mm of one problem y x - GitHub Pages

RECITATION 12: REVIEW OF CHAPTERS 3 & 4For this worksheet, you want to focus on more than merely “getting the right answer” especially

since you have no reason to expect these particular problems will appear on the midterm. What elseshould you do while working through these problems?

1. Find the points on the curve y = 2x3 � 3x2 � 12x+ 1 where the tangent is horizontal.

2. A particle moves on a vertical line so that it’s coordinate at time t is y = t

3 � 12t + 3 for t � 0.Assume t is in seconds and s is in meters.

(a) Find the velocity and the acceleration of the particle at time t.

(b) When will the velocity be zero?

(c) Sketch a diagram of the motion of the particle.

Uses a calculator 1 Recitation 12

• Paying attention to writing mathematics correctly .

~

• Focussing on the general principle being applied - not the particularsmm

of one single problem .

o¥= 6×2-6×-12=6 ( xZx . 2) =6( x. 2) ( x+D=o when X=2 or ×= -1

And : The curve has horizontal tangents at ( t,

8 ) and (2,49 )

Velocity : y'

= 3t2 - 12

acceleration : y "= let

Set y±o .So 0=3+2-12--342 - 4) =3 Lt + 2) Ct - 2)

.

Thus, t=±2

.

( -2 is not in the domain. )

Answer : The velocity Is Zero at 2 seconds

work From the sign of y'

,we know the particle moves left,

-

:

to y=3 then, at time t=2

,it turns as goes right .

t=2, y=- 13 t=z E4

;, >;>:;-o- - - 0 t + + < • to

!#-) -13 3 -19

-20 - 10 0 10 20 Ysign of y

'

3. The angle of elevation of the sun is decreasing at a rate of 0.25 radian per hour. How fast is theshadow cast by a 400-ft-tall building increasing when the angle of elevation of the sun is ⇡/6?

4. The radius of a circular disc is given as 24 cm with a maximum arror in measurement of 0.2 cm.

(a) Use differentials to estimate the maximum error in the calculated area of the disk.

(b) What is the relative error?

5. Find the critical numbers of f(t) =pt

1+t

2 in [0, 2].


-21 4001¥76 tof=tanED= 's#

'

Fg 5=40005

•

qojfp

;sample pictures general pici . equation : tano=40g= 40051

÷¥¥f###[email protected]

timet :

as time goes on ,stshadow Seto . d¥E= - 400.52 .ddqa ,→ -02-703 gets answer

smaller t Shadow Findddsq when

p

9monms-roor3.oeYy.zsPYEirfffHy4o0orniEtszsdatE4oof@anddOTt-aA-Tr2dA-ZTrdrE2Tl24l0.DE3_ol59cn21maximwmerrovinaieaA-tTl245.daI.2Tl24k0L-O.0l6TC2472.aa.2

ftttatytpnjizh.tt#A.zEfe=CitEt4t2_i-3t2FG+t42=2E(HEI2

f'

undefined at t=0Any : The critical points are

f '=o at t=±YB to, tfgstz

6. (a) Explain how Rolle’s Theorem is a special case of the Mean Value Theorem.

(b) Verify that the function f(x) =x

x+ 2satisfies the two hypotheses of the Mean Value Theorem

on the interval [1, 4]. Then, find all numbers c that satisfy the conclusion of the Mean ValueTheorem.

7. The graph of the first derivative f

0 of a function f is shown below.

(a) On what intervals is f increasing?

(b) At what values of x does f have a localmaximum or minimum?

(c) On what intervals is f concave up or con-cave down?

(d) What are the x-coordinates of the inflectionpoints of f?


Rolle 's Thm Is the MVTHM when f( a) =f( b).

• 2 ( Find C- values ) .

ftp.cx#x(D=2Cx+2) .

we need c so that4+32

So f and F'

are defined for all fkc)=gt .That is :

X in [ 1,4 ].

So f 6) is continuous 2

and differentiable In the # =¥ °rcF÷= st

interval . Thus MVTHM applies . So 3rz=c+2 .

( Now apply it :)f ( i ) = 13 ,

f (4) =4z= } .

50 C= 3k -272.24 ✓

So fl b) - f( a)=

23 - tz-

4T= gtb:~

( + to .

) (tot ) min ath -2,4

• to •

/ WY maxatxto

min ccup = f'

increasing ccup C- x,

- 1) u (1) 2) U ( 3, 5)

mat Cc down = f'

decneasin cc down C- 1) 1) U (2) 3) U ( 5,6 )

( want intervals where f'

> o ) ×= -1,1 , 2) 3,5Aid : tz ,o)u( 4 ,x)

8. Evaluate the following limits. Show your work.

(a) limx!1

(lnx)2

e

2x(b) lim

x!0+x · lnx

(c) limx!0+

x

lnx

(d) limx!1+

✓x

x� 1� 1

lnx

◆

9. Use the Second Derivative Test to identify any local maximums or minimums of f(x) = 2x(20�x

2).


form°o° formfx

-

~ lnx ¥ 2

'Ilim 2. lnx.tt =lim

- Ellin -I

- ×→ot×

' '×→o+ - £2 x2

- x→ - ze2×form #

lnx I , = line- y×=o=lim

# ¥×tgg¥ETx x→o+x→÷form #

I

=

lim eTx(×+z×5=° forma -

•former

x→x- -

= x H= lim xlnx.CIW xtlt ( x - 1) lnx

form E.= ¥ form

not indeterminate .

lalnx+ X. lz - 1

=

lim Inx l ' 7= lim - - x

× -7 It 1. lnx + ( x-D . ,±×→l+ lnxtl - ×

' '

It

. =txi→mµI¥=t÷=E

f G) = 40×-2×3

f' 67-40-6×2=2 (20-3×2)=0 .

So ×=± # = ±2Ffr3

f"

( D= - IZX

f"

( 2T%3)= - 142¥ )< on .

fhasalocalmax at # 2%3 .

!

Sof has a local min at

f "t2r5hs)= - KEYE) > ON

×= srshg

10. Sketch the graph of f(x) =1

x

2 � 9incorporating the information below.

(a) Find the domain.

(b) Find the x and y-intercepts.

(c) Find the symmetries/ periodicity of the curve.

(d) Determine the asymptotes.

(e,f) Determine where the function is increasing/ decreasing and find the local maximum/ mini-mum values

(g) Find the intervals of concavity/inflection points.

(h) Sketch the curve.


C- a,

- 3) ut3 , 3) u ( 3,6 ).

if x=o, y =

- Yq ( y- intercept . )

Note yto .

So no x ' intercept .

All even terms . f(×) is even .

×⇒,×= - 3 vertical asymptotes . xtljmy ¥g=0 .

So y=0 is a

horizontal asymptote .

f 4) = ( xzq )" So f /

> 0 for × in to ,o) ; fkoforx

in ( op ) .

f'

4) = - ( E- 9524AAnd : f. is increasing

onto ,0),

and

= # decreasing on ( op ) .

⇐2- g)2

f "G) = 64×2+3

) + 0 - o + fisccup on Go,

- 3) UG ,x )-

T and cc down C- 3,3 ).

( *2- 9)3

-3 3

inflection points : none

f"

¥0×= -3 and ×=3 aren't in

f" undefined at x=±3 the domain .

y1

^

inI:\¥3in :

11

I IX= -3 h3

11. A cylindrical can is to be constructed with a volume of 20 cubic feet. The material for the top ofthe can costs $2 per square foot. The material for the bottom of the can costs $10 per square foot.The material used to construct the side of the can costs $ 5 per square foot. What dimensions ofthe can will minimize the costs of materials?


r2

.

so

h=¥fz=¥r'

€ p

V=Trh=20-

h C=2 . Tilt 10 . tit 5.2'Tr@|

.

... ....lu ¥ #⇒- topbottom sides

2

C(r)=12tr2t1OTr(2¥52)=12Ir

+20051

with domain ( 0,6 ).

- O + ← signofc'

Cllr )=24Ir - 20052=0 #So 24tr= 2¥ .

°32,55£

Thus r3=}y0g,- = }y÷ .

So,

r -32513€ corresponds to an

absolute minimum .

so r=3F%I is + hes°h=2¥(3¥sY3

unique critical point

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mm of one problem y x - GitHub Pages