# 10.4 OFSA Solutions

OFSA SOLUTIONS for Parametrics

This page contains peer generated solutions and error explanations to OFSA questions. As you read or view the solutions, be critical: check for accuracy, but also for more efficient solution strategies. If you have a better method or different idea/answer, post a discussion and monitor the responses.

Quick Directions
• Post answers, solutions and error explanations to each OFSA question below.
• For each "distractor" or incorrect answer choice, explain the error that would lead to that incorrect answer choice.
• You may either do the above in typed format or using a pencast.
• Separate each question with a section bar.
• After each solution, provide a hyperlink back to the corresponding OFSA page.

Question 1
Given
$$x=3sint$$
and
$$y=5cost$$
What is the shape of the x-y graph and how does the particle travel as time goes on?

a. ellipse, counterclockwise
b. line, oscillates- up then down
c. ellipse, clockwise
d. line, oscillates- down then up

Solution 1
$$x=3sint$$
and
$$y=5cost$$
so,
$$\frac{x^{2}}{9}=sin^{2}t$$
and
$$\frac{y^{2}}{25}=cos^{2}t$$
If you remember back to the trig identities,
$$sin^{2}t+cos^{2}t=1$$
so if we add together the equations we should get
$$\frac{x^{2}}{9}+\frac{y^{2}}{25}=sin^{2}t+cos^{2}t=1$$

This graph is an ellipse, so we can rule out option b and d

The orientation of the graph can be found by finding where t=0 (starting point) and substituting other t values in to visualize the direction the particle is moving.

when t=0, x=0, y=5

when t=-1, x=-.05

when t=1, x=.05

The orientation can also be found by finding the starting point, then looking at the two parametric equations we were given. on the x-t graph, x increases right after the start. On the y-t graph, y decreases just after the start. By using these facts we can conclude that the particle is moving clockwise around the ellipse.

Therefore, the correct answer is c.
*a correct, full description of the motion:
The particle begins at (0,5) and continues on the path defined by
$$\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$$
in a clockwise direction and continues indefinitely.*

Error Explanation 1

a. the student found the correct graph, but assumed that the direction was counterclockwise because the majority of circles we have dealt with in the polar, trig identities, and unit circle problem all moved counterclockwise.
b. the student mis-used the pythagorean identity, instead believing that
$$sint+cost=1$$
then they only used the direction of the x-t graph to find the orientation.
c. correct!
d. the student mis-used the pythagorean identity, instead believing that
$$sint+cost=1$$
then they only used the direction of the y-t graph to find the orientation.

Question 2
Two ships, the USS Lizzy and the USS Izzy are headed toward each other on path defined by:

USS Lizzy:
 [x=t]

[x=t]

and
 [y=t+30]

[y=t+30]

USS Izzy:
 [x=80-t]

[x=80-t]

and
 [y=t+10]

[y=t+10]

When will the two paths cross, but without colliding?

a. never
b. at (30,60) when USS Lizzy's t= 50
c. at (60,90) when USS Izzy's t= 30
d. at (30, 60) when USS Izzy's t= 50

Solution 2

1) eliminate the parameter so that:
USS Lizzy:

$$y=x+30$$

USS Izzy:

$$y=90-x$$

2) Treat this as a system of equations and solve, using substitution.

$$y=x+30=90-x$$

$$2x=60$$

$$x=30$$

$$y=60$$

3) Substitute either the x or y value back into USS Izzy's equations to find that t=50 .
Therefore the solution is d.

Error Explanation 2

a. the student tried to use the method for finding simultaneous solutions, by setting x1=x2 and y1=y2. There is no simultaneous solution.

b. the student found the correct point of intersection, but substituted either the x or y value into Izzy's equation, the wrong one. The time does not match with the boat.

c. the student made an algebraic mistake when solving the system of equations. They said:

$$y=x+30=90-x$$

$$2x=120$$

$$x=60$$

d. correct!

Question 3

Given:
$$x=\frac{2}{t-3}$$

and

$$y=\frac{1}{t+5}$$

eliminate the parameter, and choose the correct graph.

a.

b.

c.

d.

Solution 3

1) To eliminate the parameter, the most efficient way is to isolate the t in the x-t equation.

$$x=\frac{2}{t-3}$$

$$t=\frac{2}{x}+3$$

2) Then substitute the t in the y-t equation

$$y=\frac{1}{t+5}$$

$$y=\frac{1}{\frac{2}{x}+8}$$

3) simplify to:

$y=\frac{x}{\2+8x}$

This is a review on graphing rational functions

VA=
$\frac{-1}{4}$
(set denominator equal to 0 and solve)

HA=
$$\frac{1}{8}$$
(set the numerator's x term with the highest degree over the denominator's x term with the highest degree)

Intercept= (0,0)
(set equation equal to zero and find x value; this is the x and the y intercept)

Therefore the correct answer is d.

Error Explanation 3
a. the student graphed only the x-t equation

b. the student made an error and added the x-t and y-t equations, a method used to eliminate the parameter with trig functions. The student graphed

$$y=\frac{3x+7}{(x-3)(x+5)}$$

c. the student graphed only the y-t equation
d. correct!

Question 4
A soccer ball is traveling across the ice and is parametrically described by: x=2t+1and y=t-4. Ali is running across the field in hopes of intercepting the ball. The position of the tip of Ali's cleat is parametrically defined by: x=3t-10 and y= -t+3
Does Ali intercept the soccer ball? If yes, when?

Solution 4
SIMULTANEOUS SOLUTION/COLLISION OF BALL AND FOOT
1. let X=X
2t+1=3t-10
t=11
2. let Y=Y
t-4=-t+3
t=
$$\frac{7}{2}$$
3. t:
$$11\neq \frac{7}{2}$$
NON-SIMULTANEOUS SOLUTION
Since the two paths intersect at different times, there is no collision and the paths cross at a non-silmultaneous solution. But where?
1. ball:
$$y=\frac{1}{2}x-\frac{9}{2}$$
2. foot:
$$y=-\frac{1}{3}x-\frac{1}{3}$$
3.
$$\frac{1}{2}x-\frac{9}{2}=-\frac{1}{3}x-\frac{1}{3}$$
4. x=5 and y=-2
non-simlutaneous solution occurs at (5, -2)
Error Explanation 4
a. After solving for a simultaneous solution,only a non-simultaneous solution was found.
$$11\neq \frac{7}{2}$$
so there are no simultaneous solutions.
b. After solving for a simultaneous solution,only a non-simultaneous solution was found.
$$11\neq \frac{7}{2}$$
so there are no simultaneous solutions.
c. Because there is no simultaneous solution, there is no collision. By subtracting the time the foot reaches the same spot the ball had been from the time the ball was there, it still does not create a simultaneous solution.
d. Correct!

Question 5
Given
$$x=\sqrt{t+4}+1$$
and
$$y=3\sqrt{t+4}-4$$
Eliminate the parameter to find any restrictions of t, x, and y (if any).

Solution 5
1. solve for t for one of the equations:
$$x=\sqrt{t+4}+1$$
$$(x-1)^{2}-4=t$$
$$y=3\sqrt{((x-1)^{2}-4))+4}-4$$
$$y=3\sqrt{(x-1)^{2}}-4$$
$$y=3x-7$$
2. you know the final number under the radical cannot exceed lower than -4 so substitute -4 into both the x=t and y=t equations
$$x=\sqrt{-4+4}+1$$
$$x=1$$

$$y=3\sqrt{-4+4}-4$$
$$y=-4$$

Error Explanation 5
a. Because the parameter 't' is inside a square root, it is impossible that t can extend beyond -4. Under the square root, the number cannot be negative, and since t is not squared, a negative will stay negative.
b.When the parameter ''t' has been limited to t:[-4,∞), the x and y values must coordinate with the t values. By substituting the value of -4 for t, the restriction on the x value becomes 1 and on y it becomes -4.
c. Correct!
d. The parameter 't' can extend until -4, making t:[4,∞) an answer forgetting the range of numbers from [-4,4]

Question 6
A football is thrown from a rooftop, 53 feet above the ground with an initial velocity of 26 ft/s and at an initial angle of 33° above the horizontal.
Create parametric equations to model the position of the ball in terms of the time 't' to find the horizontal distance to ball will travel before hitting the ground. (Calc. allowed!)

Solution 6
1. create parametric equations to model the football's position
$$x(t)=(26\cos 33)t$$
and
$$y(t)=(26\sin33)t+53-16t^{2}$$
2. make
$$y(t)=0$$
$$0=-16t^{2}+14.16t+53$$
$$t=-1.43$$
$$t=-2.32$$
because time can't be negative, the only option for the time when the football reaches the ground is 2.32 seconds.
3. plug 't' into the parametric 'x' equation to solve for the horizontal distance
$$x= (26\cos33)2.32$$
$$x= 50.59$$

Error Explanation 6
a. While on Parametric mode, this is the negative vertical height the calculator's table reads when the maximum horizontal distance has been reached.
b. The process was done in reverse, rather than plugging 0 into the y equation and 't' into the x equation, 0 was plugged into the x equation and the newfound 't' into the y.
c. This answer is not only saying the football was thrown backwards, but also that it was thrown a VERY short instance. This answer could be achieved through using the graphing mode in the calculator while in radians instead of degrees. Obviously the answer is impractical, but this shows a clear difference between radians and degrees.
WHEN A PROBLEM USES DEGREES FOR THETA MAKE SURE YOUR CALCULATOR IS IN DEGREE MODE!!
d. Correct!

Question 7 - Simultaneous Solutions
$\\ \text{Find the values of x and y at each simultaneous intersection along with the value of t. Use radians.} \\ t \in \left[0, \pi\right] \\ x_1 = 10\sin(6t), \; y_1 = 4\cos(4t) + 3\\ x_2 = 10\cos(6t), \; y_2 = 3\cos(4t) + 3\\$

Solution 7
$\\ \text{1) let } x_1 = x_2 \\ 10\sin(6t) = 10\cos(6t) \\ \sin(6t) = \cos(6t) \\ \tan(6t) = 1 \\ \\ \text{2) let } y_1 = y_2 \\ 4\cos(4t) = 3\cos(4t) \\ cos(4t) = 0 \\ \\ \text{3) list possible t values for both x and y in the restriction of t.} \\ \tan(6t) = 1 \\ 6t \in \frac{\pi}{4} + \pi k \\ t \in \frac{\pi}{24} + \frac{\pi/6} k \\ t = {\frac{\pi}{24}, \; \frac{5\pi}{24}, \; \frac{3\pi}{8}, \; \frac{13\pi}{24}, \; \frac{17\pi}{24}, \; \frac{7\pi}{8}} \\ \\ \cos(4t) = 0 \\ 4t \in \frac{\pi}{2} + \pi k \\ t \in \frac{\pi}{8} + \frac{\pi}{4} k \\ t = {\frac{\pi}{8}, \; \frac{3\pi}{8}, \; \frac{5\pi}{8}, \; \frac{7\pi}{8}} \\ \\ \text{4) Find t values that work for both the x-equations and y-equations and find the corresponding (x,y) coordinates by plugging it back into the x-t and y-t equations} \\ t = {\frac{3\pi}{8}, \frac{7\pi}{8}} \\ x = 10\sin(6t) = {5\sqrt{2}, -5\sqrt{2}} \\ y = 3\cos(4t) + 3 = {3, 3} \\ \\ \boxed{\text{C) two solutions: } (5\sqrt{2}, 3) \text{ at } t=\frac{3\pi}{8} \text{ and } (-5\sqrt{2}, 3) \text{ at } t=\frac{7\pi}{8}}} \\ \\$
Error Explanation 7
$\\ \text{a) Check for errors with trig functions} \\ \text{b) Check for all solutions within the restrictions given, don't stop at just the first simultaneous solutions} \\ \text{c) Correct Answer} \\ \text{d) Check for errors with trig functions} \\$

Question 8
Minimum Distance
Tom and Jerry are at it again! Jerry is trying to cross a square room, which has a diagonal from (0,0) to (100,100). Tom's motion is described by the following equations: x = t + 17 and y = 5t^2 - 2t. Jerry's movement is described by: x = 3t - 1and y = 36t + 63. Will Jerry be captured by Tom? If so, find the x and y coordinates at the intersection.

Solution 8
$\\ \text{1) Let x1 = x2} \\ t + 17 = 3t - 1 \\ 18 = 2t \\ t = 9 \\ \\ \text{2) Let y1 = y2} \\ 5t^2 - 2t = 36t + 63 \\ 5t^2 - 38t - 63 = 0 \\ (t - 9)(5t + 7) = 0 \\ t = 9, \; \frac{-7}{5} \\ \\ \text{3) Find t values that solve both the x equations and the y equations} \\ \box{t = 9} \\ \\ \text{4) Find the intersection} \\ x = t + 17 = 26 \\ y = 5t^2 - 2t = 387 \\ \\ \text{5) Answer the question. Will Jerry be captured by Tom?} \\ \text{NO because the only simultaneous solution occurs outside of the dimensions of the room}$
Error Explanation 8
$\\ \text{a) Yes, at (26, 27) when t = 9 Error: removing the square from the first y equation}\\ \text{b) Yes, at (26, 387) when t = 9 Error: check restrictions on each variable!} \\ \text{c) No Correct!} \\ \text{d) Yes, at (387, 26) when t = 9 Error: flipping x and y AND checking restrictions on each variable}$

Question 9
An object's path is defined as x = t(t - 3)^2, y = 3. Find the interval of t where the object is moving to the right (increasing).
 Graph Plot

Solution 9
1) Graph the x - t function
See graph to right

2) Use calculator to determine maximums and minimums
maximum (t,x) = (1, 4)
minimum (t,x) = (3, 0)

3) Recognize which parts of the x-t equation are increasing

Error Explanation 9
$\\ \text{a) } t \in \left(-\infty, 1\right) \cup \left(3, \infty\right) \text{Correct} \\ \text{b) } t \in \left(3, \infty\right) \text{Error: misread x-t function as (t-3)^2} \\$
c) Never, it's always constant Error: Using t-y (vertical position) rather than t-x (horizontal position)
d) Never, the object is always moving down Error: Graph does not see the entire picture; 1 <= domain <= 3

Question 10
Solution 10

$\bg_white \fn_phv x_{1}=x_{2}$
3t-4 = 2t+5
t = 9

$\bg_white \fn_phv y_{1}=y_{2}$
2t-7 = t+2
t = 9

x = 3(9) - 4 = 23
y = 2(9) - 7 = 11

Yes, the man will step on the bug at (23, 11) when t = 9 seconds.

Error Explanation 10
a) Incorrect...There is a simultaneous solution
• I decided this could be a mistake a person can make if they have minor calculation errors
b) Incorrect...The simultaneous solution does not occur at (-13, -13)
• This mistake would be that a person accidentally set x1 = y1 and x2=y2 and solved for t
• t still came out to be the same (t = -3) and the mistake was that the person plugged -3 into the first equations they saw without checking the other pair
c) Incorrect...The simultaneous solution does not occur at (9, 9)
• The mistake for this would be that after solving for t by setting x1 = x2 and y1 = y2, the person got t = 9 but thought that this was the answer since the answer came out the same
d) Correct...Following the procedure for solving for simultaneous solutions produces a SS at (23, 11)

Question 11
Solution 11
Graph the parametric equations first to know the directional changes in motion.

t-x graph
 Graph Plot

The x-position is constant, which produces rectilinear motion. The object will always stay on the line x = 7.

t-y graph
 Graph Plot

The y-position varies, but it is directional change that stays on the same straight path.

x-y graph
 Graph Plot

The object travels vertically upward from (7, 0) to (7, 32), back down to (7, 0), and back up to (7, 7)

Error Explanation 11
a) Correct description of motion of the object, with correct direction and specificity of the turning points
b) Incorrect...Since x = 7 (and is therefore a constant), the object must travel vertically
• The mistake for this would be that the person saw the t-x graph and mistakenly thought that the object moved horizontally since the vertical x axis on the t-x graph showed a horizontal straight line
• The correct key points were switched around (such as from (x,y) to (y,x) because of the previous mistake)
c) Incorrect...The t-y graph shows that the y-position of the object changes direction multiple times
• This is a mistake from only plugging in the restrictions for t (that is, -2 and 5) into the y-equation and assuming that there are no directional changes in between the starting and end points
d) Incorrect...The object travels a total distance of 71 units (32 up + 32 down + 7 up) but is displaced only 7 units
• It is easy to mix up distance and displacement, which makes this a possible mistake that can be made

Question 12
Solution 12
Oscillates between (1, 1) and (5, 7)
x position oscillates between 1 and 5
y position oscillates between 1 and 7

t-x graph
 Graph Plot

t-y graph
 Graph Plot

x = -2cos((pi/4)t) + 3
y = -3cos((pi/4)t) +4

period

# (2pi)/B

(2pi)/(pi/4) = 8

Distance between (1, 1) and (5, 7)
=
$\small \bg_white \fn_cm \sqrt{(5-1)^2+(7-1)^2}$
=
$\small \bg_white \fn_cm 2\sqrt{13}$

Since the restriction is t: [0, 8] and the object travels one whole length in t = 4, we need to multiply $\small \bg_white \fn_cm 2\sqrt{13}$ by 2

So the correct answer is $\small \bg_white \fn_cm 4\sqrt{13}$

Error Explanation 12
a) Incorrect...
• The mistake would be that the person forgot to multiply the distance by 2 because they figure they found the distance traveled by finding the distance between the two points
b) Incorrect...
• The person is thinking of displacement and thinks that since the object ended where it started it did not travel any distance
c) Correct...
• Took into account the two-way trip
d) Incorrect...
• Either mathematical error or accidentally thinking that it asks for the distance traveled between the interval [0, 8] and that since t = 4 is half of 8 the answer is double, or 2

Question 13
Which description of motion below is not supported by the given parametric equations?
$\left\{\begin{matrix} x(t) = 2 \\ y(t)=(t-1)(t+2)^2 \end{matrix}\right. ; -5 \leq t \leq 5 \\ \\ \text{A) increases in y-coord to 0} \\ \text{B) increases in y-coord to 149} \\ \text{C) decreases in y-coord to -4} \\ \text{D) increases in y-coord to -4} \\$

Solution 13

• graphing xt yields a domain of [-5,5] and a range of [2].
• when graphing yt, make sure you calculate the y-values for every significant t-value, in this case they are t=-5, -2, 0, 1, and 5.
• once those values are calculated, you now know that the domain is [-5,5] and the range is [-54,149].
• with these two graphs, you can now graph the motion of the particle:

• using the significant y-values, you can properly describe the motion of the particle.
• as for the answer, choices A, B, and C are supported, while choice D, "increases in y-coord to -4", is not supported. rather, the particle decreases in y-coordinate to -4.

Error Explanation 13
A) Correct! From t: [-5,-2], the particle is increasing in y-coordinates to 0.
B) Correct! From t: [0,5], the particle is increasing in y-coordinates to 149.
C) Correct! From t: [-2,0], the particle is decreasing in y-coordinates to -4.
D) Error. Student must have seen first increase in y-coord from -54 to 0 and assumed it ended at -4 instead.

Question 14
Robbie Gould, a place kicker for the Chicago Bears, kicks a football in a game with the Detroit Lions. The ball leaves the ground with a velocity of 89 ft./s at an angle of 63 degrees above the horizontal. If the ball is kicked from the Bears' 30 yard line and is aimed straight down the field, where does it land?

 photo.JPG

a) Between the 1 and 2 yard line
b) Between the 3 and 4 yard line
c) Between the 4 and 5 yard line.
d) Off the field!

Solution Explanation:
 boxed{x = x_0 + (|v_0|cos(theta))t}

 boxed{y = y_0 + (|v_0|sin(theta))t + frac{1}{2}(g)(t)^2}

1) Write two parametric equations to model this situation.

x=89cos63t
y=89sin63t - 16t^2

2) We want to solve for t, the common variable between both equations. T represents the time the object is a projectile, or is in flight. We can substitute 0 in for y in the
y=89sin63t - 16t^2 equation because that is the final position of the ball, on the ground at a height of zero.
0=89sin63t - 16t^2
16t^2 = 89sin63t
16t = 89sin63
t is approximately 4.956 seconds.

3) To figure out where the ball lands, we must find how far the ball travels horizontally. This can be determined using the x= equation. Substitute the t value we just calculated, 4.956 seconds. We know that distance= rate x time, and that is what this equation represents. The length traveled is equal to the horizontal speed of the ball x the time the ball travels.

x= 89cos63(4.956)
x=200.257 ft

The field is measured in yards, so to convert this to the correct unit we divide by 3, as a yard consists of 3 feet. We arrive at the conclusion that the ball travels approximately 66.75 yards.

4) The seperation between the lines of the field is 10 yards. Since the ball travels 66.75 yards, we know it crosses about 6.675 yard lines or between 6 and 7 yard lines. The ball begins at the Bears' 30 yard line, so we count from there 6-7 yard lines (For the non-football experts, refer to the picture!). The ball lands between the 40 and 30 yard line on the Lions' side, aka the 3 and 4 yard lines.

Therefore the correct answer is B, the ball will land between the 3 and 4 yard lines.

Errors Explained
a) Calculation error. An error of 7.42 seconds for t or 300 ft. for distance traveled would have resulted in this answer.
b) CORRECT, No errors :)
c) If you did this problem in Radian mode on the calculator, the time traveled will be significantly smaller which therefore decreases the distance traveled to about 22 yards.
d) If you used the wrong units in the y= equation and used 9.8 as g instead of 32, you will have gotten that the ball was in the air for 16.5 seconds! Therefore the ball would have traveled a whopping 222 yards, taking it off the football field.

Question 15
 second problem.JPG

Write the parametric equations, t-x and t-y, that produce this rectangular graph.

A) x=3sin(π/6t), y= -3cos((π/6t)
B) x= 3sin(π/8t), y= 3cos(π/8t)
C) x= 3sin(π/8t), y= -3cos(π/8t)
D) x=-3cos(π/8t), y=3sin(π/8t)

Solution explanation:
1) First write an equation that models the x behavior of the particle in relation to time. This will be your t-x graph.
• We see that the particle at t=0, begins its motion at x=0. As time progresses it moves from zero to its max value, back to zero, to its min value, then back to zero. This oscillation reflects the shape of a sin curve. (*Tip: It's helpful to draw out the movement of the t-x graph to help determine what function it represents!)
• A generic sin curve oscillates between -1 and 1, but this graph oscillates between -3 and 3, so we know that the amplitude is 3 to stretch the curve vertically.
• Define the cycle, or B. B= 2π/Period. The period of the motion is 16 seconds because it takes 16 seconds for the particle to return back to its starting position. Therefore B is 2π/16, or π/8.
• Put together all these components of the equation...x=3sin(π/8t).

2) Next, write an equation that models the y behavior of the particle in relation to time. This will be your t-y graph.
• We see that the particle begins at y=-3, the minimum value. As time progresses it moves from the min value, to zero, to the max value, to zero, then back to the min value. This oscillation reflects the shape of a negative cosine curve. (*Tip: It's helpful to draw out the movement of the t-y graph to help determine what function it represents!)
• A generic sin curve oscillates between -1 and 1, but this graph oscillates between -3 and 3, so we know that the amplitude is 3 to stretch the curve vertically.
• Define the cycle, or B. B= 2π/Period. The period of the motion is 16 seconds because it takes 16 seconds for the particle to return back to its starting position. Therefore B is 2π/16, or π/8.
• Put together all these components of the equation....y= -3cos(π/8t).

Errors Explained:
a) Incorrect period. May have seen t=12 being the largest t value and assumed that was the total period but remember, it must continue back to t=0.
b) May not have realized that the y movement is a NEGATIVE cosine curve as it begins from the min vs. the max.
d) The x and y are switched...read carefully!

Question 16
Solution 16
Error Explanation 16

Question 17
Find the simultaneous solution for the following:
{x = 5t -6
{y= t^2 + 1

{x= 2t + 9
{y= T^2 + 1

Solution 17
a) (19, 26) when t= 5.

5t-6= 2t + 9
t= 5

and

T^2 + 1 = 31 - T
T^2 + t - 30 = 0
(t-5) (t+6) = 0
t= 5, -6
From setting the x equations and y equations equal to each other, both gave the solution t =5. But wait, now find the point. When subbing back in the point is (19, 26).

Error Explanation 17
b) the t= 5 is correct but the x and y in the coordinate are flipped. Be careful!
c) t= -6 is one of the solutions in the procedure but it is not simultaneous since both of the problems did not give a solution of -6.
d) This is the answer to the non-simultaneous intersection. If this is your answer you are using the non-simultaneous solution procedure instead of the simultaneous solution procedure.

Question 18

For t values [0, 5], a particle travels along the following path:
y= ( t- 5)( t- 1)^2

What is the total displacement and total distance traveled by the particle?

a) Displacement 5, Distance 25
b) Displacement 0, Distance 20
c) Displacement 25, Distance 5
d) Displacement 5, Distance 5

Solution Explanation:
Remember the meaning of both displacement and distance!
Displacement=the difference between the end position and beginning position
Distance=the total units the particle travels under the time restrictions

The graph of this particle is this:

(*The horizontal axis is defined by t and the vertical by y.)

• Displacement: At t=0, the particle begins at -5. At t=5, the particle ends at 0. Displacement is 5, because it is the difference in beginning and end positions.
• Distance: At t=0 the particle begins at -5. During this time restriction, the particle travels 5 units to go back to zero, 10 units to -10, and 10 units back up to zero. If you add up the total units the particle moved (5+ 10+ 10), the distance is 25.

Answer A is correct: Displacement 5, Distance 25.

Errors Explained:
b) Calculated amounts over the wrong time restriction [1,5]
c) Does not understand the definitions of displacement and distance. Probably got the two confused in meaning because the answers are switched.
d) Correct displacement but incorrect distance. May have calculated distance along horizontal axis but remember, that defines time. The motion is described by the vertical or y axis.

Question 19 USE CALC
Solution 19
c) the is a solution at (0,5) when t= 5
on calculator find the intersections for the following:
3 sin (πt) = t-2
You will get t= 2, 1.097
3 cos (πt) + 2 = t + 3
t= 2 will also be a solution
with t= .351
Therefore, there is a solution at t=2. Sub back in to get (0,5)

Error Explanation 19
a) if you did not get a sim solution, the graph or math was incorrect
b) the simultaneous solution can be determined with a calculator.
d) this would be the non-simultaneous solution

Question 20
Given the parametric equations below, what is the orientation of the graph and in which direction will the particle be traveling? (assume t: all real numbers)
$\left\{\begin{matrix}x=3cos(\frac{\Pi }{6}t) \\ y=-sin(\frac{\Pi }{6}t) \end{matrix}\right. \\ \\ \text{A) line; oscillating left to right} \\ \text{B) line; oscillating right to left} \\ \text{C) ellipse; CW} \\ \text{D) ellipse; CCW} \\$

Solution 20

• graphing both parametric equations yields these domains and ranges:
• x: [-3,3], t: all real numbers
• y: [-1,1], t: all real numbers
• note the trends of each graph to figure out the orientation!
• as t increases, x decreases to -3 then increases to 3 then decreases again
• as t increases, y decreases to -1 then increases to 1 then again
• graph the x-y equation; eliminating the parameter may take too long, so just make a small table of points:
• you can plot points and follow through with a full graph based on the xt and yt graphs

• as you can see, this graph correctly follows the trends we stated earlier, with x decreasing then increasing and y decreasing then increasing then decreasing again, all starting at t=0 @ (3,0)
• therefore, the correct answer is choice C) CW.

Error Explanation 20
A) Error. The graph is an ellipse, not a line, or the student chose this because the direction made sense.
B) Error. The graph is an ellipse, not a line, or the student chose this because of the direction, but incorrectly graphed the xt / xy equations.
C) Correct!
D) Error. Student must have incorrectly graphed the xt / xy equations.

Question 21
Calculator allowed! Determine the non-simultaneous point of intersection of the paths.
$\left\{\begin{matrix}x=\sqrt{t} \\ y=2t+3 \end{matrix}\right.$
and
$\left\{\begin{matrix}x=\sqrt{6-t} \\ y=t^2 \end{matrix}$
$\\ A) \ t=3, (\sqrt3,9) \\ B) \ (i, 1) \\ C) \ (3.32,25) \\ D) \ \text{There are no non-simultaneous solutions.}$

Solution 21

• set both xt equations equal to each other and solve for t.
• set both yt equations equal to each other and solve for t.
• since both solutions yielded t=3, this means t=3 is part of a simultaneous solution.
• since we are solving for non-simultaneous solutions, we will resume and ignore the t=-1.

• eliminate the parameter for both sets of xt / xy equations then graph both equations and find the intersection!
• note that the non-simultaneous solution is located at the point where:

$x \neq \pm \sqrt3 \\$
• this is because the simultaneous solution lies there and we are only looking for the non-simultaneous solution!

• therefore, the correct answer is choice C) (3.32, 25)

Error Explanation 21
A) Error. This is the simultaneous solution. Correct answer if question was asking for the simultaneous solution!
B) Error. If a student truly did not know what they were doing, or they were running out of time and had to guess, they had a greater chance of choosing this answer. The t=-1 seems like it would be considered as a "non-simultaneous solution" and with correct algebra plus mindless substitution, this choice would, theoretically, make a good, logical choice! (or at least, that's what I'm thinking)
C) Correct!
D) Error. If the student was choosing between this and choice B, this answer would be chosen because you are unable to graph imaginary numbers, (and it would seem that graphing imaginary numbers wouldn't make sense and is totally out of the context of this unit).