Current Score : 80 / 100 Due : Tuesday, January 21 2014 10:00 PM CST

1. 25/25 points | Previous Answers SCalcET7 2.1.001.MI.SA.

This question has several parts that must be completed sequentially. If you skip a part of the question,you will not receive any points for the skipped part, and you will not be able to come back to the skippedpart.

A tank holds 4000 gallons of water, which drains from the bottom of the tank in half an hour. The valuesin 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) 2752 1776 1000 404 116 0

Exercise (a)If P is the point (15, 1000) on the graph of V, find the slopes of the secant lines PQ when Q isthe point on the graph with the following values.

Q

(5, 2752)

(10, 1776)

(20, 404)

(25, 116)

(30, 0)

Part 1 of 5Slopes of secant lines passing through P(15, 1000) and can be calculated using

We begin by calculating the slope when t = 5.

Section 2.1 (Homework)Zachary FergerSpring 2014 Math 1550, section 18, Spring 2014Instructor: Jesse Levitt

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Q(t, V(t))

msec = .V(t) − 1000t − 15

msec =

=

= -175.2 (rounded to the nearest tenth)

V(5) − 10005 − 15

2752 − 1000-10

Part 2 of 5When t = 10, we get

Part 3 of 5When t = 20, we get

Part 4 of 5When t = 25, we get

Part 5 of 5And finally, when t = 30, we get

Exercise (b)Estimate the slope of the tangent line at P(15, 1000) by averaging the slopes of two adjacentsecant lines.

Part 1 of 2Looking at the table, the t-values adjacent to t = 15 are

t = 10 (smaller value)

t = 20 (larger value).

Part 2 of 2To estimate the slope of the tangent line at t = 15, we average the slopes of the adjacent secantlines for t = 10 and t = 20. As obtained in part (a), those slopes are

respectively.

msec =

=

= -155.2 (rounded to the nearest tenth).

V(10) − 100010 − 15

1776 − 1000-5

msec =

= -119.2 (rounded to the nearest tenth).

404 − 100020 − 15

msec =

= -88.4 (rounded to the nearest tenth).

116 − 100025 − 15

msec =

= -66.67 (rounded to the nearest tenth).

0 − 100030 − 15

msec = −155.2 and msec = −119.2,

Therefore, the slope of the tangent line at t = 15 is as follows. (In the last step, round youranswer to one decimal place.)

If you use a graph of the function to estimate the slope of the tangent line at P(15, 1000), youwill see that it matches the above result.You have now completed the Master It.

= -137.2 −155.2 + -119.2

2

2. 25/25 points | Previous Answers SCalcET7 2.1.003.

The point lies on the curve

(a) If Q is the point use your calculator to find the slope of the secant linePQ (correct to six decimal places) for the following values of x.

(i) 4.9

(ii) 4.99

(iii) 4.999

(iv) 4.9999

(v) 5.1

(vi) 5.01

(vii) 5.001

(viii) 5.0001

(b) Using the results of part (a), guess the value of the slope m of the tangent line to the curveat

(c) Using the slope from part (b), find an equation of the tangent line to the curve at

P(5, −4) y = 4/(4 − x).

(x, 4/(4 − x)), mPQ

mPQ = 4.444444

mPQ = 4.040404

mPQ = 4.004004

mPQ = 4.000400

mPQ = 3.636363

mPQ = 3.960396

mPQ = 3.996004

mPQ = 3.999600

P(5, −4).m = 4

P(5, −4).

3. 25/25 points | Previous Answers SCalcET7 2.1.005.

If a ball is thrown in the air with a velocity 40 ft/s, its height in feet t seconds later is given by y = 40t −

16t2.

(a) Find the average velocity for the time period beginning when t = 2 and lasting

(i) 0.5 second.-32.2 ft/s

(ii) 0.1 second.-25.6 ft/s

(iii) 0.05 second.-24.8 ft/s

(iv) 0.01 second.-24.16 ft/s

(b) Estimate the instantaneous velocity when t = 2.-24 ft/s

4. 5/25 points | Previous Answers SCalcET7 2.1.007.

The table shows the position of a cyclist.

t (seconds) 0 1 2 3 4 5

s (meters) 0 1.6 4.6 10.1 17.1 25.5

(a) Find the average velocity for each time period.

(i) [1, 3]2.5 m/s

(ii) [2, 3]2.83 m/s

(iii) [3, 5]4.23 m/s

(iv) [3, 4]3.82 m/s

(b) Estimate the instantaneous velocity when t = 3.6.3 m/s

Enhanced Feedback

Please try again. When calculating the average velocity, you are actually calculating the slope of thesecant line between the two given points. Use the table to calculate the difference in distance for thegiven time period, and divide by the change in time.

To estimate the instantaneous velocity, look for a trend in average velocity and find a value matchingthe trend for the given time.