MthS 1040
Test 3
Fall 2013
Calc 3.2 – 3.7
Version A
Student’s Printed Name: __Key__&_Grading Guidelines___ CUID:___________________
Instructor: ______________________
Section # :_________
You are not permitted to use a calculator on any portion of this test. You are not allowed to use
any textbook, notes, cell phone, laptop, PDA, or any technology on either portion of this test. All
devices must be turned off while you are in the testing room.
During this test, any communication with any person (other than the instructor or his designated
proctor) in any form, including written, signed, verbal, or digital, is understood to be a violation
of academic integrity.
No part of this test may be removed from the testing room.
Read each question very carefully. In order to receive full credit for the free response portion of
the test, you must:
1.
Show legible and logical (relevant) justification which supports your final answer.
2.
Use complete and correct mathematical notation.
3.
Include proper units, if necessary.
4.
Give exact numerical values whenever possible.
You have 90 minutes to complete the entire test.
On my honor, I have neither given nor received inappropriate or unauthorized information
at any time before or during this test.
Student’s Signature: ________________________________________________
Do not write below this line.
Points
Earned
Free
Response
Problem
Possible
Points
Free Response
Problem
Possible
Points
1a
4
4c
2
1b
5
4d
3
1c
6
5
5
1d
5
6a
6
2a
5
6b
4
2b
4
1
3
6
61
4a
2
7
Free
Response
Multiple
Choice
4b
3
Test Total
100
Points
Earned
39
Page 1 of 12
MthS 1040
Fall 2013
Version A
Calc 3.2 – 3.7
Multiple Choice. There are 13 multiple choice questions. Each question is worth 3 points
and has one correct answer. The multiple choice problems will count as 39% of the total
grade. Use a number 2 pencil and bubble in the letter of your response on the scantron
sheet for problems 1 – 13. For your own record, also circle your choice on your test since
the scantron will not be returned to you. Only the responses recorded on your scantron
sheet will be graded. You are NOT permitted to use a calculator on any portion of this test.
1.
(3 pts.)
Test 3
−2sin3x
x→0
5x
Evaluate lim
a)
3
5
b) −
6
5
c) −
2
15
d) DNE
2.
(3 pts.)
Find the first derivative of y = −
a) y ′ = −
2
x
.
1
x
b) y′ = x
c) y ′ = x 3
d) y′ =
3.
(3 pts.)
1
x3
Given that y = 2x + y 2 , which of the following points is NOT on the curve?
a) (–1, –1)
b) (0,1)
c) (1, –1)
d) (–1,2)
Page 2 of 12
MthS 1040
Test 3
Fall 2013
Calc 3.2 – 3.7
Version A
Use the following table to answer questions 4 - 6.
𝑥 = −1
𝑥=0
𝑥=1
𝑓(𝑥)
2
−1
2
𝑓′(𝑥)
5
3
4
𝑔(𝑥)
1
2
1
𝑔′(𝑥)
−5
−2
−3
4.
(3 pts.)
(
𝑥=2
1
0
5
−4
)
3
d ⎡
f ( x) ⎤
⎥⎦
dx ⎢⎣
x = −1
a) 8
b) 12
c) 75
d) 60
5.
(3 pts.)
d ⎡ x ⋅ f ( x) ⎤
⎢
⎥
dx ⎢⎣ g ( x ) ⎥⎦
x = 0
a) −2
b) −
1
2
c) −
3
2
d) 0
6.
(3 pts.)
d
⎡ 2 f ( x ) g ( x ) ⎤⎦
dx ⎣
x =1
a) −20
b) −24
c) −8
d) −4
Page 3 of 12
MthS 1040
Test 3
Version A
7.
(3 pts.)
8.
(3 pts.)
Fall 2013
Calc 3.2 – 3.7
d2y
Consider y = cot x . Find
.
dx 2
a)
d2y
= 2csc 2 x cot x
2
dx
b)
d2y
= 2csc x cot x
dx 2
c)
d2y
= − csc 2 x
2
dx
d)
d2y
=0
dx 2
Let 𝑔 𝑥 =
!!!
!
. Find 𝑔! 4 .
a) 1
b) −
1
16
c) −
1
4
d)
9.
(3 pts.)
3
2
()
Find f ′ ( x ) if f x =
9
5
−
.
sin x cot x
a) f ′ ( x ) = 9csc x − 5tan x
b) f ′ ( x ) = 9csc x cot x + 5csc 2 x
c) f ′ ( x ) = −9csc x cot x − 5sec 2 x
()
d) f ′ x =
9
5
+
cos x csc 2 x
Page 4 of 12
MthS 1040
10.
(3 pts.)
Test 3
Fall 2013
Version A
Calc 3.2 – 3.7
The position function 𝑠(𝑡) of an object moving on a horizontal line is given by
𝑠 𝑡 = −16𝑡 ! + 80𝑡 − 1 where t is measured in seconds and 𝑠(𝑡) is measured in
feet with s > 0 corresponding to positions right of the origin and 0 ≤ t ≤ 6 .
Determine when the object is moving to the left.
a) 0 < t < 6
b)
5
<t <6
2
c) 0 < t <
5
2
d) The object never moves left.
11.
(3 pts.)
12.
(3 pts.)
Find
dy y − 1
d2y
=
for x + y = xy given that
.
2
dx 1− x
dx
a)
d2y
y−x
=
2
2
dx
(1− x )
b)
d2y
= y−x
dx 2
c)
d 2 y 2 ( y − 1)
=
dx 2 (1− x )2
d)
d2y
= 2y − 2
dx 2
Find the equation for the line tangent to f (x) = 2 − sin x at x = 2π .
a) y = −x + 2π + 2
b)
y = −x + 2
c) y = x − 2
d)
y = x + 2π − 2
Page 5 of 12
MthS 1040
13.
(3 pts.)
Test 3
Fall 2013
Version A
Calc 3.2 – 3.7
3
2
Let f ( x ) = 2x − 3x − 12x + 4 . Find all points ( x -values only) on the graph of f at
which the tangent line is horizontal.
a) x = −3, x = 4
b) x = 2, x = 3
c) x = −1, x = 2, x = 6
d) x = −1, x = 2
The Free Response section follows.
PLEASE TURN OVER YOUR SCANTRON while you work on the Free Response questions. You
are welcome to return to the Multiple Choice section at any time.
Page 6 of 12
MthS 1040
Test 3
Fall 2013
Version A
Calc 3.2 – 3.7
Free Response. The Free Response questions will count as 61% of the total grade. Read
each question carefully. In order to receive full credit you must show legible and logical
(relevant) justification which supports your final answer. Give answers as exact answers.
You are NOT permitted to use a calculator on any portion of this test.
1. Find the derivative of the following functions. Use appropriate notation to denote the
derivative. Simplify by combining like terms, reducing fractions, and removing negative
exponents from answers.
Work on Problem
⎛2
⎞
Derivative of first term
a) (4 pts.) f ( x ) = π 10 + x ⎜ − x −23 ⎟ = π 10 + 2 − x −22
⎝x
⎠
Distribute, derivative of next two terms
f ′ ( x ) = 0 + 0 + 22x
−23
Points Awarded
1
2
(OK with product rule if correct)
Simplify
1
Notes:
-2 incorrectly labeling derivative or not labeling derivative
22
= 23
x
can be done without rewrite and with product rule but this is NOT the best method
⎛2
⎞
f ( x ) = π 10 + x ⎜ − x −23 ⎟
⎝x
⎠
⎛2
⎞
f ( x ) = 0 + x −2x −2 + 23x −24 + 1⎜ − x −23 ⎟
⎝x
⎠
(
)
= −2x −1 + 23x −23 +
2
22
− x −23 = 22x −23 = 23
x
x
7
⎛ 3 1 3⎞ ⎛ 3
1 3⎞
b) (5 pts.) g ( x ) = ⎜ 4 − x 5 ⎟ = ⎜ x −4 − x 5 ⎟
3 ⎠
⎝ 2x 3 ⎠ ⎝ 2
6
⎛ 3 1 3⎞ ⎛ 3
1 ⎛ 3⎞ − 2 ⎞
g ′ ( x ) = 7 ⎜ 4 − x 5 ⎟ ⎜ ( −4 ) x −5 − ⎜ ⎟ x 5 ⎟
3 ⎝ 5⎠
⎝ 2x 3 ⎠ ⎝ 2
⎠
⎛ 3 1 53 ⎞
= 7⎜ 4 − x ⎟
⎝ 2x 3 ⎠
6
7
Work on Problem
Points Awarded
Derivative of outside, keep the inside
2
Derivative of inside
2
Simplify
1
Notes:
-2 incorrectly labeling derivative or not labeling derivative
-4 for derivative of outside at derivative of inside
-0.5 missing ( )
⎛
⎞
6
1
⎜− 5 − 2 ⎟
⎜⎝ x
⎟
5x 5 ⎠
Page 7 of 12
MthS 1040
Test 3
Fall 2013
Calc 3.2 – 3.7
Version A
(
)
(
)
y′ = 9 ⎡⎣sec ( tan ( 4x )) ⎤⎦ ⋅sec ( tan ( 4x )) tan ( tan ( 4x )) ⋅sec ( 4x ) ⋅ 4
= 36sec ( tan ( 4x )) tan ( tan ( 4x )) sec ( 4x )
c) (6 pts.) y = 3sec3 tan ( 4x ) = 3 ⎡⎣sec tan ( 4x ) ⎤⎦
2
3
3
2
2
Work on Problem
Points Awarded
Derivative of outside, keep the inside
1
Derivative of first inside
2
Derivative of second inside
1
Derivative of third inside
1
Simplification
1
Notes:
-6 for trying to do a product
-4 to 6 for derivative of outside at derivative of inside, depending on
correctness to the “derivative”
-2 for incorrectly labeling derivative or not labeling derivative
-1 missing exponent
-0.5 exponent on wrong parentheses
-0.5 missing parentheses
d) (5 pts.) y = 7x sin x 4
y′ = 7x cos x 4 ⋅ 4x 3 + 7sin x 4
= 28x 4 cos x 4 + 7sin x 4
Work on Problem
Points Awarded
Hold first
0.5
Correct derivative of second w chain rule
2
Correct derivative of first
1
Hold second
0.5
Simplify
1
Notes:
-2 incorrectly labeling derivative or not labeling derivative
-5 f’*g’
-1 for missing half of chain rule
-1 missing plus sign for product rule
-0.5 slight multiplication error in simplification
-1 correct simplification but then incorrect rewrite
-0.5 for copy error
2. Let y = −3x 4 + 7x 3 − x + 15 .
a. (5 pts.) Find the slope of the normal line when 𝑥 = 1.
Work on Problem
y ′ = −12x 3 + 21x 2 − 1
y ′ (1) = −12 + 21− 1 = 8
mnor = −
1
8
b. (4 pts.) Find the third derivative of y
y ′′ = −36x 2 + 42x
y ′′′ = −72x + 42
Derivative
Evaluation at given x value
Slope of normal
Points Awarded
1
2
2
Notes:
-1 poor notation
-2 incorrectly labeling derivative or not labeling
derivative
-1 incorrectly labeling slope of normal line
-1 incorrect sign on slope of normal line
-0.5 arithmetic error
.
Work on Problem
Points Awarded
Second derivative
1
Third derivative 1st term
1.5
Third derivative 2nd term
1.5
Notes:
-1 poor notation
-1 incorrect second derivative, but followed work
for third derivative
Page 8 of 12
MthS 1040
Test 3
Fall 2013
Version A
Calc 3.2 – 3.7
x3
3. (6 pts.) Find the x -values of all the points where the tangent line to f (x) = − 3x 2 + 2x is
3
parallel to the line y = −3x − 7 .
()
need to get slope of f x to be same as slope of given line
i.e need f ′
( x ) = −3
f ′ ( x ) = x 2 − 6x + 2
x 2 − 6x + 2 = −3
x 2 − 6x + 5 = 0
( x − 5)( x − 1) = 0
x = 1, 5
Work on Problem
Points Awarded
Correct derivative
2
Derivative = proper slope
1
Solves for x
3
Notes:
-0.5 to -1 algebra mistakes or not simplifying answer
-6 setting function = slope
-4 derivative = line
-4 derivative = 0
-3 second derivative = slope
-3 factoring with non-zero on right
only received last point of the 3 points for solving if correct
Page 9 of 12
MthS 1040
Test 3
Fall 2013
Version A
Calc 3.2 – 3.7
4. A car is traveling at 100 ft/s when the driver suddenly applies the brakes. The position (in feet)
of the skidding car after t seconds is s ( t ) = 100t − 5t 2 where the position is measured from the
point where the brakes are applied. Assume the car is travelling on a straight, flat road.
Be sure to include units on your final answers for each part.
a) (2 pts.) What is the displacement of the skidding car during the first 4 seconds?
s ( 4 ) − s ( 0 ) = 400 − 80 − (0 − 0) = 320 ft
Work on Problem
Points Awarded
s(b)
1
s(a)
0.5
Simplify
0.5
Notes:
-0.5 to -1 algebra mistakes
-0.5 for missing or incorrect units
-0.5 once for meters instead of feet thoughout the problem
-2 average velocity
b) (3 pts.) What is the speed of the skidding car after 4 seconds?
v ( t ) = 100 − 10t ft
v ( 4 ) = 100 − 40 = 60 ft/s
speed at t = 4 is 60 ft/s
Work on Problem
Points Awarded
Velocity function (not required)
1
Evaluation
2
Notes:
-0.5 for missing or incorrect units
-0.5 notation equating s and v
-0.5 notation equating v(t) and v(4)
a statement that speed is absolute value of velocity is not
required
-1 answers without supporting work
c) (2 pts.) What is the acceleration of the skidding car at any time t?
Work on Problem
a ( t ) = −10 ft/s 2
Correct derivative
Units
Notes:
-0.5 for missing or incorrect units
-1 lack of negative
-1 answers without supporting work
Points Awarded
1.5
0.5
d) (3 pts.) How long does the car skid before it comes to a stop?
()
looking for time where v t = 0
100 − 10t = 0
100 = 10t
t = 10 sec
Work on Problem
Points Awarded
Recognizes that stop means that
0.5
v(t)=0
Correctly solves for t
2
Units
0.5
Notes:
-2 no work shown but correct answer stated
-0.5 to -1.5 algebra mistakes (egregious errors)
-0.5 for missing or incorrect units
Page 10 of 12
MthS 1040
Test 3
Version A
( x + 3) . DO NOT SIMPLIFY.
5. (5 pts.) Find f ′ ( x ) for f ( x ) =
( x − x)
( x − x ) ⋅8( x + 3) ( 2x ) − ( x + 3) ⋅ 4 ( x − x ) ( 2x − 1)
f ′( x) =
( x − x)
8
2
4
2
2
4
2
7
8
2
2
Fall 2013
Calc 3.2 – 3.7
2
3
8
Work on Problem
Points Awarded
Correct derivative of first function with chain
1.5
rule
Hold the second function
0.5
Correct derivative of second function with chain
1.5
rule
Hold the first function
0.5
Square the denominator
1
Notes:
-5 for f’ / g’, even if f’ and g’ are correct
-2 missing denominator of quotient rule
-0.5 for numerator of quotient rule reversed
-3 appears tried product rule and is “correct”
-2 for incorrectly labeling derivative or not labeling derivative
-1 missing exponents or incorrect exponents
-1/2 each missing parentheses
-1 each missing piece of a chain rule
-1.5 f’(g”(x))#
-4 egregious algebra errors
Page 11 of 12
MthS 1040
Test 3
Fall 2013
Calc 3.2 – 3.7
Version A
6. Consider the function x cos ( y ) + y = 3x + y 3 .
3
a) (6 pts.) Find
dy
.
dx
dy
dy
dy
+ 3x 2 cos y +
= 3+ 3y 2
dx
dx
dx
dy dy
dy
−x 3 sin y +
− 3y 2
= 3− 3x 2 cos y
dx dx
dx
dy
−x 3 sin y + 1− 3y 2 = 3− 3x 2 cos y
dx
dy
3− 3x 2 cos y
= 3
dx −x sin y + 1− 3y 2
x 3 ( − sin y )
(
)
Work on Problem
Points Awarded
Product Rule on left
2.5
Derivative of second term on
1
left
Derivative of first term on right
0.5
Derivative of second term on
1
right
Isolate dy/dx
1
Notes:
OK to use y ′
-1 notational error
-1 copy error
-2.5 f’ * g’ on 1st term on left
-0.5 notation equates derivative and derivative
evaluation
b) (4 pts.) Find the equation of the tangent line at the point (0,1).
dy
3− 3⋅0 ⋅cos1
3− 0
3
3
=
=
=
=−
dx (0,1) −0 ⋅sin1+ 1− 3⋅1 0 + 1− 3 −2
2
so equation of tangent at ( 0,1) is
y −1= −
3
( x − 0)
2
Work on Problem
Points Awarded
Correct slope of tangent from a)
2
Correct substitution into
2
point/slope form
Notes:
-1 converts to slope/intercept with error
-3 function instead of number for slope
3
y = − x +1
2
7. (1 pt.) Check to make sure your Scantron form meets the following criteria. If any of the items
are NOT satisfied when your Scantron is handed in and/or when your Scantron is processed one
point will be subtracted from your test total.
My scantron:
□ is bubbled with firm marks so that the form can be machine read;
□ is not damaged and has no stray marks (the form can be machine read);
□ has 13 bubbled in answers;
□ has MthS 1040 and my Section number written at the top;
□ has my Instructor’s name written at the top;
□ has Test No. 3 written at the top;
□ has Test Version A both written at the top and bubbled in below my CUID;
□ and shows my correct CUID both written and bubbled in (bubble in a 0 in place of the C).
Page 12 of 12