06 particle motion presenter notes - wag &...

Copyright © 2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org

AP Calculus

ParticleMotion

PresenterNotes

2016‐2017 EDITION

Copyright © 2016 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org 1

StudentStudySession‐PresenterNotes

ThankyouforagreeingtopresentatoneofNMSI’sSaturdayStudySessions.Wearegratefulyouaresharingyourtimeandexpertisewithourstudents.Saturdaymorningscanbea“toughsell”forstudents,soweencourageyoutoincorporatestrategiesandtechniquestoencouragestudentmovementandengagement.Suggestionsfordifferentpresentationoptionsareincludedinthisdocument.Ifyouhaveanyquestionsaboutthecontentoraboutpresentationstrategies,pleasecontactMathematicsDirectorCharlaHolzbogatcholzbog@nms.orgorAPCalculusContentSpecialistKarenMikschatkmiksch@nms.org.ThematerialprovidedcontainsmanyreleasedAPmultiplechoiceandfreeresponsequestionsaswellassomeAP‐likequestionsthatwehavecreated.ThegoalforthesessionistoletthestudentsexperienceavarietyofbothtypesofquestionstogaininsightonhowthetopicwillbepresentedontheAPexam.Itisalsobeneficialforthestudentstohearavoiceotherthantheirteacherinordertohelpclarifytheirunderstandingoftheconcepts.Suggestionsforpresenting:ThevastmajorityofthestudysessionsareonSaturdayandstudentsandteachersarecomingtobeWOWed!WewantactivitiestoengagethestudentsaswellaspreparethemfortheAPExam.Thefollowingpresenternotesincludepacingsuggestions(youonlyhave50minutes!),solutions,andrecommendedengagementstrategies.Suggestionsonhowtoprepare: Thenotes/summariesonthelastpage(s)areforreference.Wewantthestudents’timeduring

thesessiontobefocusedonthequestionsasmuchaspossibleandnottakingorreadingthenotes.Asthequestionsarepresentedduringthesession,youmaywishtoreferthestudentsbacktothosepagesasneeded.Itisnotourintentforthesessionstobeginwithalectureoverthesepages.

Asyouprepare,workthroughthequestionsinthepacketnotingthelevelofdifficultyandtopicorskillrequiredforthequestions.

Designaplanforwhatquestionsyouwouldliketocoverwiththegroupdependingontheirlevelofexpertise.Somegroupswillbereadyforthetougherquestionswhileothergroupswillneedmoreguidanceandpracticeontheeasierones.Createaneasy,medium,andhardlistingofthequestionspriortothesession.Thiswillallowyoutoadjustontheflyasyougettoknowthegroups.Inmostinstances,therewillnotbeenoughtimetocoverallthequestionsinthepacket.Useyourjudgementontheamountofquestionstocoverbasedonthestudents’interactions.Remembertoincludebothmultiplechoiceandfreeresponsetypequestions.Discussionsontesttakingstrategiesandscoringofthefreeresponsequestionsarealwaysgreattoincludeduringtheday.

Theconceptsshouldhavebeenpreviouslytaught;however,bepreparedto“teach”thetopicifyoufindoutthestudentshavenotcoveredtheconceptpriorinclass.Insessionswheremultipleschoolscometogether,youmighthaveamixtureofstudentswithandwithoutpriorknowledgeonthetopic.Youwillhavetouseyourbestjudgementinthissituation.

Considerworkingthroughsomefreeresponsequestionsbeforethemultiplechoicequestions,orflippingbackandforthbetweenthetwotypesofquestions.Sometimes,iffreeresponsequestionsaresavedforthelastpartofthesession,itispossiblestudentsonlygetpracticewithoneortwoofthemandmoststudentsneedadditionalpracticewithfreeresponsequestions.


Particle Motion Student Study Session Presenter Notes

This session includes a reference sheet at the back of the packet. We suggest that the presenter spends a few minutes only on the differentiation rules (note the instructions for students to highlight and study for the introductory activity) and does not spend time going over the whole reference sheet, but may point it out to students that it is available to refer to if needed throughout the session. We suggest that students will work in pairs (depending on the size of the class)- arrange the desks prior to the start of the session. We have intentionally included more material than can be covered in most Student Study Sessions to account for groups that are able to answer the questions at a faster rate. Use our presenter notes and your own judgment, based on the group of students, to determine the order and selection of questions to work in the session. Be sure to include a variety of types of questions (multiple choice, free response, calculator, and non-calculator) in the time allotted. Notice in the solutions guide the questions are categorized as 3, 4, or 5 indicating a typical question of the difficulty level (DL) for a student earning these qualifying scores on the AP exam.

I. 10 minutes Card Match/checking for understanding and knowledge of group Allow students a few minutes to underline or highlight key words and phrases on the

back page reference sheet with definitions-let them know we will “play cards” in 5 minutes. Cards are located at the back of this presenter packet-please be sure to have the cards cut and mixed up before presentation begins so they are ready for distribution to the students.

Randomly distribute cards for particle motion—depending on size of the class perhaps distribute two to each student

Have students either walk around to find the match or start with one student holding up card to find the match, then proceed with others. This activity can be used to put students in pairs or groups for remainder of the session. Quickly point out the pairs at the front of the room or around the room to reinforce the main ideas.

II. 20 minutes Particle Motion with Derivatives Model several multiple choice questions involving derivatives, then assign certain

questions to the group or pairs. #1, 3, and 6 (average velocity) suggested as models, then 2, 9, and 11.

Play “pass the pen” with a friend to go over each with students explaining how they arrive at the answer: Pass the Pen: Ask for a student volunteer to begin pass the pen where the multiple choice questions are projected on a smart board or under a document camera to go over the answers. The student selects another student from the group to display another question, allowing that student to go up with the partner if desired. Continue the process engaging a different student or pair to present each question

Model one free response question with derivative parts only-suggest 16 a and b, then use the integration parts to transition to the integral MC particle motion questions.


III. 20 minutes Particle Motion with Integrals Model free response part of FR question (used with derivatives) for integrals-#16 c, d. Model one or two multiple choice questions for particle motion with integrals then

continue pass the pen activity or think/pair/share with selected questions. Suggest model #4 then 8, assign 7, 12, then FR 17 if time allows.

Return to FR questions and choose one to do all parts, derivatives and integrals-suggest the 2013 FR #15.


Particle Motion Solutions Multiple Choice Solutions 1. E (2003 AB25) DL: 3

3 2

2

2

( ) 2 21 72 3

( ) ( ) 6 42 72 0

6 7 12 0

3 ( 4) 0 3,4

x t t t t

v t x t t t

t t

t t t

2. C (AP-like) DL: 3 2( ) ( 5) 2( 2) ( 2)a t t t t ( ) ( 2)(2 10 2) ( 2)(3 12) 0a t t t t t t when 2, 4t t 3. A (2008 AB21/BC21) DL: 4

V is increasing when ( ) 0 ( ) 0v t a t which occurs when ( )x t is concave up, so 0 2t .

4. B (2008 AB7) DL: 4 Using Fundamental Theorem of Calculus:

1 2

0(1) (0) (3 6 )x x t t dt

13 2

0(1) 2 ( 3 )

t

tx t t

(1) 2 (4 0) 6x Alternatively:

2

3 2

3 2

3 2

( ) 3 6

( ) 3

(0) 0 6(0 ) 2

2

( ) 3 2

(1) 1 3 2 6

v t t t

x t t t c

x c

c

x t t t

x

5. D (1985 AB14) DL: 4 ( ) 0v t for all 0t therefore,

1 3

4 42 2

0 0( ) ( ) 3 5x t v t dt t t dt

43 5

2 2

0

2 2

t

t

t t

16 64 80 meters


6. C (1985 AB28) DL: 3

Average velocity of the particle is 25(3) 5(0)

153 0

s

t

.

7. B (1988 BC12 appropriate for AB) DL: 4 ( ) 3 3v t dt t C and (2) 10v

10 3(2)

4

C

C

Distance traveled from (0) 4v and (2) 10v

2

0(3 4)t dt

2

2

0

34

2

t

t

t t

6 8 14 meters 8. D (2012 AB6) DL: 3

Evaluate 3 2

06 18t t dt

9. B (2012 AB16) DL: 4 The particle is at rest when the velocity v(t) = 0. Since 2( ) ,x t t tb at ab

( ) 2 . 0 2 2 .2

b av t t b a t b a b a t t

10. C (2008 AB86) DL: 3 (3) (3) 0v x , so ( )x t has a horizontal tangent at 3t ; therefore, the only possible graphs

are C and E. From the table, (1) (1) 2v x , so ( )x t is increasing at 1t , so the answer is C. 11. C (2003 AB76) DL: 3 Using the derivative function on the calculator: ( ) ( )v t a t (4) 1.633a 12. E (2003 AB91/BC91) DL: 4 Using the Fundamental Theorem of Calculus and the integral function on the calculator:

2

1(2) (1) ln 1 2tv v dt

2

1(2) 2 ln 1 2 3.346tv dt

13. A (2003 AB83) DL: 3 Average velocity of a function on [0, 3]:

3

0

1 feet( ) 20.086

3 0 secondt te te dt

c


14. B (AP-like) DL: 4 Let the total area above the t-axis be A and the area below the t-axis be B. The particle’s

total distance on the time interval is 6

0( ) 16v t dt A B and the particle’s displacement

is 6

0( ) 4v t dt A B . Solving the system gives A = 10 and B = 6. Since the graph of

v(t)is below the axis for the interval 2 < t < 4, 4

2( ) 6.v t dt

Free Response 15. 2013 AB2

(a) Solve ( ) 2v t on 2 4t .

3.128( 3.127)t or and 3.473t

1: considers ( ) 2v t

2: 1: answer

(b)

1

0

5

0

( ) 10 ( )

(5) 10 ( ) 9.207

s t v x dx

s v x dx

1: s(t) 2: 1: s(5)

(c) ( ) 0v t when 0.536033,3.317756t

V(t) changes sign from negative to positive at time

0.536033t . V(t) changes sign from positive to negative at time

3.317756t . Therefore, the particle changes direction at time

0.536t and time 3.318( 3.317)t or .

1: considers ( ) 0v t 3: 2: answers with justification

(d) (4) 11.475758 0, (4) '(4) 22.295714 0v a v

The speed is increasing at time 4t because velocity and acceleration have the same sign.

2: conclusion with reason


2

3

3

16. (1999 AB1) (a) 2(1.5) 1.5sin(1.5 ) 1.167v Up, because (1.5) 0v

1: answer and reason

(b) 2 2 2( ) ( ) sin 2 cosa t v t t t t (1.5) (1.5) 2.048 or 2.049a v No, v is decreasing at 1.5

because (1.5) 0v

1: (1.5)a 1: conclusion and reason

(c) ( ) ( )y t v t dt

2

2 cossin

2

tt t dt C

1 7

(0) 32 2

y C C

21 7( ) cos

2 2y t t

1 7

(2) cos 4 3.826 or 3.8272 2

y

1: ( ) ( )y t v t dt

1: 21( ) cos

2y t t C

1: (2)y

(d) distance = 2

0( ) 1.173v t dt

or 2( ) sin 0v t t t

0t or 1.772t

(0) 3; 4; (2) 3.826 or 3.827y y y

(0) (2)y y y y

1.173 or 1.174

1: limits of 0 and 2 on an integral of ( )v t or | ( ) |v t

or uses (0)y and (2)y to compute

distance 1: handles change of direction at

student’s turning point 1: answer 0/1 if incorrect turning point


3

3

2

17. (2005 Form B AB3)

(a)5

(4) (4)7

a v

1: answer

(b) ( ) 0v t

2 3 3 1t t 2 3 2 0t t ( 2)( 1) 0t t 1, 2t ( ) 0v t for0 1t ( ) 0v t for1 2t ( ) 0v t for2 5t

1: sets ( ) 0v t 1: directionchangeat 1, 2t 1: intervalwithreason

(c) 2

0( ) (0) ln( 3 3)

ts t s u u du

2 2

0(2) 8 ln( 3 3)s u u du

8.368 or 8.369

1:2 2

0ln( 3 3)u u du

1: handlesinitialcondition1: answer

(d)2

0

1| ( ) | 0.370 or 0.371

2v t dt

1: integral1: answer


2

2

3

3

2

18. 2008 AB/BC 4 (a) Since ( ) 0v t for 0 3t and 5 6t ,

and ( ) 0v t for 3 5t , we consider 3t and 6t .

3

0(3) 2 ( ) 2 8 10x v t dt

6

0(6) 2 ( ) 2 8 3 2 9x v t dt

Therefore, the particle is farthest left at time 3t when its position is (3) 10x

1: identifies 3t as a candidate

1: considers 6

0( )v t dt

1: conclusion

(b) The particle moves continuously and monotonically from (0) 2x to

(3) 10x . Similarly, the particle moves continuously and monotonically from

(3) 10x to (5) 7x and also from (5) 7x to (6) 9x .

By the Intermediate Value Theorem, there

are three values of t for which the particle is at ( ) 8x t .

1: position at 3t , 5t , and 6t 1: description of motion 1: conclusion

(c) The speed is decreasing on the interval 2 3t since on this interval 0v and v is increasing.

1: answer with reason

(d) The acceleration is negative on the intervals 0 1t and 4 6t since velocity is decreasing on these intervals.

1: answer 1: justification

19. (2011B AB/BC5)

(a) 210 0 0.35 0.03 meters/sec

10 0 10

v va

1: answer

(b) 60

0v t dt is the total distance, in meters, that Ben rides over

the 60-second interval 0t to 60t .

60

02.0 10 2.3 40 10 2.5 60 40 139 metersv t dt

1: meaning of integral 1: approximation

(c) Because 60 40 49 9

260 40 20

B B

, the Mean Value

Theorem implies there is a time t, 40 60t , such that

2v t .

1: difference quotient 1: conclusion with

justification


Initiallymeans

whentimet=0.

Attheoriginmeans

x(t)=0.

Atrestmeans

v(t)=0.


Averagevelocityoftheparticleis

2 1

2 1

( ) ( )s s t s t

t t t

Ifthevelocityoftheparticleispositive,

thentheparticleis

movingtotheright.

Ifthevelocityoftheparticleisnegative,

thentheparticleis

movingtotheleft.


Iftheaccelerationoftheparticleispositive,

thenthevelocityisincreasing.

Iftheaccelerationoftheparticleisnegative,

thenthevelocityisdecreasing.

Speedis

theabsolutevalueofvelocity.


Ifvelocityandaccelerationhavethesame

sign,

thenthespeedisincreasing.

Ifvelocityandaccelerationhaveopposite

signs,

thenthespeedisdecreasing.

Todeterminethetotaldistance

traveledoveratimeinterval,

calculatetheareaundertheabsolutevalueofthevelocity

curve, 2

1

( )t

tv t dt .


Displacementcanbedetermined

using

2

1

( )t

tv t dt

Todeterminefinalpositionofaparticleaftermotionuse

2

112 )( ) ( ( )

t

ts t s t v t dt

06 particle motion presenter notes - wag &...

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