APPM 1350 Final Exam Solutions - Summer 2006, Exams of Calculus for Engineers

The solutions to the final exam of appm 1350, a mathematics course focusing on calculus, for the summer semester of 2006. It includes answers to various calculus problems, such as determining limits, finding derivatives, and evaluating integrals. Students can use this document as a reference to check their understanding of the concepts covered in the exam.

Typology: Exams

2012/2013

Uploaded on 02/25/2013

digambar
digambar 🇮🇳

4.2

(14)

30 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
APPM 1350 FINAL EXAM SOLUTIONS SUMMER 2006
1. (15 points) For each of the following unrelated questions, answer either ALWAYS TRUE,
ALWAYS FALSE or NEITHER. No justification is necessary.
(a) If f(x)>1 for all xand lim
x0f(x) exists, then lim
x0f(x)>1. (NEITHER)
(b) All continuous functions have derivatives. (ALWAYS FALSE)
(c) The derivative of the function tan2xis the derivative of sec2x. (ALWAYS TRUE)
(d) Z5
5
(ax2+bx +c)dx = 2 Z5
0
(ax2+c)dx (ALWAYS TRUE)
(e) If limx6f(x)g(x) exists, then the limit is f(6)g(6). (NEITHER)
2. (21 points) Find dy
dx in each case. No simplification is necessary.
(a) y=xln (arccos x)
dy
dx = (x)1
arccos x 1
1x2+ (1)(ln (arccos x))
=x
arccos x1x2+ ln (arccos x)
(b) y=xex
ln y=exln x
1
y
dy
dx = (ex)1
x+ (ex)(ln x)
dy
dx =y ex1
x+ ln x
=xexex1
x+ ln x
(c) x ey= ln xy + arctan y
(x)ey·dy
dx+ (1)(ey) = 1
xy xdy
dx + 1 ·y+1
1 + y2·dy
dx
dy
dx x ey1
y1
1 + y2=1
xey
dy
dx =
1
xey
xey1
y1
1+y2
pf3
pf4
pf5

Partial preview of the text

Download APPM 1350 Final Exam Solutions - Summer 2006 and more Exams Calculus for Engineers in PDF only on Docsity!

APPM 1350 FINAL EXAM SOLUTIONS SUMMER 2006

  1. (15 points) For each of the following unrelated questions, answer either ALWAYS TRUE, ALWAYS FALSE or NEITHER. No justification is necessary. (a) If f (x) > 1 for all x and lim x→ 0 f (x) exists, then lim x→ 0 f (x) > 1. (NEITHER) (b) All continuous functions have derivatives. (ALWAYS FALSE) (c) The derivative of the function tan^2 x is the derivative of sec^2 x. (ALWAYS TRUE)

(d)

− 5

(ax^2 + bx + c) dx = 2

0

(ax^2 + c) dx (ALWAYS TRUE)

(e) If limx→ 6 f (x)g(x) exists, then the limit is f (6)g(6). (NEITHER)

  1. (21 points) Find dydx in each case. No simplification is necessary.

(a) y = x ln (arccos x)

dy dx =^ (x)

arccos x

√−^1

1 − x^2

  • (1)(ln (arccos x))

=

−x arccos x

1 − x^2

  • ln (arccos x)

(b) y = xex

ln y = ex^ ln x ⇒ 1 y

dy dx

= (ex)

x

  • (ex)(ln x)

⇒ dy dx

= y ex

x

  • ln x

= xe

x ex

x + ln^ x

(c) x ey^ = ln xy + arctan y

(x)

ey^ · dy dx

  • (1)(ey) = 1 xy

x dy dx

  • 1 · y

1 + y^2

· dy dx dy dx

x ey^ − 1 y

1 + y^2

x

− ey

dy dx

1 x −^ e

y xey^ − (^1) y − (^) 1+^1 y 2

  1. (28 points) Evaluate each of the following limits, if it exists. If the limit does not exist, state this and state your justification. Show all your work.

(a) (^) t→lim 0 + tt

2

lim t→ 0 +^

tt

2 = lim t→ 0 +^

et

(^2) ln t = elimt→^0 +^ t

(^2) ln t

lim t→ 0 +^

t^2 ln t = lim t→ 0 +

ln t t−^2

L =′H lim t→ 0 +

t−^1 − 2 t−^3

= lim t→ 0 +^

t^2 = 0

⇒ lim t→ 0 +^

tt

2 = e^0 = 1

(b) (^) rlim→∞

re^1 /r^ − r

r^ lim→∞

re^1 /r^ − r

= (^) rlim→∞ r

e^1 /r^ − 1

= lim r→∞

e^1 /r^ − 1 1 r L =′H lim r→∞

e^1 /r^ · − r 21 − 1 r^2

= lim r→∞ e^1 /r^ = e^0

= 1

(c) (^) xlim→ 0 x arccot^1 x

x^ lim→ 0 x^ arccot x^1 =^ xlim→ 0

arccot (^1) x 1 x

L =′H lim x→ 0

− 1 1+ (^) x^12 ·^

− 1 x^2 − 1 x^2 = (^) xlim→ 0

1 + (^) x^12

= (^) xlim→ 0

−x^2 x^2 + 1 = 0

(d) (^) hlim→ 0

h

∫ (^) 2+h

2

1 + t^3 dt

h^ lim→ 0

h

∫ (^) 2+h

2

1 + t^3 dt = (^) hlim→ 0

∫ (^) 2+h a

1 + t^3 dt −

a

1 + t^3 dt h = F ′(2) where F (x) =

∫ (^) x

a

1 + t^3 dt

F ′(x) =

1 + t^3 (By FTC pt 1) ⇒ F ′(2) =

1 + 2^3 =

  1. (25 points) Do ONE of the following two problems. State clearly which problem you have chosen in your bluebook. Only work for the chosen problem will be graded. (a) What is the area of the largest rectangle in the first quadrant with two sides on the axes and one vertex on the curve y = e−x? Solution: First, draw a picture:

0.5 1 1.5^2 2.

1

y=e-x

A

A = (x)(y) = x e−x ⇒ dA dx

= (x)(−e−x) + (1)(e−x) = e−x(1 − x) = 0 ⇒ x = 1 ⇒ A = 1 · e−^1 =^1 e

(b) Plot the function f (x) = x ln x on (0, e]. Find and label all critical points, local and absolute extrema, and inflection points.

f ′(x) = 1 + ln x = 0 ⇒ x =^1 e f ′′(x) = (^1) x > 0 for all x in (0, e]

-1 -0.5 0.5 1 1.5 2 2.5 3

-0.

1

2

3

fHxL=xlnHxL

H €€€€^1 e , €€€€€€€- e^1 L

H0,0L

He,eL

  1. (40 points)

(a) What does it mean for f (x) to be continuous at x = a? ⇒ f is continuous at x = a if lim x→a f (x) = f (a).

(b) What does it mean for f (x) to be differentiable at x = a? ⇒ f is differentiable at x = a if lim h→ 0

f (a + h) − f (a) h exists or, alternatively, lim^ x→a

f (x) − f (a) x − a exists.

(c) State both parts of the Fundamental Theorem of Calculus.

FTC Part 1: If f is continuous on [a, b] and F (x) =

∫ (^) x

a

f (t) dt, then for a ≤ x ≤ b

d dx F^ (x) =^

d dx

∫ (^) x

a

f (t) dt = f (x)

FTC Part 2: If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then ∫ (^) b

a

f (x) dx = F (b) − F (a)

(d) If f is a continuous function such that ∫ (^) x

0

f (t) dt = x sin x +

∫ (^) x

0

f (t) 1 + t^2 dt

for all x, find an explicit formula for f (x).

d dx

∫ (^) x

0

f (t) dt = d dx

x sin x +

∫ (^) x

0

f (t) 1 + t^2

dt

f (x) = x cos x + sin x + f^ (x) 1 + x^2 ⇒ f (x) = x^ cos^ x^ + sin^ x 1 − (^) 1+^1 x 2