Problems to practice for FINAL
1. Below is the graph of a function  () 
At which of the marked − values (     and  ) is:
(a)  () greatest
=
(b)  () least
=
(c)  0 () the greatest
(d)  0 () the least
=
=
(e)  0 () = 0
 =  
(f) 
=
00
() = 0
Identify the interval(s) in terms of (     and  ) on which
(g)  0 () is positive
( ) ∪ (  )
(h)  0 () is negative
( )
(i)  0 () is positive
(  )
(j)  0 () is negative
( )
(k)  () is increasing and concave up
(l)  () is increasing and concave down
(m)  () is decreasing and concave up
(n)  () is decreasing and concave down
(  )
( )
( )
( )
2. Evaluate (if infinity, specify whether it is +∞ or −∞)

(a) lim sin−1
 =1
→0
3 −1
=3
→1 −1
√
4 +52 +3−6
lim
2 −1
→−∞
(b) lim
(c)
(d)
√
lim
→−∞
46 +53 +3−6
3 −1
(e) lim
sin 

→∞ 
(f)
=1
= −2
=0
2
=∞
→−∞ ln 2
lim
3
(g) lim (1 − 2)  = −6
→0
3. Evaluate (if infinity, specify whether it is +∞ or −∞)
2
2
=
(a) lim sin
→0 sin 5
5
(b) lim − = 0
→0
(c)
lim tan  ln  = 0
→0+
¡
¢
(d) lim  − 2 = −∞
→∞
(e) lim
→0
(f)
2 +5−6
−1
lim−
→1
2 +5
−1
=6
= −∞
√
4 −32 +−298
=1
2 +17
→∞
√
46 +54 +3−6
lim
= −2
3 −1
→−∞
(g) lim
(h)
(i)
(j)
(k)
lim
→−∞
√
92 +3−6
−1
lim  () = 17, where  () =
→0+
lim  () = 14, where  () =
→1−
(l) lim

−
→∞ 
(m)
= −3
=∞
2
2 = 0
→−∞ 
lim
2
=2
→0 1 − cos 
 − 1 − 
= 12
(o) lim
→0
2
(n) lim
(ln )2
=0
→∞

(p) lim
2
(q) lim 3 − = 0
→0
½
½
|3 − 17| if  ≤ 1
√
2 + 1 if   1
|3 − 17| if  ≤ 1
√
2 + 1 if   1
(r)
(s)
lim+
→0
µ
1
1
− 
  −1
√

lim+ 
=1
(t) lim
µ
2 − 3
2 + 5
(u) lim
µ
→0
→∞
→ 2
¶
¶2+1
1 − sin 
1 + cos 2
¶
=
1
2
= −8
=
£0¤
0
= lim
→ 2
³
− cos 
−2 sin 2
´
=
£0¤
0
sin 
→ 2 −4 cos 2
=
= lim
1
4
tan 
(v) lim (sin )
→0
ln  =
ln sin 
= lim
→0 cot 
→0
lim ln (sin )tan  = lim tan  ln sin  = lim
→0
0
→0
1
sin 
cos 
= lim sin  cos  = 0
→0
csc2 
 =  =1

if
4. Find 
3−
(a)  − 3 +  = 3 Solution:  +  0 +  0 = 3, Solution is:  0 =
.
1+
(b) 2  − 3 3 −  = 3 Solution: 2 + 2  0 − 9 2  0 − 1 = 3 0 . , Solution is:  0 =
1 − 2
2 − 9 2 − 3
(c) 2  + 3 3 −  = 3 Solution: 2 + 3 3 + 2  0 + 9 2  0 − 1 = 0. , Solution is:  0 =
1 − 2 − 3 3
2 + 9 2

5. Find the derivative 
for:
¢
¡
(a)  =  tan  Solution:  0 =  tan  tan  +  sec2 
(b)  = sin ( ) Solution:  0 =  cos ( )
3
(c)  = 4
−1
(d)  = sin−1
3
Solution:  0 = 32 (ln 4) 4
−1
¡1 ¢
1
0
r
3  Solution:  =
1
3 1 − 2
9

 + 1
(e)  = ln ( + 1) Solution:  0 =
(f)  = sin (cos ) Solution:  0 = − sin  cos (cos )
sin 
(g)  = sin−1 (cos ) Solution:  0 = − √
1 − cos2 
−1
sin 
Solution:  0 = √
1 − 2
¢
1¡ 2
(i)  = tan (ln ) Solution:  0 =
tan (ln ) + 1

(h)  = sin
−1

(j)  = 2sin  Solution:  0 = (cos  ln 2) 2sin 
(k)  =  sin  Solution:  0 = (cos )  + (sin ) 
(l)  =
sin 
ln 
Solution:  0 =
1
1
sin 
cos  −
ln 
 ln2 
(m)  = 4 3 Solution:  0 = 43 3 + 34 3
(n)  = 13  cos (3) Solution:  0 =
(o)  =
3
√
3 2

Solution:  0 = −2
1
cos 3 −  sin 3
3

4
(2 ) 3
µ
¶
(sin ) (cos )
1 −
1
1
0
(p)  =
Solution:  = 2 
 cos 2 − sin 2 −  sin 2


2
2
³
´
 cos 
(q)  = (sin ) Solution:  0 = ((sin ) ) ln (sin ) +
sin 
5
(r)  = sin−1 (5) Solution:  0 = √
1 − 252
¡ ¢
(s)  = cos 2 Solution:  0 = −2 sin 2
¶
µ
³ 2 ´
¢

1 ¡
(t)  = ln  +1 Solution:  0 = 2
2 − 2 2 + 1
 +1

√



(u)  = 2 + 5 Solution: 
= 0 while 
=√
2 + 5

−1
(v)  = ( + sin ) Solution:  0 =  ( + sin )
(cos  + 1)
√
√
2
(w)  =  2 + 1 Solution:  0 = 2 + 1 + √
2 + 1
(x)  = sin3 () Solution:  0 = 3 cos  sin2 
¡p  p  ¢
6. Find the tangent line to the curve cos () = 2 −  2 at the point
2
2 .
r
r


cos () = 2 −  2 :  =
; =
2
2
− sin () ( +  0 ) = 2 − 2 0
2 0 − sin ()  0 = 2 + sin () 
r
r ¶
µ
2 + sin () 


0 =
: =
; =
2 − sin () 
2
2
¡p  p  ¢ p 
¡ ¢ p
p
p
p
p
2 2 + sin 2
2 2 + 2
2 2 + sin
2
2
2
2
¡p  p  ¢ p  = p 
¡ ¢ p = p p = 3
p
 =
2 2 − sin
2 2 − sin 2
2 2− 2
2
2
2
2
r
r ¶
µ


−
= 3 −
2
2
r

 = 3 − 2
2
√
2 3 7 − 14
7. Find the derivative of  =
(HINT: Easier if you use logarithmic differentiation).
(1 + 2 )4
¢
¡
1
1
ln  = 2 ln  + ln 7 + ln ( − 2) − 4 ln 1 + 2
3
3
√
µ
¶
2 3 7 − 14 2
1
8
0
 =
+
−
4
 3 ( − 2) 1 + 2
(1 + 2 )
8. Suppose the function  () is defined as  () =
½
2 + 5 + 
√
2 + 4
function continuous on the interval (−∞ +∞)?
lim−  () = 
→0
lim  () = 2
→0+

9. Evaluate:
Z
1
(a)
 = tan−1  + 
2 + 1
Z
¢

1 ¡
(b)
 = ln 2 + 1 + 
2
 +1
2
(c)
Z1
√
2√
1
 2 + 1 =
2 − = 0609
3
3
0
(d)
Z1
(e)
Z
0
(f)
(g)
Z
Z
¢
¡ 2
5
 + 3 − 1  = = 083
6
¢
¡
2
 = ln 2 + 1 + 
+1
2
2 − 3 3
2 3
 = − −  2 + 
2

 2
√
 
2 √
√  =   + 
3
3 
= 2
if
if
0
. For what value of  is this
≥0
(h)
Z
(i)
Z
43
(4 + 1)

2
(j)
9
4

(k)

=−
1
+
4 + 1
1
cos 
 =
3
3
0
Z
2 
√
1−2 
√
 = −8

Z
1
1
 =


Z2
2
1
 =


1

(l)

1

(m)

Z
sin 
sin 
 = 2
+1
 +1
2
0
(n)


√
Z0 √
sin 
sin 

=
−
2 + 1
2 + 1


(o)

Z
0
(p)
Z1
0
(q)
sin−1 
sin−1 
√
 = √
1 − 2
1 − 2
1
sin−1 
√
 =  2
8
1 − 2
√
1
Z 3
1
 = 19 
1 + 92
0
(r)
√
2
Z 3
1
1
 =

2
4 + 9
18
0
2
2
10. Graph the function  = 2 = 2 −  Determine the asymptotes (if any), intercepts (if nice), where the

function is increasing/decreasing and concavity. Identify the points of extrema and in ection points (if any).
LABEL ALL OF THESE IMPORTANT POINTS.
(a) Horizontal asymptote is ?
(b) Explain (in your own words why there is no vertical asymptote)
2
(c) Show that the first derivative can be simplified to  0 = 2−
¡
¢
1 − 2
(d) Identify the points where the first derivative is zero
(e) Identify the interval(s) where the function is increasing
(f) Identify the interval(s) where the function is decreasing
(g) Show that the second derivative can behsimplified to
i
¢
¡ ¢
¡ ¢2
2 ¡
2
 00 = − 44 − 102 + 2 = − 4 2 − 10 2 + 2
(h) Show that the points where the second derivative is zero are approximately: −1 51 −047 047 and1 51
(i) Identify the interval(s) where the function is concave up
(j) Identify the interval(s) where the function is concave down
(k) Graph the function
y
0.5
0.4
0.3
0.2
0.1
-3
-2
-1
1
-0.1
2
3
x
11. For the functions below:
(a) Find the roots of  () 
(b) Find the horizontal and vertical asymptotes of  ()
(c) Determine the critical points and the inflection points of the function  () 
(d) Identify the interval(s) where the function is increasing or decreasing.
(e) Where is the function  () concave up, and where is it concave down?
(f) What are the relative extrema of the function  ()?
(g) Sketch the graph of  () showing the most important features.
1
 () =
y

1−2
2
2
 () = −
0.4
y
20
0.2
10
-4
-2
2
-10
4
-5
x
-4
-3
-2
-1
-0.2
 () =
2
3
4
1
2
3
4
2
4
6
8
5
x
-0.4
-20
3
1
2

(−1)(+2)
4
 () = 4 − 82 − 10
y
20
y
20
10
-5
-4
-3
-2
-1
-10
1
2
3
4
5
x
-5
-4
-3
-2
-1
-20
5
x
-20
5
 () =
22 −8
2 −16
6
 () =
+1
2 +1
y 1.0
10
y
0.5
-14 -12 -10 -8 -6 -4 -2
2
4
6
8 10 12 14
x
-10
-8
-6
-4
-2
-10
7
22 + 8 + 6
 () =
2 − 4
8
 () = 34 − 22 − 2
20
y
10
x
y
2
10
-5
-4
-3
-2
-1
1
2
3
4
-10
5
-3
-2
x
-20
9
 () =  () =
√
3
2 − 1
y
2
1
-5
-4
-3
-2
-1
1
2
3
4
5
x
12. Position function of a particle is  () = 12 4 − 4 + 10 for  ≥ 0
 () = 23 − 4
 () = 62
:
(a) Find the position, the velocity and the accceleration at the time  = 2
 (2) = 10 ft,  (2) = 12 ft/s and  (2) = 24 ft/s2 
-1
1
-2
2
3
x
(b) At what time did the particle stop?
√
, Solution is:  = 3 2 = 1 26 seconds
 () = 23 − 4 = 0
(c) At what time interval(s) is the particle’s velocity increasing?
0 () =  ()  0 : Always
¡
¢2
13. Find the absolute maximum and the absolute minimum of  () = 2 +  3 on the interval [3 7] 
2 (2 + 1)
1
: Critical points:  = −   = 0  = −1
 0 () = √
3
2
2
3 +
√
√
3
3
2
Between 3 and 7 :  (3) = 12 = 5 241 (Min) :  (7) = 562 = 14 637 (Max)
14. A rectangle is to be inscribed in a right triangle having sides of length 6 in, 8 in, and 10 in. Find the dimensions
of the rectangle with greatest area assuming the rectangle is positioned as in the accompanying figure. (see
below)
8
8−
6
=
:  = (8 − ) :  in [0 8]
6

8
¢
6
6¡
 () =  =  (8 − ) =
8 − 2
8
8
6
0
 =
(8 − 2) : Critical point:  = 4;  = 3
8
 :  (0) = 0;  (4) = 12;  (8) = 0
  :  = 4;  = 3
15. A box-shaped wire frame consists of two identical wire squares whose vertices are connected by four straight
wires of equal length. If the frame is to be made from a wire of length 96 in, what should the dimensions be to
obtain a box frame of greatest volume? (see figure above)
96 − 8
length = 4 + 4 + 4 = 8 + 4 = 96 :  =
= 24 − 2 where  is in [0 12]
4
Volume =  = 2  = 2 (24 − 2) :  () = 242 − 23
 0 () = 48 − 62 = 6 (8 − )
Critical points:
 = 0;  = 8 in
 (0) = 0; MAX:  (8) = 512 in3 ; and  (12) = 0
16. A closed cylindrical can (with top and bottom) is to hold 100 cm3 of liquid. How should we choose the height
and the radius to minimize the amount of material needed to manufacture the can?
100
 = 2  = 100 :  = 2 :  is in (0 ∞)

100
200
 = 22 + 2 : Surface  () = 22 + 2 2 =
+ 22


¢
¡
4 3 − 50
200
 0 () = 4 − 2 =
2

r 
3 50
Critical points:
 = 0; and  =
= 2 515

µ
¶
200
 = 0 : lim
+ 22 = ∞
→0

Ãr !
r
3 50
3 50
 =
:
= 19 955 :  


µ
¶
200
2
 = ∞ : lim
+ 2 = ∞
→∞

17. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance
around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?
 = 2 · 
 = 108 = 4 +  :  = 108 − 4
 () = 2 (108 − 4)

18. Express the number 100 as a sum of two nonnegative numbers whose product is as large as possible.
100 =  +  :  = 100 − 
 =  =  (100 − )

19. Two nonnegative numbers multiply to 25. How small can their sum be?
25
 = 25 :  =

25
 = + =+


20. Find the largest√possible value of 2 + , if  and  are the lengths of the sides of a right triangle whose
hypothenuse is 5 inches long.
 = 2 + 
h √ i
p
2 +  2 = 5 :  = 5 − 2 :  is in 0 5
h √ i
p
Find the max of  () = 2 + 5 − 2 on the interval 0 5

21. A conical water tank with vertex down has a radius of 10 ft at the top and is 24 ft high. If water flows into the
tank at a rate of 20 ft3  min, how fast is the depth of the water increasing when the water is 16 ft deep?
related rates see your book/notes
22. An open box is to be made from a 16-inch by 30-inch piece of cardboard by cutting out squares of equal size
from the four corners and bending up the sides. What size should the squares be to obtain a box with the largest
volume?
= (16 − 2) (30 − 2)  = 43 − 922 + 480 for 0 ≤  ≤ 8
10
 0 () = 122 − 184 + 480 = 0 for  = 12
3
 = 0 :  (0) = 0
µ ¶ µ
µ ¶¶ µ
µ ¶¶ µ ¶
10
10
10
10
10
 =
:
= 16 − 2
30 − 2
= 725 93 :  
3
3
3
3
3
 = 8 :  (8) = 0
¡
¡ ¢¢
¡
¡ ¢¢
Answer: box with largest volume is of the size 16 − 2 10
= 9 33 by 30 − 2 10
= 23 33.
3
3

23. A farmer has 4000 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs
no fence along the river. What are the dimensions of the field that has the largest area?
2 +  = 4000 :  = 4000 − 2
 =  =  (4000 − 2) = 4000 − 22 : 0 ≤  ≤ 2000
4000
0 () = 4000 − 4 :  =
= 1000 :
4
 (0) = 0
 (750) = (1000) (4000 − 2 (1000)) = 2 000 000 :  rectangle is 1000 × 2000 × 1000
 (2000) = 0
24. Find the point on the line  = 2 − 4 that is closest to the origin!
2
2
2
2
 between ( ) and (0 0) = () = ( − 0) + ( − 0) = 2 +  2 = 2 + (2 − 4)
0 = 2 + 2 (2 − 4) (2) = 10 − 16 = 0 for  = 16 :  = 2 (16) − 4 = −08
q
2
2
(16 − 0) + (−08 − 0) ≈ 179
 =
25. Find the point on the line  = 2 + 6 that is closest to the origin! What is the shortest distance (up to two
decimal places)?
2
2
2
2
 between ( ) and (0 0) = () = ( − 0) + ( − 0) = 2 +  2 = 2 + (2 + 6)
0 = 2 + 2 (2 + 6) 2 = 10 + 24 = 0 for  = −24 :  = 2 (−24) + 6 = 12
q
2
2
(−24 − 0) + (12 − 0) ≈ 2 68
 =
26. A piece of wire 100 cm long is going to be cut into several pieces and used to construct the skeleton of
rectangular box with a square base. What are the dimensions of the box with the largest volume.
see previous prbs.
27. Find the largest√possible value of 3 + 5, if  and  are the lengths of the sides of a right triangle whose
hypothenuse is 5 inches long.
see previous prbs
28. How should two nonnegative numbers be chosen so that their sum is 100 and the sum of their squares is as small
as possible?
 +  = 100 :  = 100 −  in [0 100]
 = 2 +  2 = 2 + (100 − )2

29. Sand pouring from a chute forms a conical pile whose height is always equal to the diameter. If the height
increases at a constant rate of 5 ft/min, at what rate is sand pouring from the chute when the pile is 8 ft high?
related rates see your book/notes
30. A truck driving over a flat interstate at a constant speed of 50 mph gets 4 miles per gallon. Fuel costs $1.19 per
gallon. The truck loses a tenth of a mile per gallon in fuel efficiency for each mile per hour increase in speed.
Drivers are paid $27.50 per hour in wages and benefits. Fixed costs for running the truck are $11.33 per hour. A
trip of 300 miles is planned. What speed minimizes operating expenses?
 =  :  =  : distance =  =  ·  = 300 miles
300
300
 at  = 50 mph  (50) =
· 119 +
· (2750 + 1133)
4
50
300
300
3570
11649
 at   50 mph  () =
(119) +
(2750 + 1133) =
+
1

90 − 

4−
( − 50)
10
Find minimum of  () for speed  in interval [50 90]

31. Sand is being dumped on a pile in such a way that it always forms a cone whose radius equals its height. If the
sand is being dumped at a rate of 10 cubic feet per minute, at what rate is the height of the pile increasing when
there is 1000 cubic feet of sand in the pile?
related rates see your book/notes
32. A 10-ft plank is leaning against a wall. If a certain instant the bottom of the plank is 2 ft from the wall and
is being pushed toward the wall at the rate of 6 in/s, how fast is the acute angle that the plank makes with the
ground increasing?
related rates see your book/notes
33. A stone dropped into a still pond sends out circular ripple whose radius increases at a constant rate of 3 ft/s. How
fast is the area enclosed by the ripple increasing at the end of 10 seconds?
related rates see your book/notes
34. A baseball diamond is a square 90 ft on a side. A runner travels from home plate to first base at 20 ft/sec. How
fast is runner’s distance from second base changing when the runner is halfway to first base?
related rates see your book/notes
35. A highway patrol plane flies one mile above a straight section of rural interstate highway (speed limit 55 mph) at
a steady ground speed of 120 miles per hour. The pilot sees an oncoming car and, with radar, determines that the
line-of-sight distance from the plane to the car is 1.5 miles and that the distance is decreasing at a rate 136 miles
per hour. Should the driver of the car be given a ticket for speeding? Explain it to the judge.
related rates see your book/notes
36. A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 35 ft/sec.
How fast must she let out her string when the kite is 500 feet away from her?

2 + 3002 =  2 :
= 35 ft/s

p
√
When  = 500  = 3002 + 5002 = 100 34 = 583 1 ft
:
2
37. Let  () =
Z



 
500
= 2
or
=
=
(35) = 30 ft/s



 
5831
 ()  where  is the function graphed below (NOTE: The graph of  is made up of straight
−1
lines and semicircles.)
4
3
2
1
-5 -4 -3 -2 -1
-1
1
-2
-3
Evaluate  (−5),  (2), and  (4) 
see previous prbs
2
3
4
5
38. Let  () =
Z
 ()  where  is the function graphed below (NOTE: The graph of  is made up of straight
−5
lines and a semicircle.)
6
5
4
3
2
1
-5 -4 -3 -2 -1
1
2
3
4
5
-1
-2
-3
Which is the smallest:  (−4)   (−2) or  (5)? Justify your answer.
see previous prbs
39. Let  () =
Z
 ()  where  is the function graphed below (NOTE: The graph of  is made up of straight
2
lines and a semicircle.)
6
5
4
3
2
1
-5 -4 -3 -2 -1
1
2
3
4
5
-1
-2
-3
Which is the smallest:  (−5)   (−2) or  (5)? Justify your answer.
see previous prbs
40. What is the total area between the function  () =  − 1 and the − axis on the interval −1 ≤  ≤ 3
Z 1
Z 3
Total Area =
| − 1|  +
| − 1| 
−1
1
=
Z
−1
1
(1 − )  +
Z
1
3
( − 1) 
µ
µ
¶3
¶1
1 2
1
 − 2
+
 −
2
2
−1
1
µ
¶ µ
¶ µ
¶ µ
¶
1
1 2
1
1
=
1−
− −1 −
+
3 −3 −
−1 =4
2
2
2
2
=
41. Let  () =
Z
 ()  where  is the function graphed below (NOTE: The graph of  is made up of straight
−3
lines and semicircles.)
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2
3
4
5
-2
-3
Evaluate  (−5),  (2), and  (4) 
µZ −35
¶
Z −3
Z −3
Z −5
 ()  = −
 ()  = −
 ()  +
 () 
 (−5) =
−3
−5
−5
−35
µ
¶
1
1
= −
(15) (−3) + (05) (1) = 2
2
2
Z −1
Z 2
Z 2
 ()  =
 ()  +
 () 
 (2) =
−3
−3
−1
¶
µ
1
1
=
(2) (1) − 12 + (3) (1) = 5 − 
2
2
Z 2
Z 3
Z 4
Z 4
 ()  =
 ()  +
 ()  +
 () 
 (4) =
−3
−3
2
3
µ
¶
¶ µ
¶ µ
1
1
1
1 2
=
5 −  + 1 + (1) (2) + 3 + 1 = 10 − 
2
2
4
4
42. A particle moves with acceleration  () =  − 2 2 along an  - axis and has velocity  = 0  at time
 = 0 sec. Find the displacement and the distance traveled by the particle during the interval 1 ≤  ≤ 5
 () =  − 2
Z
1
 () =
 ()  = 2 − 2 +  :  (0) = 0 so  = 0
2
¶
Z
Z µ
1 2
1
 () =
 ()  =
 − 2  = 3 − 2 + 
2
6
µ
¶
µ
¶
1 3
1 3
2
2
dsiplacement is  (5) −  (1) =
−
 − +
 − +
6
6
µ
¶=5 µ
¶ =1
1 3
1
10
=
5 − 52 +  −
13 − 12 +  = −
6
6
3
1
1
velocity is zero:
 () = 2 − 2 = 0  ( − 4) = 0 :  = 0 and  = 4 :
2
2
Change of direction at  = 4
total distance =
| (5) −  (4)| + | (4) −  (1)| =
¯µ
¶ µ
¶¯ ¯µ
¶ µ
¶¯
¯ 1 3
¯ ¯ 1 3
¯ 17
1 3
1 3
2
2
2
2
¯
¯
¯
= ¯ 5 −5 + −
4 −4 + ¯+¯ 4 −4 + −
1 − 1 +  ¯¯ =
6
6
6
6
3
43. A model rocket is fired vertically at ground level upward from rest. Its acceleration for the first three seconds is
() = 60 at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the
rocket’s parachute opens, and the (downward) velocity slows linearly to −18 ft/s in 5 s. The rocket then "floats"
to the ground at that rate.
(a) Determine the position function s and the velocity function v (for all times t). Sketch the graphs of s and v.
During the first 3 seconds  () = 60;  () = 302 + , but since
¡ ¢ (0) = 0 (from rest);
So : the velocity function is  () = 302 ; at  = 3 is  (3) = 30 32 = 270 ft/s
The position function is then:  () = 103 +  , and given that it is fired from the ground  (0) = 0 so
 = 0 and  () = 103
 () = 60 ft/s2
 () = 302 ft/s
 () = 103 ft
:
:
:
0≤≤3
 (0) = 0 ft/s2
 (0) = 0 ft/s
 (0) = 0 ft
:
:
:
 (3) = 180 ft/s2
 (3) = 270 ft/s
 (3) = 270 ft
Next fourteen seconds: from  = 3 to  = 17 with acceleration = gravity, so we have  () = −32
ft/s2 and  () = −32 +  so 270 = −32 (3) + , so  = 366, or  () = 366 − 32 ft/s, and at
 = 17 :  (17) = 366 − 32 (17) = −178 ft/s
2
The position function
 (3) = 270, we have
¡ 2is¢ then  () = 366 − 16 + ; and given that
270 = 366 (3) − 16 3 + , so,  = −684 and  () = 336 − 162 − 684 Then:
 () = −32 ft/s2
 () = 366 − 32 ft/s
 () = 366 − 162 − 684 ft
:
:
:
3 ≤  ≤ 17
 (3) = −32 ft/s2
 (3) = 270 ft/s
 (3) = 270 ft
:
:
:
 (17) = −32 ft/s2
 (17) = −178 ft/s
 (17) = 914 ft
From  = 17 to  = 22 (5 s), rocket’s parachute opens, and the (downward) velocity slows linearly from −178 to −18 ft/s. For 17   ≤ 22 :  (17) = −178 ft/s and  (22) = −18 ft/s: so the slope of
this linear function (derivative of velocity) is acceleration  () = −18−(−178)
= 32 and  () = 32 +  (a
22−17
linear function). Plug it in: −178 = 32 (17) + , Solution is:  = −722 or  () = 32 − 722 ft/s
The position function is then  () = 162 − 722 + , and since we know that  (17) = 914 we have
914 = 16 (17)2 − 722 (17) + , Solution is:  = 8564 and therefore  () = 162 − 722 + 8564
17 ≤  ≤ 22
 () = 32 ft/s2
:
 (17) = 32 ft/s2
:  (22) = 32 ft/s2
 () = 32 − 722 ft/s
:  (17) = −178 ft/s :  (22) = −18 ft/s
 () = 162 − 722 + 8564 ft :
 (17) = 914 ft
:
 (22) = 424 ft
After  = 22 seconds, the velocity is constant  () = −18 ft/s., hence  () = 0, and  () = −18 + 
Plug in  = 22 to get  : 424 = −18 (22) + , so  = 820 and  () = −18 + 820
22 ≤ 
 () = 0 ft/s2
:
 (22) = 0 ft/s2
 () = −18 ft/s
:  (22) = −18 ft/s
 () = −18 + 820 ft :
 (22) = 424 ft
All together:
180
y 160
⎧
60 for
⎪
⎪
⎨
−32 for
 () =
for
⎪ 32
⎪
⎩
0
for
0≤≤3
3   ≤ 17
17   ≤ 22
22  
140
120
100
80
60
40
20
0
-20
10
20
30
x
y
200
⎧
⎪
⎪
⎨
302
for
366 − 32 for
 () =
32 − 722 for
⎪
⎪
⎩
−18
for
0≤≤3
3   ≤ 17
17   ≤ 22
22  
100
0
10
20
30
x
-100
y
1500
1000
⎧
⎪
⎪
⎨
103
for
366 − 162 − 684 for
 () =
⎪ 162 − 722 + 8564 for
⎪
⎩
−18 + 820
for
0≤≤3
3   ≤ 17
17   ≤ 22
22  
500
0
0
10
20
30
x
(b) At what time does the rocket reach its maximum height? (Give your answer correct to one decimal place.)
obviously - the height is maximum when  () = 0 : (between 3    17)
183
366 − 32 = 0 so  =
= 11 437 5 ≈ 114 s
16
(c) What is that height? (Give your answer correct to the nearest whole number.)
2
 (114) = 366 (114) − 16 (114) − 684 = 1409 04 ≈ 1409 ft
(d) At what time does the rocket land? (Give your answer correct to one decimal place.)
410
rocket lands when  () = 0 (after  = 22 s), so − 18 + 820 = 0, so  =
= 45 555 5 ≈ 456 s
9
44. A projectile is launched upward from ground level with an initial speed of 256 ft/s.
 (0) = 0 ft (ground level),  (0) = 256 ft/s (initial velocity), and  () = −32 ft/s2 (gravity)
 () = −32 + 256
 () = −162 + 256
(a) How long does it take for the projectile to reach its highest point?
at highest point  () = 0 so − 32 + 256 = 0 or  = 8 s
(b) How high does the projectile go?
¡ ¢
 (8) = −16 82 + 256 (8) = 1024 ft
(c) When will the projectile reach the height of 880 feet?
 () = 880 for 880 = −162 + 256
twice  = 5 (on the way up) and  = 11 (on the way down)
(d) What is the speed of the projectile when it reaches 880 feet?
 (5) = 96 ft/s (Up)and  (11) = −96 ft/s (Down)
(e) What is the acceleration of the projectile when it reaches 880 feet?
 (5) =  (11) = −32 ft/s2 ()
45. Position function of a particle is  () = 12 4 − 4 + 10 for  ≥ 0
(a) Find the position, the velocity and the accceleration at the time  = 2
1 4
 () =
 − 4 + 10 :  (2) = 10
2
 () = 23 − 4 :  (2) = 12
 () = 62 :  (2) = 24
(b) At what time did the particle stop?
√
3
 () = 0 for 23 − 4 = 0 or  = 2
(c) At what time interval(s) is the particle’s velocity increasing?
√
3
 ()  0 when   2
46. The position (in feet) of a particle is given by  =  () = 3 − 62 + 9 , where  is in seconds.
(a) Find the velocity function
¡
¢
 () = 32 − 12 + 9 = 3 2 − 4 + 3 = 3 ( − 3) ( − 1)
(b) What is the velocity after 4 s?
2
 (4) = 3 (4) − 12 (4) + 9 = 9 ft/s
(c) When is the particle at rest?
 () = 0 :  = 3 s or  = 1 s
(d) When is the particle moving forward (that is in the positive direction)?
 () ≥ 0; for 0 ≤  ≤ 1 and for 3 ≤ 
(e) Find the total distance traveled by the particle during the first five seconds?
³
´ ³
´
3
2
3
2
distance for  = 0 to  = 1 :  (1) −  (0) = (1) − 6 (1) + 9 (1) − (0) − 6 (0) + 9 (0) = 4
³
´ ³
´
distance for  = 1 to  = 3 :  (3) −  (1) = (3)3 − 6 (3)2 + 9 (3) − (1)3 − 6 (1)2 + 9 (1) = −4
³
´ ³
´
distance for  = 0 to  = 1 :  (5) −  (3) = (5)3 − 6 (5)2 + 9 (5) − (3)3 − 6 (3)2 + 9 (3) = 20
total distance = (4 ft in positive) + (4 ft in negative) + (20 ft positive) = 28 ft
(f) Find the acceleration at time  = 4 seconds.
 () = 6 − 12
 (4) = 6 (4) − 12 = 12 ft/s2