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Lesson 4.1: lnterpreting Derivatives in Context

lnterp.rgtation$ of Derivalives

There are three primary interpretations of derivatives;

1. The instantaneous rate of change with respect to its SdNeClr VXrdsk,

2. Rates of change in applied contexts.

3. Straight-Line Motion

lnstantaneous Ratq of CharlFe (Review)

The instantaneous rate of change is given by the slope of theto a function at a given point:

.\tngacN \is,,e

\e \tehan\aneo\rt Ya\e ot\$ SNen bS:

$'(o)

ehange of {x) @ X' o,

Example: Find the instantaneous rate of change of the following function at z = 3

R(z)=r6;l:E = (st-t)""

t, (re _t)-Y' (t)q'(t\ =

h'(1) =

5z ffi;i

_5t \sst-tbE\1=ts€4-zS fi

FS:\\

Rates of Chanee (ln Real-World Context)

Using derivatives, we can see when a function is rncte atinq or*)

This is often useful when using functions to model real world situations.

Example:

Suppose the amount of water in a holding tank at t minutes is given by 7(t) =2* - L6t + 35. Determine each of the following:

a. ls the volume of water in the tank increasing or decreasing at t = L

minutd?\t(t\ = qt -\u\'(\) = \(\\ -\{o

= -\1- e nne Vo\rrrne \* decveat\nt a\ t: \

b. ls the volume of water in the tank increasing or decreasing at f = 5minutes?

v\(t) = t+(t)*\t= t0-\t= \ F rilne {o\,}Tne \5 \ncYe at\c\t a\ t = 5

ls the volume of water in the tank changing faster at t = 1 or t = 5minutes?

\he \ o\us1s of vlt\ er \t c\rtngrfi$ t a*.r\eva\ ! =\ becauSe \tre t\ope- of \Y'e\angen\ \Yre \\ g\ eeQeY a\ \ha\ ?urt*

d. ls the volume of water in the tank ever not changing? lf so, when?

\{hen v\(t\ = 0 : \e \anL \a rro\\t -\b =$

+tt +\b\t=\e.'.\\

t=\

c\nang'nX ,*,y*-\H6\r\\ren t=S, \he

oq L^rO\ ev \n\ank \% fte\

Vo\u$ne\he

ch'*ntr ft*

Straieht Line Motion: Position. Velocitv & Accelg.rqtion

XeyVocabulary:

Position Function tive$ \ \ocq\\on oti an obiec+ e\ \rrne t,utsa\\ $(t\ r ".(t) oY \(t\ .

Velocity Functiontrhe va\e of chahqe (deviva\we) o* gs\\orr (v(t\).* pob$Ne . $P, y\S\{ itnNon * matn.te = doon/\eQ\trrshu

Acceleration Functiontine ya\e o$ chasqe

usrrA\\rt a (t) -

(Aevirr ahve) og ve\oe\ r

lnitial Position t\arhng godr\ron (t\ t.$) r $olnitialVelocity $\arhng Ye\oc\\ (a\ t.,0) ! !o

Speed \{ne abto\q\e \ a\ue o$ rre\oc\\Displacement the Re\ dnanqe \n po$,\tm (Sna\ ro$.- rn\\a\?6.)Total Distance tota\ dshancn tvave\e{ bg \hc ob\ec\ tn rftre \sre

ir*ewa\,(taLes slro tc.cqrrr\ trtr\ d\rec\on drranqeS)

Example: lf s(t) = t3 + t, find u(t) and a(t).

t'(t\ * I ($

',(t\=j{"+\\'(t\ = 6 t$

a(\'\ = kt

4,ractice Prohlems:

1. Use the position function s(t) = 76t3 - 36tz + 24 of an object moving on ahorizontal line for the following problems. Distance units are measured in feetand time units are measured in seconds.

a. What is the initial position of the object?s(0) =\b(0\6- 36(CI)r +?\s(0\ " t\ teel

b. What is the velocity of the object at f = I second?\(t)-- qtsz -11tv(\\ = \t(\)"-11"[\) = \t*-11 * *1\ \\lrec.

c. What is the speed of the object 0t t = t second?

WeeC"\-t\\ = t\ \\ I Sec

d. What is the acceleration of the object at f = t second?

e.

f.

a(t\ * qbt "1'l-0(\\ = q{"(\1 -1L * Qb -11 = t\ *\ I *ca

When is the obJect at rest?v (\.\ = o - t-- O qt \- tt =*1., \\-ut{i = \%tt -t-r\ / t--a/t Gr *t}v(\\=t(\'t[-]U r '\- \].t=E \=

When is the object moving right? v tt\ > s tt \b

v(t\ t - *t \\hen \ ) t/r-r7hWhen is the object moving left? v{kb (-t

0( t ( jA-

When is the velocity of the object equal to 5* {t\ttt -11t * B\r(t\ \N\en 1= llt$la

-G-qe c.

i. What is the displacement of the object between f = 0 and t = 2 seconds?

tr!l la l.llE VgM,lly \lt LllE LrrrJGr,L sYrac.r lv *rT _.,\ttt -tlt = bt| d G( tt" -\t\.*q\ ="d" p t = \tlN\a"qttl-t1t -r\=0 J 1= - (-t'i\t {fiff::Ti,i"ilT\ J .ur*.^, =k

=b(u= \-t -: \-tt=1"1 t

- r(01 + [s(r-) - t(iA)Ur-\\ + \t-(-$\\ + \u\\\=m

j. What is the total distance traveled by the object between f = 0 and t * 2seconds?to\a\' drttgnce

\r$ sqah \e("

The graph shows the position of a radio controlled model car. Answer thesequestions and explain.

Vertical Motion Examoles

3. Suppose s(t) = -L6* + 48t + 160 gives the position (in feet) above theground for a ball thrown into the air from the top of a high cliff {where time ismeasured in seconds).

a. Find the initialvelocity.\(t\ * _?r\. *.q!i,

v (0\ * -tt(b\ 4\% *qqq% q\ ltec.

b **n1,1;tflig1{ffiIru"'nll-* ha\\ hr*

\b ( {.. r\ (\" *"1\ . {} 1\* gt5'e*SA\ t"%'%ecov:&q.

t=5 oS,, -1"-*Ac. At what time does the ball reach its maximum height?

a. When was the car stopped?Be\ween g an* t becau+"e t(t\

ia no\ r$C\eBtrYrq o(decre aarns

b. At which point was the ca/s velocity the greatest?

At porn\ B hQ( tsrl'rQ \he t\opeo\ \he \anqtn\ \\ne e\ Po\ft\ &\t \he t\eeptt\ psBr\rv e t\o$e .

c. At which point was the ca/s speed the greatest?

\\ eo\n\ E be cauta \kre s\streo\ the \anqent \irrQ a\ Bors\ tit \h,e gtedpet\

-\b t1 +$ tt = *\6$- \G(t'- t\ + 1&) * -\r*0 t (-\txvs\- \(,(t - t[)' = -\b0 -rG-U"( \-b/t\ ? +\1t = t(t)

\ey\ex: (n1r-)\qr")

N\. a/t geqqr\St , $neba\\ veache\ a nt\Ax\erqh\ oQ \q. t +\.