Math 131 Form 222 Test 2 With Answers - La Sierra University
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1
2
3
4Math 131.
Test 2, Form 222, Friday, October 23
Name.
Instructions. Do each of the following 8 questions. Please show all appropriate work, and do your best!
Express answers in exact form, unless asked to round. You may use a scientific non-graphing calculator.
1. (10 pts) Find the derivatives of the following functions. Simplify the functions first if it will make the
derivative easier to find.
4
6
(a) ( ) =
(6
5
)
(b) ( ) =
10
6
(c) ( ) =
4
(d) ( ) = cos( )
(e)
( ) = sec(6 )
4
10
3
9
Solution: (a) Write ( ) = 6
5
and so
( ) = 24
50
.
10
1 2
1 2
(b) Write ( ) =
10
and so
( ) =
.
2
4
5
(c) Write ( ) = 6
and so
( ) =
24
(d)
( ) =
sin( )
(e)
( ) = 6 sec(6 ) tan(6 )
2. (10 pts) A ball is thrown straight down from the top of a 680-foot building with an initial velocity of
14
2
feet per second. Use the position function
( ) =
16
14 + 680 and calculus to answer the following
questions.
(a) What is the ball’s velocity after 3 seconds.
(b) How long did it take the ball to fall 135 feet?
(c) What was the ball’s speed after falling 135 feet?
2
Solution: Since ( ) =
16
14 + 680, the velocity is given by ( ) = ( ) =
32
14.
(a) After 3 seconds, the velocity is (3) =
32(3)
14 =
110 feet per second.
(b) The height of the ball will be 680
135 = 545 feet when it has fallen 135 feet. Therefore, we solve
( ) = 545 for to find how long it took the ball to fall 135 feet. Thus
2
2
2
16
14 + 680 = 545
16
14 + 135 = 0
16
+ 14
135 = 0
and we want the positive solution to the given quadratic (since the position we found function is valid for
0). Using the quadratic formula, we find
2
14 +
(14)
4(16)( 135)
=
= 2 5 seconds
2(16)
(c) After 2 5 seconds, the speed of the rock was (2 5) =
32(2 5)
14 =
94 feet per second.
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