Answers and Solutions to Section 2.1 Homework Problems S. F. Ellermeyer

[Pages:12]Answers and Solutions to Section 2.1 Homework Problems S. F. Ellermeyer

1. A tank holds 1000 gallons of water, which drains from the bottom of the tank

in half an hour. The values in the table show the volume, V, of water remaining in the tank (in gallons) after t minutes.

t (min) 5 10 15 20 25 30

V (gal) 694 444 250 111 28 0

a. If P is the point 15, 250 on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with t 5, 10, 20, 25, and 30.

i. The slope of the secant line passing through the points P15, 250 and Q5, 694 is

mPQ

694 250 5 15

444 10

44. 4.

ii. The slope of the secant line passing through the points

P15, 250 and Q10, 444 is

mPQ

444 250 10 15

194 5

38. 8.

iii. The slope of the secant line passing through the points

1

P15, 250 and Q20, 111 is

mPQ

111 250 20 15

139 5

27. 8.

iv. The slope of the secant line passing through the points

P15, 250 and Q25, 28 is

mPQ

28 250 25 15

222 10

22. 2.

v. The slope of the secant line passing through the points

P15, 250 and Q30, 0 is

mPQ

0 250 30 15

250 15

16. 6 .

b. Estimate the slope of the tangent line at P by averaging the slopes

of two secant lines.

The slope of the secant line passing through the points P15, 250

and Q10, 444 is 38. 8 and the slope of the secant line passing

through the points P15, 250 and Q20, 111 is 27. 8. The average

of these slopes is

38. 8 27. 8 2

33. 3.

We can thus estimate that the slope of the tangent line at the point P15, 250 is 33. 3. However, we certainly don't know that this is the exact value of the slope of the tangent line and, furthermore, there

is no way to find the exact value because we are only given a

discrete set of data points. In order to find the exact value of the slope of the tangent line at P15, 250, we would need to be given the values of V at all values of t in some (at least small) time interval containing t 15.

c. Use a graph of the function to estimate the slope of the tangent line at P. (This slope represents the rate at which the water is flowing from the tank at the instant t 15 min. In other words, it is the instantaneous flow rate at t 15 min.)

Once again (just as in part b), our estimate is bound to be rather

crude because it relies on guessing from incomplete information.

Using the hand?drawn guess at a tangent line shown in the picture

below, we arrive at

m

170 315 17 12. 5

32. 2 .

This is remarkably close to the answer that we got in part b!

2

3.

The point P

1,

1 2

lies on the curve y

x 1x

.

a. Use your calculator to find the slope of the secant line passing

through P and the point Q

x,

x 1x

where x is each of the following

values given in parts i through viii below.

i. We will do this one "by hand" without using a calculator: If x 1/2, then

1

y

2

1

1 2

1 3

so the point Q is Q

1 2

,

1 3

. The slope of the secant line

passing through P and Q (pictured below) is

mPQ

1 3

1 2

1 2

1

1 6

1 2

1 3

0. 333333.

3

ii. If x 0. 9, then

y

0. 9 1 0. 9

9 10 19

9 19

10

so the point Q is Q

9 10

,

9 19

. The slope of the secant line

passing through P and Q is

mPQ

9 19

1 2

9 10

1

0. 26315 8.

iii. If x 0. 99, then

y

0. 99 1 0. 99

99 100 199 100

99 199

so the point Q is Q

99 100

,

99 199

. The slope of the secant line

passing through P and Q is

mPQ

99 199

1 2

99 100

1

0. 251256.

iv. If x 0. 999, then

y

0. 999 1 0. 999

999 1000 1999

999 1999

1000

so the point Q is Q

999 1000

,

999 1999

. The slope of the secant

line passing through P and Q is

4

mPQ

999 1999

1 2

999 1000

1

0. 250125.

v. If x 1. 5, then

y

1. 5 1 1. 5

15 10 25 10

3 5

so the point Q is Q

15 10

,

3 5

. The slope of the secant line

passing through P and Q is

mPQ

3 5

1 2

15 10

1

1 5

0. 2.

vi. If x 1. 1, then

y

1. 1 1 1. 1

11 10 21 10

11 21

so the point Q is Q

11 10

,

11 21

. The slope of the secant line

passing through P and Q is

mPQ

11 21

1 2

11 10

1

0. 238095.

vii. If x 1. 01, then

y

1. 01 1 1. 01

101 100 201

101 201

100

so the point Q is Q

101 100

,

101 201

. The slope of the secant line

passing through P and Q is

mPQ

101 201

1 2

101 100

1

0. 248756.

viii. If x 1. 001, then

y

1. 001 1 1. 001

1001 1000 2001 1000

1001 2001

so the point Q is Q

1001 1000

,

1001 2001

. The slope of the secant

line passing through P and Q is

mPQ

1001 2001

1 2

1001 1000

1

0. 249875.

b. Using the results in part (a), guess the slope of the tangent line to

the curve y

x 1x

at the point P

1,

1 2

.

Based on these calculations, I guess that the slope of the tangent

line at P is 0. 25 (which is the same as 1/4)

c.

The point in question is P

1,

1 2

and I am guessing that the slope of

the tangent line is 1/4. Therefore (assuming that my guess for the

slope is correct), an equation of the tangent line is

5

y

1 2

1 4

x

1

which can also be written (in slope?intercept form) as

y

1 4

x

1 4

.

When I graph this line along with the curve below, it looks to be

reasonable!

5. If a ball is thrown into the air with a velocity of 40 ft/sec, its height (in feet) after t seconds is given by y 40t 16t2. a. Find the average velocity for the time period beginning when t 2

and lasting

i. 0.5 sec When t 2, we have y 40 2 16 22 16. This means that 2 seconds after the ball is thrown, it is 16 feet above

the ground. When t 2. 5, we have y 40 2. 5 16 2. 52 0. This means that 2. 5 seconds after the ball is thrown, it is on the

ground.

The average velocity of the ball during the time period

2 t 2. 5 is

0 ft 16 ft 2. 5 s 2 s

16 ft 0. 5 s

32 ft / s .

6

This means that, on average, the ball was travelling downward at a speed of 32 ft / s during the final 1/2 second

of its flight. Note that the velocity of the ball is always

changing. The velocity that we have computed is an average velocity over a particular time interval.

ii. 0.1 sec

When t 2. 1, we have y 40 2. 1 16 2. 12 13. 44. This means that 2. 1 seconds after the ball is thrown, it is 13. 44 ft above the ground.

The average velocity of the ball during the time period

2 t 2. 1 is

13. 44 ft 16 ft 2. 1 s 2 s

25. 6 fst .

iii. 0.05 sec

When t 2. 05, we have y 40 2. 05 16 2. 052 14. 76. This means that 2. 05 seconds after the ball is thrown, it is 14. 76 ft above the ground.

The average velocity of the ball during the time period

2 t 2. 05 is

14. 76 ft 16 ft 2. 05 s 2 s

24. 8 fst .

iv. 0.01 sec

When t 2. 01, we have y 40 2. 01 16 2. 012 15. 758 4. This means that 2. 05 seconds after the ball is thrown, it is 15. 7584 ft above the ground.

The average velocity of the ball during the time period

2 t 2. 01 is

15. 7584 ft 16 ft 2. 01 s 2 s

24. 16 fst .

b. Find the instantaneous velocity when t 2.

Based on the calculations done above, I guess that the

instantaneous velocity of the ball at the instant t 2 is 24

ft s

.

This

means that exactly 2 seconds after the ball is thrown, it is travelling

downward at a speed of 24

ft s

.

Below

is

shown

a

graph

of

the

function y 40t 16t2 along with the line y 16 24t 2, which

is the tangent line at the point P2, 16.

7

7. The displacement (in feet) of a certain particle moving in a straight line is given by s t3/6, where t is measured in seconds.

a. Find the average velocity over the following time periods:

i. 1, 3

The average velocity of the particle over the time period from t 1 to t 3 is

33

13

6

6

31

2. 166667 ft/sec.

ii. 1, 2

The average velocity of the particle over the time period from t 1 to t 2 is

23

13

6

6

21

1. 166667 ft/sec.

iii. 1, 1. 5

The average velocity of the particle over the time period from t 1 to t 1. 5 is

1.53

13

6

6

1. 5 1

0. 791667 ft/sec.

iv. 1, 1. 1

The average velocity of the particle over the time period from t 1 to t 1. 1 is

8

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