Calculus: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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Question

Chapter 3.1, Problem 21E

(a)

To determine

Tocalculate:The time rate and the date on which the amount of daylight increasing at the fastest rate.

(a)

Expert Solution

Answer to Problem 21E

The required date is April 1^s^t; and the rate is 0.15 hours per day.

Explanation of Solution

Given information:

Starting period is January 1.

End period is December 31.

Each vertical ticks corresponds to 3 hours.

Each month has 30 days.

Daylight graph

Calculus: Graphical, Numerical, Algebraic, Chapter 3.1, Problem 21E , additional homework tip 1

Formula Used:

Slop formula of a straight line:

Slop between two points (x1,y1) and (x2,y2) is

m=y2−y1x2−x1 , where m represent the slop of the line connecting the points (x1,y1) and (x2,y2) .

Calculation:

At the highest point where the graph has the steepest positive curve, the sum of daylight would grow. From the given graph steepest points is at x=90 .

Given that each month has 30 days, represented by the horizontal ticks.

Thus, after 90 days the month will be April 1.

Now, find the slop at that point,

Here each vertical tick corresponds to 3 hours.

Consider the points of the graph 30 days before and 30 days later than April 1.

Consider 30 days before

Thus the points are (60, 6) and (120, 15).

Calculate the slop between these points. Use the slop formula of a straight line.

m=15−6120−60=960=320=0.15

Hence, the required date is April 1^s^t; and the rate is 0.15 hours per day.

(b)

To determine

The days on which the rate of change in the amount of daylight is zero.

(b)

Expert Solution

Answer to Problem 21E

The required days are January 1 and July 1.

Explanation of Solution

Given information:

Starting period is January 1.

End period is December 31.

Each vertical ticks corresponds to 3 hours.

Each month has 30 days.

Daylight graph

Calculus: Graphical, Numerical, Algebraic, Chapter 3.1, Problem 21E , additional homework tip 2

If the slop of the graph is zero, the rate of change will be zero.

In the graph the points where the graphstops increasing and starts decreasing or stop decreasing and starts increasing, slop will zero.

There are two points where the slop is zero. These are at x=0 and x=180 .

Now, find the month at x=0 and x=180 .

From the graph at x=0 , date is January 1.

At x=180 , date is July 1.

Hence, the required days are January 1 and July 1.

(c)

To determine

The dates on which the rate of change in the number of daylight hours positive.

(c)

Expert Solution

Answer to Problem 21E

The rate of change is positive from January 1 to 30 June and the negative rate of change is from July 1 to 31 December.

Explanation of Solution

Given information:

Starting period is January 1.

End period is December 31.

Each vertical ticks corresponds to 3 hours.

Each month has 30 days.

Daylight graph

Calculus: Graphical, Numerical, Algebraic, Chapter 3.1, Problem 21E , additional homework tip 3

Find the positive rate of change.

The exchange rate is positive where the graph increases. From the graph it can be seen that the curve is increases from 1 January until 30 June.

Thus, the rate is positive from 1 January to 30 June.

Similarly find the negative rate of change.

The exchange rate is negative if the graph reduces. From the graph it can be seen that the curve is decreases from 1 July to 31 December.

Hence, the rate of change is positive from January 1 to 30 June and the negative rate of change is from July 1 to 31 December.

Conclusion:

Chapter 3.1, Problem 20E

Chapter 3.1, Problem 22E

Chapter 3 Solutions

Calculus: Graphical, Numerical, Algebraic

Ch. 3.1 - Prob. 1QR Ch. 3.1 - Prob. 2QR Ch. 3.1 - Prob. 3QR Ch. 3.1 - Prob. 4QR Ch. 3.1 - Prob. 5QR Ch. 3.1 - Prob. 6QR Ch. 3.1 - Prob. 7QR Ch. 3.1 - Prob. 8QR Ch. 3.1 - Prob. 9QR Ch. 3.1 - Prob. 10QR

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