Answer Key
Calculus 1 Homework Answer Key
-
University:
Northeastern University -
Course:
MATH 1341 | Calculus 1 For Science And Engineering Academic year:
2023
-
Views:
144
Pages:
70
Author:
Michelle Hensley
Exam review 1
1.A Find open intervals on which f is increasing or
decreasing
Find intervals when g is
concave up and down Find df/ft when x=3 and dx
/ dt=-2 F(x) is invertible if it is one to
one. So for every y value
there is one x value
Ex: Use inverse
function
theorem to
calculate the
derivative of
inverse of f at b Critical Points (Exponentials and Logs) Worksheet
Findandclassifythecritical pointsofthefollowingfunctions.
1)f (x)=(x2 +3x +1)ex
2)f (x)=
e2x
x
3)f (x)=x2 ln(x) Solutions
1)f 0 (x)=(2x +3)ex +(x2 +3x +1)ex
= ex (x2 +5x +4) = ex (x +4)(x +1) = 0 ! x = 1, 4
f 0 (5)=e5 (5+4)(5+1) = (+)()() = +
! increasingon(1, 4)
f 0 (2)=e2 (2+4)(2+1) = (+)(+)()=
! decreasingon(4, 1)
f 0 (0)=e0 (0+4)(0+1) = (+)(+)(+) = +
! increasingon(1, 1)
Localmaxatx = 4,localminat x = 1
2)f 0 (x)=
2e2x ·xe2x ·1
x2
=
e2x (2x1)
x2
=0! x =
1
2
When testing the first derivative, note that the original function has a
verticalasymptoteat x =0. Sowhentestingavaluetotheleftofx = 1
makesurethatitisalsotherightofx =0.
f 0 ( 14 )=
e1/4 ( 1 2 1)
(1/4)2
=
f 0 (1)=
e(21)
12
(+)(+)
(+)
=
Localminatx =
(+)()
(+)
= ! decreasingon(0, 1
=+ increasing on ( 1
2)
2 , 1)
1
2
3)f 0 (x) = 2x · ln(x)+x2 ·
1
x
=2x ln(x)+x
= x(2ln(x)+1) = 0 ! x =0,e1/2
However,sincethedomainoftheoriginalfunctionis(0, 1),becauseyou
can’tplug0ornegativenumbersintoalogarithm,wehavetodisregard
x =0.
f 0 (e1 )=e1 (2ln(e1 )+1) = e1 (2(1)+1) = (+)()=()
! decreasingon(0,e1/2 )
f 0 (e0 )=f 0 (1)=e1 (2ln(1) + 1) = e(0 + 1)=(+)(+)=+
! increasingon(e1/2 , 1)
Localminatx = e1/2
2, Logarithmic Differentiation Worksheet
Uselogarithmicdifferentiationtofindthederivative, respecttox, ofthe
followingfunctions.
1.) y = xln(x)
2.) y =
3. y =
3
p
2x (5x + 10)ln(x)
(x+1)2
(4x+1)(x2)4
4.) y = x
p
x Solutions
å
å
1.) ln(y) = ln xln(x) =ln(x) · ln(x)=ln2 (x)
1
1
0
y · y = 2ln(x) · x
y 0 = xln(x)
2.) ln(y)=
1
13
3
=
å 2ln(x) å
x
å
å
ln 2x (5x1+ 10)ln(x)
ln(2x )+ 3 ln(5x +10)+
1
ln ln(x)
ln(5x +10)+
=
ln ln(x)
31
1
1
1
1 ln(2)+ 1
0
·y =3
3 · 5x+10 · 5+ 3 ln(x) · x
1 3
1
1
3 x ln(2)+ 3
1
y
.
y 0 = p3
ÿ
2x (5x + 10)ln(x)
1
3
ln(2)+·
5
15x+30 +
3x ln(x)
3.) ln(y)=2 ln(x +1) ln(4x +1) 4ln(x 2)
1
y
· y0 =
1
2
x+1 4x+1 ·
(x+1)2
y0 =
å
(4x+1)(x2)4
4
4
x2
4
x+1 4x+1
2
4
å
x2
4.) ln(y)=x1/2 ln(x)
1
y
· y 0 = 21 x1/2 · ln(x)+x1/2 ·
y0 = x
p
å 2p x
x ln(x)
+
1
p
å
x
1
1
x
=
ln(x)
p
2 x
+
1
p
x
Trigonometric Functions Worksheet
1)Differentiatethefollowingfunctions.
p
(b)f (x)= x tan(x)
(a)f (x)=sin(x)cos(x)
(c)f (x)=sin1
cos(6x)
(d)f (x)=x2 tan1 (10x)
2)Considerthefunctionf (θ)=sin2 (θ)+cos(θ).
(a)Findallofthecriticalpointsofthisfunctionontheinterval[0, 2π ]
(b)Findallintervalsonwhichthisfunctionisincreasingandonwhichitisdecreasing.
3)TrueorFalse(nocalculators)
(a) sin sin1 (1) =1
1
(b) sin sin (2) =2
(c)sin1 sin(−1) = −1
(d)cos1 cos(−1) = −1
(e)cos1 cos(2) =2 Solutions
1)(a)f 0 (x)=cos(x) · cos(x)+sin(x) · − sin(x)
2
2
=cos (x) − sin (x)=cos(2x)(butyoudon’tneedtoknowthistrigidentity)
(b)f 0 (x)=
å
1
(c)f 0 (x)=
2
√
x tan(x)
1
1cos2 (6x)
å1/2 å
å
· 1 · tan(x)+x sec2 (x)
·− sin(6x) · 6=
1
(d)f 0 (x)=2x · tan1 (10x)+x2 ·
6sin(6x)
√
1+(10x)2
1cos2 (6x)
· 10=2x tan1 (10x)+
10x2
å
å
2)(a)f 0 (θ) = 2sin(θ) · cos(θ) − sin(θ)=sin(θ) 2cos(θ) − 1 =0
sin(θ)=0→ θ =0,π,2π ,
cos(θ)= 1 2 → θ = π3 , 5π3
(b)Intervals: [0,π/3],[π/3,π],[π,5π/3],[5 π/3, 2π ]
θ = π/6:f 0 (π/6)=
1
2
(2·
p
3
2
− 1)=+→ Increasingon(0,π/3)
θ = π/2:f 0 (π/2) = 1· (2· 0 − 1]=−→ Decreasingon(π/3,π)
θ =3π/2:f 0 (3π/2)=−1 · (2· 0 − 1)=+→ Increasingon(π,5π/3)
θ =11π/6:f 0 (11π/6)=− 1
3)(a)True
(b)False
(c)True
(d)False
(e)True
2
(2·
p
3
2
− 1)=−→ Decreasingon(5π/3, 2π )
1+100x2 Optimization and l’Hôpital’s Rule
1.) Findthefollowinglimits.
(a) lim
x!1
3x2
2x
(b) lim (1− x)2/x
+
x!0
(c) lim+ x · [ln(x)]2
x!0
2) An open-top box is to be made from a 24 inch by 36 inch piece of cardboard by
removing a square from each corner and folding up the flaps on each side. What size
squareshouldbecutoutofeachcornertogetaboxwiththemaximumvolume?
x
24-2x
x
x
x
x
36-2x
36-2x
24-2x
x
x
x
x 3) You wish to manufacture a cylindrical cup with no lid, which is to hold a volume
of 100 cm3 . The material for the bottom costs 3¢/cm3 , and material for the upright
sidecosts1¢/cm3 . Findthedimensionsofthecupthatresultintheminimumcostof
manufacturingthecup.
Reminder: area of upright side=2πrh
area of base=πr2
volume of cylinder=πr 2 h
4) Find the maximum and minimum values of the function f (x)=x3 (1− x)4 on the
interval[−1, 1]. Solutions
1.) (a) 1
1
→ indeterminate
6x
2x ln(2)
l’Hopital: = lim
x!1
6
2x [ln(2)]2
l’Hopital: lim
x!1
1
=
1
6
1
=
→ indeterminate
=0
(b)11 → indeterminate
Takethe log: lim+
x!0
l’Hopital: = lim+
2
x
2ln(1x)
x
ln(1− x)= limx!0+
2
1−x
0
→ indeterminate
0
Chain Rule
·(1)
x!0
=
= −2
1
Limitofthelogofthefunctionis −2,sothelimitoftheoriginalfunctionise2
(c) 0·∞ → indeterminate
Rewritetogetitreadyforl’Hopital: lim
l’Hopital: = lim+
x!0
2ln(x)· 1
1
x2
Chain Rule
x
[ln(x)]2
x!0+
l’Hopital: = lim+
x!0
x12
1
1
→ indeterminate
= lim −2x · ln(x) = 0·−∞ → indeterminate
+
x!0
Rewritetogetitreadyforl’Hopital: = lim
2x
=
1
x
2ln(x)
x!0+
1
x
=
1
1
→ indeterminate
== lim 2x =0
+
x!0
2.) V = x(24− 2x)(36− 2x)=4x3 − 120x2 +864x
Keepinmindthatthedomainofthisfunctionis0 0 → increasing
V 0 (10)< 0 → decreasing
Sothemaximumvolumeisachievedwhenwecutoutasquareofsidelength
p
x = 10− 2 (7) 3)V = πr 2 h =100→ h =
100
πr2
C =3(πr 2 )+1(2πrh)
=3πr 2 +2πr( 100 πr2 ) = 3πr 2 +
0
C =6πr−
200
6πr − 200=0→ r3 =
0
C (5)=6π(5)−
r
=0
r2
3
C 0 (1)=6π(1)−
200
200
100
3π
→ r = 3q
3π
100
≈ 2.197
12
< 0 → decreasing
52
=30π − 8 > 0 → increasing
200
Thus,wegetaminimumcost.
Dimensionsforminimumcost:
r =2.197
100
h ≈ π(2.197)
2 =6.595
Chain Rule
4)f 0 (x)=3x2 (1− x)4 +4x3 (1− x)3 · (−1)=3x2 (1− x)4 − 4x3 (1− x)3
x2 (1− x)3 [3(1− x) − 4x]=x2 (1− x)3 (3− 7x)=0→ x =0, 3
7
,1
Testcriticalpointsandendpointsoftheintervalintheoriginalfunction(notthefirst
derivative).
4
3
f (−1)=(−1)
1 − (−1) = −(2)4 = −16
f (0)=03 (1− 0)4 =0
f ( 37 )=( 3 7 )3 (1−
3
3
7
)4 =( 3 7 )3 ( 47 )4 ≈ 0.0084
4
f (1)=1 (1− 0) =0
Maximumvalue: 0.0084
Minimumvalue: -16 Math1341
Test2Review
Test 2 Review
Directions: Below are some review questions for the second exam. This is neither a sample
examnorapracticeexam: Thesequestionsareintendedtogiveyoumorepracticeandassessyour
readiness for the exam.
1. Considerthefunctionf (x)=3e2x +ln (1+x)forx>−1.
(a) Findf (0).
(b) Find(f 1 )0 (3).
2. Findtheequationofthetangentlinetothegraphf (x)=2x cosx at x = π .
3. Uselogarithmicdifferentiation(noothermethod!) tofindthederivativeofthefunction
f (x)=
xx esinx
.
cosx
4. Differentiate thefollowingfunctions. DoNOTsimplifyyouranswer.
q
x3 )
1
(a) f (x)=eln(2xcos
1+ sin (3x)
x
(b) f (x)=
1 + tan(2x)
5. Supposeaparticleismovinginthexy-plane,anditspositionattimet (inseconds)isgiven
byp(t)=(t2 +1, 4 − t3 ),where theunitsinthe x andy directionsare giveninfeet.
(a) Findthevelocityoftheparticleattime t.
(b) Findthespeedoftheparticleattimet =2.
(c) Findtheaccelerationoftheparticleattimet =2.
(d) Findtheslopeofthetangentlinetothetrajectoryatthepoint(2, 3).
6. Findthefollowinglimits.
(a) lim
lnx
sin(πx)
(b) lim
x3 − 2x
e2x − 1
x!1
x!0
7. Weconstructaclosedrectangularboxwithasquarebaseandvolume80cubicmeters. The
costofthematerialsforthebaseandtopis$20persquaremeter. Thecostofthematerials
for the sides is $10 per square meter. What is the cost of the least expensive box we can
build?
8. A box with an open top is constructed with 600 in2 of material. The length of the base of
theboxistwicethewidth. Findthedimensionsoftheboxthatmaximizevolume.
9. Calculatethefollowingindefiniteintegrals.
Page1 Math1341
Test2Review
Z
(a)
Z
(b)
Z
(c)
√
3x +x2x
dx
(4−xx2 )3
x
exe +3
Z
(d)
dx
2t
5/2
−
dx
6
dt
1+t2
Z
(e)
(7+8x)49 dx
8. A box with an open top is
constructed with 600 in2 of
material. The length of the base
of the box is twice the width.
Find the dimensions of the box
that maximize volume.
7. We construct a closed rectangular box with a square base and
volume 80 cubic meters. The cost of the materials for the base
and top is $20 per square meter. The cost of the materials for the
sides is $10 per square meter. What is the cost of the least
expensive box we can build?
Page2 Sample Test 2
Math 1341
You will see *different* problems on our test. The actual test may be slightly longer or slightly
shorterinlength. However,thedifficultyandgeneraltypesofquestionsarethesameasthissample
test.
1. Wewanttobuildalargestorageboxwithasquarebase. Thevolumeoftheboxis490ft3 .
• Thematerialforthebasecosts$ 7persquarefoot.
• Thematerialforthetopcosts$ 3persquarefoot.
• Thematerialforthesidescosts$ 3.50persquarefoot.
(18points)Findthedimensionsoftheboxthatminimizethecost. Show your work
Answer:
ft ×
ft ×
ft
2. Calculateeachlimit. IfyouuseL’Hôpital’sRule,youneedtoclearlystatewhereitisused.
Show your work.
(a) (8points) lim
x!1
10x − x2 − 9
3lnx
Answer: (b) (8points) lim
x!1
e5/x − 1
5/x
Answer:
3. Considerf (x)=42x+1 +tan1 (x)on(−∞, ∞). Show your work. Give exact answers.
(a) (3points)Findf (0).
f (0)=
(b) (6points)Calculatef 0 (x).
f 0 (x)=
(c) (6points)Find(f 1 )0 (4).
(f 1 )0 (4)=
2 4. Aparticleismoving onxy-planewithpositiongivenby~p(t) = (sin(t), sin(2t)).
Show your work.
(a) (3points)Findthepositionoftheparticleattime t = á .
~p(á)=
(b) (5points)Findthevelocityoftheparticleattime t.
~v (t)=
(c) (5points)Findthespeedoftheparticleat t = á .
Answer:
(d) (6points)Findanequationofthetangentlinetothetrajectoryat t = á .
Answer:
5. Findeachindefiniteintegral. Show your work.
Z
(a) (6points)
t4 t3 + t8 dt
Answer:
3 Z 7
(b) (6points)
√ +
1 3y
2
ç
y 2 +1
dy
Answer:
Z
(c) (8points)
x4 sin x5 +9 dx
Answer:
6. (12points)UseLogarithmicDifferentiationtocalculatey 0 where y =
Answer:
4
√4
x +6 .
x7tanx
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Calculus 1 Homework Answer Key
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