Download Unit 1 - 3 Review and Key

MIDTERM
MATH 31
1  3, 18, 20
Text page 157
1.
REVIEW
p 344
CHAP. 1 to 8
NAME_______________
1, 7 b, g, h 8 14
Fill in the chart and use the limiting value of the slope of a sequence of secants (m ABn )
 1 
to find the slope of y = sin x at A  ,  accurate to 2 decimal places.
 6 2
A
B1
B2
B3
x
/6
1
0.6
0.53
sinx
1/2
_______
_______
_______
_______
_______
_______
_______
_______
slope ABn 
y
x
B4
0.523
limiting value of slope of secantsABn = slope of tangent at /6 = ___________
dy
dy
 f ( x  h)  f ( x ) 
 lim 
as the definition of the derivative of y = f(x) find
for

dx
dx h0 
h

y = ax2 + b where a and b are constants.
2.
Using
3.
If a closed circular cylinder is to have a volume of 1000 cm3, find accurate to 2 decimal
places, the radius and height of the cylinder which will have a minimum surface area.
1
4.
Find and simplify
dy
for each of the following expressions.
dx
3
a) y = x 3 - 8x 2 + 8x - 5
c) y =
3x  5
x7
b) y = 5sin( 2x  3)
d) y = ( e 3x )( 5x 2 + 7 )
e) y = 3 log8 (4x + 1)
f) y = 5 cos ( e 8x )
g) siny = x lny
h) y = (2x  3)3(3  4x)  4
i) x2 + 3xy2  2y2 + 5x = 8
2
5.
Water is running out of a conical funnel at 100 cm3/s. If the radius of the funnel is 10 cm at
the top and the height is 20 cm find the rate at which the radius is decreasing, accurate to 2
decimal places, when the water is 5 cm below the top?
10 cm
20 cm
6.
a)
c)
e)
lim  x  9
=
b)
lim e x cos x
=
d)
x 13
x 0
lim
 
x  
2

ln(cos x)
g) lim 5 x  3 =
x  
=
 x3  8 

lim  2
x 2
 x  4
=
 x 2  5x  6 

lim 
x 2
 3x  6 
 x3  8 

f) lim  2
x  2 x  4 


h) lim  3 cos x
x 
=
=
=
3
7.
Using the Newton-Raphson Approximation,
to
ex = x
xn+1 = xn 
f ( x)
f / ( x)
to find a solution
accurate to 6 decimal places.
8.
If s = t4  2t2 where s is displacement in meters after t seconds, find:
a) the average acceleration in the third second
b) the instantaneous acceleration at t = 5s
9.
What is the equation of the tangent line to y =
10.
Determine the concavity and find the points of inflection for
1 3
x  x2 + 7x  3 at its point of inflection?
6
y = 4 sin 2 x  1
for [0,].
4
11.
Clearly indicate the x- intercepts, critical points and points of inflection for
Y the graph of
20
5
3
y = 3x  10x .
10
X
-4
-3
-2
-1
0
1
2
3
4
-10
-20
Created with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agraphe
12.
Determine the intervals of increase (Given f / (2) = 0; use synthetic division) and the intervals
of downward concavity for the graph of f(x) = x 4  18 x 2 + 40 x + 3.
5
13.
Using
y=
14.
dy
dy
 f ( x)  f (a ) 
as the definition of the derivative of y = f(x) find
for
lim 

dx
dx x a 
xa

3x  2 .
Use calculus to find the equation of the tangent to the circle
x2 + (y  4)2 = 25 at (3, 0).
0.864 0.866
= mtangent = 0.87
2. y / = 2ax
4. a) 3x2  12x1/2 + 8
b) 10 cos(2x  3)
- 3
1
12
c) (x + 7) 2 (3x + 47)
d) e3x (15x2 + 10x + 21)
e)
2
(4 x  1) ln 8
ln y
y ln y
f)  40 e8x sin e8x
g)
h) 2(2x  3)2(3  4x)5(4x  15)

x y cos y  x
cos y 
y
Answers:
1. mAB = 0.717 0.846
3. r = 5.42 h = 10.84
2x  3 y 2  5
 dr 
 0.28 cm/s
5.  
6. a) 2
b) 3
c) 1
2 y (2  3 x)
 dt  r 7.5,h15
6. d) 1/3
e)  
f) does not exist
g) 3
h) 1/3
7. 0.567143
Y
20
2
8. a) 72 m/s
b) 296 m/s
9. y = p.i.at (2,25/3) tangent is y = 5x  5/3
  3 
 
 3 
10. concave down  ,  p.i. at  ,1 and  ,1
4 4 
4 
 4 
i)
  30 
,0  c.p. (0,0), ( 2 ,11.31), ( 2 ,11.31)
11. (0,0), 
3


p.i, (1,7), (0,0), (1, 7) 12. concave down ( 3 , 3 ) -4
increasing for (1  6 ,  1 + 6 ) and (2, )
dy
3
13.
14. 3x  4y  9 = 0

dx 2 3a  2
10
X
-3
-2
-1
0
1
2
3
4
-10
6
-20
Created with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ a
MIDTERM
REVIEW
1  3, 18, 20
Text page 157
1.
MATH 31
CHAP. 1 to 8
p 344
NAME_______________
1, 7 b, g, h 8 14
Fill in the chart and use the limiting value of the slope of a sequence of secants (m ABn )
 1 
to find the slope of y = sin x at A  ,  accurate to 2 decimal places.
 6 2
.


sin 1  sin   

y
 6    0.71677
eg. slope AB1  
Note that x is in radians and slope ABn 

x


1


6


A
B1
B2
B3
x
/6
1
0.6
0.53
0.523
sinx
1/2
0.841
0.565
0.505
0.499
0.717
0.846
0.864
0.866
slope ABn 
y
x
limiting value of slope of secantsABn = slope of tangent at
2.
B4

= 0.87
6
dy
dy
 f ( x  h)  f ( x ) 
 lim 
for
 as the definition of the derivative of y = f(x) find
h

0
dx
dx
h


y = f(x) = ax2 + b where a and b are constants.
Using
dy
 f ( x  h)  f ( x ) 
 lim 

h

0
dx
h



 

 a(x  h) 2  b  ax 2  b 
dy

 lim 
dx h 0 
h



2
2
2
 ax  2axh  ah  ax 
dy
 lim 

dx h 0 
h

 h  2ax  ah  
dy
 lim 
  2ax  0  2ax
dx h 0 
h

7
3.
If a closed circular cylinder is to have a volume of 1000 cm3, find accurate to 2 decimal
places, the radius and height of the cylinder which will have a minimum surface area.
V cylinder = p r 2h = 1000 so h =
1000
pr 2
Acylinder = 2pr 2 + 2prh
ж1000 ц
ж2000 ц
ч = 2pr 2 + 2000r - 1
A = 2pr 2 + 2pr зз 2 ч
= 2pr 2 + зз
ч
ч
и pr ш
и r ш
dA
= (2)2p r 1 + (- 1)2000r - 1- 1 = 4p r - 2000r - 2
dr
dA
c.p. at
= 0 = 4p r - 2000r - 2 = 4r - 2 (p r 3 - 500) r  0
dr
1
500
ж500 ц
3
ч
and r = зз
or 5.42
(p r 3 = 500), so r 3 =
ч
иp ш
p
3
1000
h=
=
pr 2
1
1000
2 (500)
2 (500)3
2 (500)3
=
=
=
= 2r or 10.84
2
2
3
2
1
500 3
500
й 500 1 щ
3
3ъ
p
p3
p3
pк
p
p
кл p
ъ
ы
( )
0
dA
dr
( )
_─
__
( )
5.42
+
c.p. is a minimum
4.
Find and simplify
dy
for each of the following expressions.
dx
3
1
a) y = x 3 - 8x 2 + 8x - 5
y / = 3x 2 - 12x 2 + 8
b) y = 5 sin (2x - 3)
y / = 5 cos (2x - 3)
c) y 
1
3x  5
  3x  5 x  7  2
x7
d
(2x - 3) = 10 cos (2x - 3)
dx
1
1
d
d
 x  7  2   x  7  2  3x  5 
dx
dx
3
1
d
 1 
  3x  5     x  7  2
 x  7    x  7  2  3
dx
 2 
y /   3x  5 
3
1
 1 
  3x  5     x  7  2 1   x  7  2  3 
 2 
3
1
  x  7  2  3x  5   3 2  x  7  
2
3
1
  x  7  2 3x  47 
2
8
4.
y /   e3x 
d
d
5x 2  7    5x 2  7   e3x 

dx
dx
/
3x
2
3x
y   e  10x    5x  7   e   3
d) y  e3x  5x 2  7 


y /   e3x  10x  3  5x 2  7 
y /   e3x 15x 2  10x  21
e) y  3log8  4x  1 
3  1  d

  4x  1
ln 8  4x  1  dx
12  1 
y/ 


ln 8  4x  1 
3
ln  4x  1
ln 8
y/ 
y /  5   sin   e8x 
f) y  5cos  e8x 
d 8x
e 
dx  
y /  5sin  e8x   e8x  8 
y /  40e8x sin  e8x 
d
d
d
sin y  x  ln y   ln y  x 
dx
dx
dx
d
1 d
cos y y  x
y  ln y 1
dx
y dx
dy x dy
cos y 
 ln y
dx y dx
dy 
x
 cos y    ln y
dx 
y
g) sin y  x ln y
dy
ln y
y ln y


dx 
x  y cos y  x
 cos y  y 


h) y   2x  3 3  4x 
3
4
d
4
4 d
3
 2x  3
3  4x   3  4x 
dx
dx
5 d
4
3
2 d
y /   2x  3  43  4x 
3  4x   3  4x  3 2x  3  2x  3
dx
dx
5
4
3
2
/
y   2x  3  43  4x   4  3  4x   3 2x  3  2 
y /   2x  3
3
161 2x  3  3  4x 61
  2x  3 3  4x  16 1 2x  3  3  4x   6 1
  2x  3 3  4x   32x  48  18  24x 
y/   2x  3 3  4x 
y/
y/
2
5
2
5
2
5
y /  2  2x  3 3  4x 
2
5
4x  15
9
4.
i) x2 + 3xy2  2y2 + 5x = 8
d 2 
d
d
d
d
d
x   3x  y 2   y 2
 3x   2  y2    5x   8 

dx
dx
dx
dx
dx
 dx

d
d
d


2x  x   3x  2y   y   y 2  3   2  2y   y   5  0
dx
dx
dx


dy
dy
2x 1  3x  2y   3y 2  4y
5  0
dx
dx
dy
dy
2x 1  3y 2  5  4y  6xy
dx
dx
dy
2x  3y 2  5 
 4y  6xy 
dx
dy 2x  3y 2  5

dx
4y  6xy
5.
Water is running out of a conical funnel at 100 cm3/s. If the radius of the funnel is 10 cm at
the top and the height is 20 cm find the rate at which the radius is decreasing, accurate to 2
decimal places, when the water is 5 cm below the top?
10 cm
20 cm
r 10

so h  2r
h 20
1
1
2
V  r 2 h  r 2  2r   r 3
3
3
3
d
2 d
2
d
dr
V   r 3    3r 2  r  2r 2
dt
3 dt
3
dt
dt
water 5 cm below top so h = 15, r = 7.5
dV
2 dr
 100  2  7.5 
dt
dt
100
8
 dr 


or  0.28
 
2
9
 dt r 7.5 2  7.5
so
6.
a)
b)
lim
x 13 
x 9
 x3  8 

lim 
x  2  x 2  4 


=
=
=
lim
x 13 
x  9  13  9  4  2
  x  2   x 2  2x  4  
 x 3  23 

lim  2
 lim 
2 
x  2 x  2
x 2 
x

2
x

2


 




2
  x 2  2x  4     2   2  2   4

lim 
x  2 
x

2
     2   2 
 

   3


10
6.
c)
d)
e)
lim e x cos x
x 0
 x 2  5x  6 

lim 
x 2
 3x  6 
lim
 
x  
2

ln(cos x)
 x3  8 

f) lim  2
x  2 x  4 


e0 cos 0  11  1
=
  x  2   x  3 
  x  3  2  3 1
  lim 
lim 


x  2 
 x  2  3 
3
3
3  x  2


=
    
lim  ln(cos x)  ln  cos     ln  small positive  

  2  
x  
2
=
  x  2   x 2  2x  4  
 x 3  23 


lim 
  xlim
x  2 x 2  2 2
 2 

x

2
x

2









=
2
  x 2  2x  4     2   2  2   4

lim 
x  2 
x  2   

  2   2 

 
=
Does not exist
=
5  3 
h) lim  3 cos x
=
3cos   31 
x 
   4


0
1
1
3 3 03 3

5

g) lim 5 x  3
x  
7.
=

1
3
x n 1  x n 
Using the Newton-Raphson Approximation,
f (x)
f / (x)
to find a solution to
ex = x accurate to 6 decimal places.
f (x)  ex  x, f / (x)  e x  1
Here x1 can be 0, ─1 or any other number. I used x1 = ─1 = ANS
f (x)
x n 1  x n  /
f (x)




e    1
e    ANS
f  1
f (x)
x 2  x1  /
  1  /
  1 
  ANS 
 0.5378828
1
ANS
f (x)
f  1
e   1
e   1
x 3  0.56698699,
1


ANS


x 4  0.5671432904  x 5
x solution  0.567143
11
8.
If s = t4  2t2 where s is displacement in meters after t seconds, find:
a) the average acceleration in the third second
b) the instantaneous acceleration at t = 5s
displacement is s = t4  2t2
ds
velocity is v   4t 3  4t
dt
v v t 3  v t 2
a average 

t
3 2
3
 4  3   4  3     4  2 3  4  2  
 

a avg  
3 2
108  12  32  8  72 m
a avg 
3 2
s2
9.
What is the equation of the tangent line to y 
ds
 4t 3  4t
dt
d 2s dv
acceleration is a  2 
 12t 2  4
dt
dt
velocity is v 
a t 5  12  5   4  296
2
1 3
x  x 2  7x  3
6
m
s2
at its point of inflection?
1 3
x  x 2  7x  3
6
dy 3 2
 x  2x  7
dx 6
d2 y
 x2
dx 2
point of inflection at x ─ 2 = 0 so x = 2
2
 1 3
  25 
p.i. is (2, f(2)) or  2,  2    2   7  2   3   2, 
 6
  3
3 2
 dy 
slope at p.i. is slope at x = 2 is     2   2  2   7  5
 dx x 2 6
25
y
y  y1
3 or 5x  10  y  25 so y  5x  5
tangent is m 
or 5 
x  x1
x2
3
3
y
10.
Determine the concavity and find the points of inflection for
y = 4 sin 2 x  1
for [0,].
y  4sin 2 x  1  4  sin x   1
2
y /  4  2  sin x 
1
d
 sin x   0  8sin x cos x  4  2sin x cos x   4sin 2x
dx
d
 2x   8cos 2x
dx
potential p.i. at y//  0  8cos 2x, cos 2x  0
 3
 3
so 2x  ,
and x  ,
2 2
4 4
y //  4 cos 2x
12
y//
0
_+_
_

4
3
4
_─
__
_+_
_

 3
 1 2
,
,
y  4
 1  2 1  1
4 4
 2
    3

    3 
  3 
so p.i. at  ,1 ,  ,1  concave up 0,  ,  ,  , concave down  , 
4   4

 4  4 
4 4 
at x 
Y
11.
20
Clearly indicate the x- intercepts, critical points and points of inflection for the
graph of
5
3
y = 3x  10x .
16
f (x)  3x 5  10x 3  x 3  3x 2  10 
12
f ( x)  3x 5  10x 3    3x 5  10x 3   f  x 
8
so f(x) is odd and symmetric about the origin
 10 
, 0  about  1.82, 0 
x─intercepts at  0, 0  ,  
3


f / (x)  15x 4  30x 2  15x 2  x 2  2 

4
-5
-4
-3
-2

2, 11.32

3
-12
2
-16
f // (x)  0  60x  x 2  1 at x  0, 1
p.i. at (0,0), (─1,7) and (1,─7)
12.
2
-8
f (x)  60x  60x  60x  x  1
3
1
-4
c.p at x  0,  2 i.e.  0,0  ,  2,11.32 ,
//
0
-1
f // (x)
─
(─1)
-20
(0) ─
+
(1)
+
Determine the intervals of increase (Given f / (2) = 0; use synthetic division) and the intervals
of downward concavity for the graph of f(x) = x 4  18 x 2 + 40 x + 3.
f / (x)  4x 3  36x  40  4  x 3  9x  10 
2
for P(x) = x 3  9x  10 , P(2) = 0; so (x ─ 2) is a factor
c.p. at f / (x)  0  4  x 3  9x  10   4  x  2   x 2  2x  5 
1
0
─9
1
2
2
4 ─10
─5
0
10
2  22  4 1 5
b  b2  4ac
2  24
2  2 6
c.p. at x  2,
 2,
 2,
 2,
 2, 1  6
2a
2 1
2
2
y/

_─
__
decrea sin g  , 1 
_+_ 1  6 _─ _2 _+
+
+
_ _
_
6  or  1  6, 2  , increa
_ sin g  1  6, 1  6 
1  6
_
or

2,  
13
4
5
f // (x)  12x 2  36  12  x 2  3 so potential p.i. at x   3
concave up
13.
Using
y=

 
 ,  3 or


3,  , concave down  3, 3

dy
dy
 f ( x)  f (a ) 
lim 
as the definition of the derivative of y = f(x) find
for

dx
dx x a 
xa

3x  2 .
 3x  2  3a  2 
dy
 f (x)  f (a) 
lim 
lim 


dx x a  x  a  x a 
x a

 3x  2  3a  2  3x  2  3a  2 
dy
lim 

 3x  2  3a  2 
dx x a 
x a




3x  2  3a  2
dy


lim
dx x a   x  a  3x  2  3a  2 




3x  a
dy

lim 
dx x a   x  a  3x  2  3a  2 




dy 
3
3


dx  3 a   2  3a  2  2 3a  2



 




14.



Use calculus to find the equation of the tangent to the circle
d 2
d
d
2
x    y  4 
 25

dx
dx
dx
d
d
2x  x   2  y  4   y  4   0
dx
dx
2x 1  2  y  4 
dy
0
dx
dy
 2x
dx
dy
2x
x


dx 2  y  4   y  4 
2  y  4
x2 + (y  4)2 = 25 at (3, 0).
slope of tangent at (3,0)
x
3
3
 dy 



 
 dx 3,0  y  4   0  4  4
y  y1
tangent is m 
x  x1
3 y0

4 x 3
3x ─ 4y ─ 9 = 0
14