1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
November 21
1. Find f (x) for the following.
√
(a) f 0 (x) = 5 x + 4 + sec(x) tan(x)
10
f (x) = x3/2 + 4x + sec(x) + C
3
00
(b) f (x) = 8 sin(3x) + e2x if f (0) = 6 and f 0 (0) = 5
−8
1
f 0 (x) =
cos(3x) + e2x + C
3
2
1
43
Since f 0 (0) = 5 this means that 5 = −8
3 + 2 + C or C = 6
−8
43
1
f (x) =
sin(3x) + e2x + x + K
9
4
6
Since f (0) = 6 this means that 6 = 0 + 14 + K or K = 5.75
−8
43
1
Answer: f (x) =
sin(3x) + e2x + x + 5.75
9
4
6
3 − 5x + 15
3
10x
10x
5x
15
10x
5
15
(c) f 0 (x) =
=
− 2+ 2 =
−
+ 2
2
2
4x
4x
4x
4x
4
4x 4x
5
5 1 15 −2
0
f (x) = x − ∗ + x
2
4 x
4
5 2 5
−15 −1
f (x) = x − ln |x| +
x +C
4
4
4
2
(d) f 0 (x) = (x3 + x2 + 1)(x + 2) + 3x = x4 + 3x3 + 2x2 + x + 2 + e−3x
e
x5 3x4 2x3 x2
−1 −3x
f (x) =
+
+
+
+ 2x +
e
5
4
3
2
3
Z
2.
Z
x9 + a2 dx =
x10
+ a2 x + C
10
a3
+C
3
→
−
4. Find the position vector if −
a (t) = t2 i + cos(2t)j, →
v (0) = j,
D 3
E
→
−
→
a = t2 , cos(2t) and −
v = t + C, 1 sin(2t) + K . Since
3.
x9 + a2 da = x9 a +
3
2
−
→
r (0) = i
→
−
v (0) =< 0, 1 > this means that
c = 0 Dand K = 1.
E
D 4
E
3
t
→
−
→
→
v = t3 , 12 sin(2t) + 1 and −
r = 12
+ M, − 14 cos(2t) + t + N . Since −
r =< 1, 0 > this means
that M = 1 and N =
−
Answer: →
r =
D
t4
12
1
4
+ 1, − 14 cos(2t) + t +
1
4
E
.
5. A ball is thrown upward with a velocity of 50ft/sec from the edge of a building. The building is
150 ft tall.
(a) Find the height of the ball after t seconds. (gravity is either 9.8m/sec2 or 32 ft/sec2
a = −32 so v = −32t + C. Since v(0) = 50 we get C = 50.
v = −32t + 50. s = −16t2 + 50t + K. Since s(0) = 150, this means that k = 150.
answer: s = −16t2 + 50t + 150
(b) When does it reach its max height? at t = 50/32 or 1.5625 seconds.
(c) When does it hit the ground? set s=0 and solve for t. take only the positive answer.
Answer= t = 5seconds.