AP Calculus AB / Mr. Hansen
9/15/2003

Name: _________________________

Solutions to Chapter 1 Practice Test, pp. 35-36

 

Instructions: Work the test first, and then check your answers. The practice test is of little value if you merely look at the problems and then read the solutions. You should be able to do all of these problems in about 40 minutes (60 minutes for extended time).

 

 

T1.


[Tick marks are optional. However, if you omit the tick marks, the coordinates of the points must be clearly labeled.]

T2.


[Again, tick marks are optional. However, if you omit the tick marks, the coordinates of the point must be clearly labeled. The open circle is required, and the left- and right-hand limits must appear to be equal.]

T3.

f ¢(1) » –1.683 by calc. [Since problem did not specify method, using calc. is fine. For full credit, you should use correct notation as shown and give answer rounded to 3 decimal places.] The concept is the derivative of a function at a point.

T4.

i

xi

yi = f (xi)

weight

weight · yi

 

 

 

0

1

1.2

1

1.2

 

 

 

1

1.5

1.31 . . .

2

2.62 . . .

 

 

 

2

2

1.44

2

2.88

 

 

 

3

2.5

1.57 . . .

2

3.15 . . .

 

 

 

4

3

1.72 . . .

2

3.45 . . .

 

 

 

5

3.5

1.89 . . .

2

3.78 . . .

 

 

 

6

4

2.07 . . .

2

4.14 . . .

 

 

 

7

4.5

2.27 . . .

2

4.54 . . .

 

 

 

8

5

2.48 . . .

1

2.48 . . .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TOTAL:

28.284 . . .

 

 

 

 

 

ò15 f (x) dx » T = ½ Dx (28.284 . . .) = 7.071

 

[You need to show approximately this much work: formula, plug-ins, and result. It is fine to write the dots as shown above while storing full precision in your calculator’s memory. Note that if you round your intermediate results, your final answer will probably not be accurate to the 3 decimal places that are required for full credit.]

T5a.


[You can create a graph very much like this—including the hole!—if you set your graphing window’s Xmin and Xmax to be centered on –2. For example, Xmin = –5 and Xmax = 1 will do the trick.]

b.

limx®–2 g(x) = –3

c.

[See above. The range of x values is very narrow, from –2.08 to –1.92. We can calculate these values precisely since the function is essentially a line with slope 5. In other words, d = 0.08 in order to guarantee that the function stays within 0.4 of –3.]

d.

c = –2
L = –3
e = 0.4
d = 0.08

e.

We want to be able to define a limit at a point even if the function itself is not defined there. [In the example, we found a limit as x approaches –2 even though g(2) is undefined.]

[To say “... but not equal to c” is not really a restriction. Your book is misleading in this wording. It is more correct to say that the phrase “... but not equal to c” is a relaxation of a restriction.]

T6a.

v(1) = 2.7181 = 2.718 in./min.

b.

v¢(1) » 2.718 in./min.2 by calc. [no work needed]
[Notice the units: v¢ must be in acceleration units, namely inches per minute per minute.]

c.

ò01 v(t) dt = 1.718 in. by calc.
[No work is required, but the integral notation is required in order to provide justification.]

T7.

Let v(t) = velocity as a fcn. of time.
ò028 v(t) dt » T = ½ Dt (v0 + 2v4 + 2v8 + 2v12 + 2v16 + 2v20 + 2v24 + v28)
 = ½(4) (16 + 2(26) + 2(30) + 2(34) + 2(37) + 2(36) + 2(32) + 25)
 = 2(431)
 = 862 ft.
[Note that since we do not have a formula for v(t), we cannot compute the exact answer to this problem. We have to settle for an approximation. The amount of work shown above is correct.]