intersects the circle at just one point. Thus there is one simultaneus solution to y = m(x 5) and x 2 + (y 1) 2 = 1.
Combining the equations we get x 2 + (m(x 5) 1) 2 = 1 x 2 + m 2 (x 2 10x + 25) 2m(x 5) + 1 = 1 (1 + m 2 )x 2 + ( 10m 2 2m)x + (25m 2 + 10m) = 0. We want to find m so that this has a unique solution.
A quadratic equation ax 2 + bx + c = 0 has solutions x = b + b 2 4ac. 2a This has a unique solution if b 2 4ac = 0. In our problem we have a = (1 + m 2 ) b = ( 10m 2 2m) c = (25m 2 + 10m) So we have a unique solution if ( 10m 2 2m) 2 4(1 + m 2 )(25m 2 + 10m) = 0
( 10m 2 2m) 2 4(1 + m 2 )(25m 2 + 10m) = 0 100m 4 + 40m 3 + 4m 2 4(25m 2 + 10m + 25m 4 + 10m 3 ) = 0 96m 2 40m = 0 8m(12m + 5) = 0 So m = 0, 5/12 and the equations of the lines we want are y = 0 and y = (5/12)(x 5).
It is always good to check your results graphically.
Announcements 1 scientific calculator required 2 graphing calculator not allowed for exams 3 notes available at http://www.math.washington.edu/ hoffman/124.html
Parametric Equations Parametric equations give the location of an object as it moves. At time t the object is at (5 cos(t), 5 sin(t), t/5).
In this class we will use parametric equations for equations of lines graphs of functions and circular motion.
Skeet Shooting At the shooting range a shell is fired. It follows the trajectory x 1 (t) = 200t and y 1 (t) = 70t 16t 2. One second later a shooter fires a bullet whose trajectory is x 2 (t) = 100+400(t 1) and y 2 (t) = 75(t 1) 16(t 1) 2. Does the bullet hit the shell? If so, where?
If they hit both the x and y coordinates must be the same. Setting x 1 = x 2 we get 200t = 100 + 400(t 1) = 500 + 400t so t = 2.5. Then we check to see if y 1 (2.5) = y 2 (2.5). but y 1 (2.5) = 70(2.5) 16(2.5) 2 = 175 100 = 75 y 2 (2.5) = 75(2.5 1) + 16(2.5 1) 2 = 112.5 36 = 75.5. The bullet misses by half a foot.
Running around a football field A football field is a rectangle that is 360 feet long and 160 feet wide. There are goalposts in the middle of each of the short sides. A football player starts at one of the goal post and jogs around the edge of the field at a rate of 10 feet/second. 1 At what times does he make it to the four corners of the field? 2 Find parametric equations for his position as a function of time. 3 Find an equation for the distance between him and the initial goalposts as a function of time.
Running around a football field First we draw a picture Then we put down a coordinate system. We are will use a multi-part function for the parametric equations. For the last part we use the distance function.
Running around a football field It takes the player 16 seconds to run the width of the field and 36 seconds to run the length of the field. He will make it to the corners after 8, 44, 60 and 96 seconds and back to the goalposts after 104 seconds. Our parametric function for the x-coordinate will look like x 1 (t), 0 t < 8; x 2 (t), 8 t < 44; x(t) = x 3 (t), 44 t < 60; x 4 (t), 60 t < 96; x 5 (t), 96 t < 104.
Running around a football field All of the functions will be linear. Plugging in values at the endpoints we see x(t) = 0, 0 t < 8; 10(t 8), 8 t < 44; 360, 44 t < 60; 360 10(t 60), 60 t < 96; 0, 96 t < 104.
Running around a football field All of the functions for y(t) will also be linear. Plugging in values at the endpoints we see y(t) = 80 10t, 0 t < 8; 0, 8 t < 44; 10(t 44), 44 t < 60; 160, 60 t < 96; 160 10(t 96), 96 t < 104.
Finally we use the distance formula d(t) = = x(t) 2 + (y(t) 80) 2 10t, 0 t < 8; (10(t 8)) 2 + 80 2, 8 t < 44; 360 2 + (10(t 44)) 2, 44 t < 60; (360 10(t 60) 2 + 80 2, 60 t < 96; 80 10(t 96), 96 t < 104.
Circular Motion The parametric equations ( ) 2π x(t) = A cos B t C + D 1 ( ) 2π y(t) = A sin B t C + D 2 describe motion around a circle of radius A and center (D 1, D 2 ).
We can check that equations trace out a circle. ( ) distance (x(t), y(t)), (D 1, D 2 ) = (x(t) D 1 ) 2 + (y(t) D 2 ) 2 ( ( )) 2π 2 ( ( )) 2π 2 = A cos B t C + A sin B t C = A 2 = A
What role do the constants have? (D 1, D 2 ) give the x and y coordinates of the center A is the radius of the circle B is the period the sign of B determines the direction of motion C determines the initial position.
Ferris wheel A ferris wheel does ten revolutions counterclockwise in 8 minutes at a constant rate. At the lowest point the seats are three feet off the ground. At the high point they 113 feet high. Diego s seat starts at the bottom when the revolutions begin. Find parametric equations for the location of Diego s seat.
First we set up a coordinate system with the ground being y = 0 and x = 0 is the middle of the ferris wheel. The middle of the circle is at x = 0 and y = 1 2 (3 + 113) = 58 so D 1 = 0 and D 2 = 58. The radius of the circle is A = 1 2 (113 3) = 55. The wheel makes ten revolutions in 8 minutes so the period B = 8/10 =.8. Thus the parametric equations for Diego s seat are (( ) ) 2π x(t) = 55 cos t C.8 and y(t) = 55 sin (( ) ) 2π t C + 58.8
To find C we use that at time t = 0 Diego s seat is at the lowest point so (( ) ) 2π y(0) = 55 sin 0 C = 55 sin ( C) + 58 = 3.8 and sin( C) = 1 and C = π/2. Thus the parametric equations for Diego s seat are (( ) ) 2π x(t) = 55 cos t π/2.8 and y(t) = 55 sin (( ) ) 2π t π/2 + 58..8
Bicycle Wheel A rock is stuck in the tire of a bicycle. The bicycle tires have a diameter of 27 inches and is moving at a rate of 10 feet per second. The rock first touches the ground after.5 seconds. Find the parametric equations for the location of the rock as a function of time.
First we need to pick a consistent set of units to work with. We will use feet for all distances so the diameter of the wheel is 2.25 feet and the radius is 1.125 feet. Next we set up a coordinate system. The problem is just like the last one except that the center of the wheel is moving The parametric equations for the center of the wheel are x(t) = 10t and y(t) = 1.125.
The wheel has done a full revolution every time it moves 2.25π feet (the circumference of the wheel). This takes B = 2.25π 10 = 9π 40 seconds. Thus the parametric equations for the rock are (( ) ) 80 x(t) = 10t + 1.125 cos t C 9 and (( ) ) 80 y(t) = 1.125 + 1.125 sin t C 9
To find C we know that y(.5) = 0. This gives us 0 = y(.5) (( ) ) 80 0 = 1.125 + 1.125 sin.5 C 9 (( ) ) 80 1 = sin.5 C 9 π/2 = 40 9 C C = π/2 40 9 C 2.873
Thus the parametric equations for the rock are (( ) ) 80 x(t) 10t + 1.125 cos t + 2.873 9 and (( ) ) 80 y(t) 1.125 + 1.125 sin t + 2.873 9
Instantaneous Velocity If an arrow is shot upward on the moon with a velocity of 58 meters per second. It height in meters is given by h(t) = 58t 0.83t 2. 1 Find the average velocity over the time intervals [1, 2], [1, 1.5], [1, 1.1], [1, 1.01], and [1, 1.001]. 2 Find the instantaneous velocity at t = 1.
Average Velocity over [1, t] h(t) h(1) t 1 t 2 55.51 1.5 55.925 1.1 56.257 1.01 56.3317 1.001 56.339 As t approaches 1 the average velocity approaches 56.34. We say the instantaneous velocity at t = 1 is 56.34 and write h(t) h(1) lim = 56.34. t 1 t 1
All of our work over the next few weeks will be trying determine instantaneous velocity for an arbitrary function. In order to do this we first must discuss what we mean by the limit of a function as we approach a point. Next week we will start with this.