section 2.1 quiz

Current Score : 63 / 69 Due : Saturday, February 1 2014 11:59 PM EST

1. 5/6 points | Previous Answers SCalcET7 2.1.001.MI.

A tank holds 5000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show thevolume V of water remaining in the tank (in gallons) after t minutes.

t (min) 5 10 15 20 25 30

V (gal) 3430 2150 1200 575 135 0

(a) If P is the point (15, 1200) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graphwith the following values. (Round your answers to one decimal place.)

Q slope

(5, 3430) -223

(10, 2150) -190

(20, 575) -106.5

(25, 135) -106.5

(30, 0) -80

Master ItA tank holds 6000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the tableshow the volume V of water remaining in the tank (in gallons) after t minutes.

t (min) 5 10 15 20 25 30

V (gal) 4104 2646 1440 678 162 0

If P is the point (15, 1440) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graphwith the following values.

Q

(5, 4104)

(10, 2646)

(20, 678)

(25, 162)

(30, 0)

Step 1 of 5Slopes of secant lines passing through P(15, 1440) and can be calculated using

We begin by calculating the slope when t = 5.

Section 2.1 QUIZ (Quiz)Frances CoronelMAT 151 Calculus I, Spring 2014, section 01, Spring 2014Instructor: Ira Walker

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Q(t, V(t))

msec = .V(t) − 1440t − 15

msec =

=

= (No Response) (rounded to the nearest tenth)

V(5) − 14405 − 15

(No Response) − 1440(No Response)

(b) Estimate the slope of the tangent line at P by averaging the slopes of two adjacent secant lines. (Round your answerto one decimal place.)-157.5 Master ItA tank holds 6000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the tableshow the volume V of water remaining in the tank (in gallons) after t minutes.

t (min) 5 10 15 20 25 30

V (gal) 4104 2646 1440 678 162 0

Estimate the slope of the tangent line at P(15, 1440) by averaging the slopes of two adjacent secant lines.Step 1 of 2Looking at the table, the t-values adjacent to t = 15 are

t = (No Response) (smaller value)

t = (No Response) (larger value).

Enhanced Feedback

Please try again. To find the slope of a secant line, use the formula .

For the best estimate of the slope in part (b), use the secant lines whose endpoints are closest to the original point.

2. 20/20 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 anypoints for the skipped part, and you will not be able to come back to the skipped part.

A tank holds 4000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show thevolume V of water remaining in the tank (in gallons) after t minutes.

t (min) 5 10 15 20 25 30

V (gal) 2764 1788 960 460 104 0

Exercise (a)If P is the point (15, 960) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph withthe following values.

Q

(5, 2764)

(10, 1788)

(20, 460)

(25, 104)

(30, 0)

Part 1 of 5Slopes of secant lines passing through P(15, 960) and can be calculated using

mPQ = V2 − V1t2 − t1

Q(t, V(t))

http://www.webassign.net/web/Student/Assignment-Responses/last?dep=8337862#

We begin by calculating the slope when t = 5.

Part 2 of 5When t = 10, we get



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

Exercise (b)Estimate the slope of the tangent line at P(15, 960) by averaging the slopes of two adjacent secant 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 secant lines for t = 10 andt = 20. As obtained in part (a), those slopes are respectively.

Therefore, the slope of the tangent line at t = 15 is as follows. (In the last step, round your answer to one decimalplace.)

msec = .V(t) − 960t − 15

msec =

=

= -180.4 (rounded to the nearest tenth)

V(5) − 9605 − 15

2764 − 960-10

msec =

=

= -165.6 (rounded to the nearest tenth).

V(10) − 96010 − 15

1788 − 960-5

msec =

= -100 (rounded to the nearest tenth).

460 − 96020 − 15

msec =

= -85.6 (rounded to the nearest tenth).

104 − 96025 − 15

msec =

= -64 (rounded to the nearest tenth).

0 − 96030 − 15

msec = −165.6 and msec = −100,

If you use a graph of the function to estimate the slope of the tangent line at P(15, 960), you will see that it matches theabove result.You have now completed the Master It.

3. 9/10 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 line PQ (correct to sixdecimal places) for the following values of x.

(i) 6.9

(ii) 6.99

(iii) 6.999

(iv) 6.9999

(v) 7.1

(vi) 7.01

(vii) 7.001

(viii) 7.0001

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

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

= -132.8 −165.6 + -100

2

P(7, −3) y = 3/(6 − x).

(x, 3/(6 − x)), mPQ

mPQ = 3.333333

mPQ = 3.030303

mPQ = 3.003003

mPQ = 3.00030

mPQ = 2.727272

mPQ = 2.970297

mPQ = 2.997002997

mPQ = 2.9997003

P(7, −3).m = 3

P(7, −3).




The table shows the position of a cyclist.

t (seconds) 0 1 2 3 4 5

s (meters) 0 1.5 4.9 10.7 17.6 26.7

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

(i) [1, 3]4.6 m/s

(ii) [2, 3]5.8 m/s

(iii) [3, 5]8 m/s

(iv) [3, 4]6.9 m/s

(b) Estimate the instantaneous velocity when t = 3.6.35 m/s


The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s =

3 sin πt + 4 cos πt, where t is measured in seconds. (Round your answers to two decimal places.)

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

(i) [1, 2]8 cm/s

(ii) [1, 1.1]-7.312770483 cm/s

(iii) [1, 1.01]-9.22585187 cm/s

(iv) [1, 1.001]-9.405023265 cm/s

(b) Estimate the instantaneous velocity of the particle when t = 1.-9 cm/s

6. 8/11 points | Previous Answers SCalcET7 2.1.AE.001.



Video Example

EXAMPLE 1 Find an equation of the tangent line to the function at the point P(1, 4).

SOLUTION We will be able to find an equation of the tangent line t as soonas we know its slope m. The difficulty is that we know only one point, P, on t,whereas we need two points to compute the slope. But observe that we can

compute an approximation to m by choosing a nearby point on thegraph (as in the figure) and computing the slope mPQ of the secant line PQ.

[A secant line, from the Latin word secans, meaning cutting, is a line thatcuts (intersects) a curve more than once.]

We choose so that Then,

For instance, for the point Q(1.5, 9) we have

The tables below show the values of mPQ for several values of x close to 1.

The closer Q is to P, the closer x is to 1 and, it appears from the tables, thecloser mPQ is to 8 . This suggests that the slope of the tangent line t

should be m = 8 .

x mPQ x mPQ

2 12 0 4

1.5 10 .5 6

1.1 8.4 .9 7.600

1.01 8.040 .99 7.960

1.001 8.004 .999 7.996

We say that the slope of the tangent line is the limit of the slopes of thesecant lines, and we express this symbolically by writing

Assuming that this is indeed the slope of the tangent line, we use the point-slope form of the equation of a line (see Appendix B) to write the equation ofthe tangent line through (1, 4) as

or The graphs below illustrate the limiting process that occurs in this example.As Q approaches P along the graph, the corresponding secant lines rotateabout P and approach the tangent line t.

y = 4x2

Q(x, 4x2)

x ≠ 1 Q ≠ P.

mPQ = .4x2 − 4x − 1

mPQ = = = 2 .5 − 41.5 − 1

1 .5

mPQ = m and = 8 .lim Q → P

lim x → 1

4x2 − 4x − 1

y − 4 = 8 (x − 1) y = 8 x − 4 .


t Q

0.00 100.00

0.02 81.89

0.04 67.05

0.06 54.89

0.08 44.95

0.10 36.78

R mPR

(0.00, 100.00)

(0.02, 81.89)

(0.06, 54.89)

(0.08, 44.95)

(0.10, 36.78)

Video Example

EXAMPLE 2 The flash unit on a camera operates by storing charge on acapacitor and releasing it suddenly when the flash is set off. The data in thetable describe the charge Q remaining on the capacitor (measured inmicrocoulombs) at time t (measured in seconds after the flash goes off). Usethe data to draw the graph of this function and estimate the slope of thetangent line at the point where t = 0.04. [Note: The slope of the tangent linerepresents the electric current flowing from the capacitor to the flash bulb(measured in microamperes).]

SOLUTION In the figure we plot the data and use it to sketch a curve thatapproximates the graph of the function.

Given the points P(0.04, 67.05) and R(0.00, 100.00) on the graph, we findthat the slope of the secant line PR is

The table at the left shows the results of similar calculations for the slopes ofother secant lines. From this table we would expect the slope of the tangentline at t = 0.04 to lie somewhere between and In fact,the average of the slopes of the two closest secant lines is

So, by this method, we estimate the slope of the tangent line to be -675 (rounded to the nearest integer).

Another method is to draw an approximation to the tangent line at P andmeasure the sides of the triangle ABC, as in the figure. This gives anestimate of the slope of the tangent line as

−823.75

−742.00

−608.00

−552.50

−504.50

mPR = = 812.5 .100.00 − 67.05 0.00 − 0.04

−742.00 −608.00.

0.5(−742.00 − 608.00) = -675 .

− ≈ − = −670.|AB||BC|

80.4 − 53.60.06 − 0.02




Timeinterval

Averagevelocity (m/s)

Video Example

EXAMPLE 3 Suppose that a ball is dropped from the upper observationdeck of the CN Tower in Toronto, 450 m above the ground. Find the velocityof the ball after 5 seconds.

SOLUTION Through experiments carried out four centuries ago, Galileodiscovered that the distance fallen by any freely falling body is proportionalto the square of the time it has been falling. (This model for free fall neglectsair resistance.) If the distance fallen after t seconds is denoted by s(t) andmeasured in meters, then Galileo's law is expressed by the equation

The difficulty in finding the velocity after 5 s is that we are dealing with asingle instant of time (t = 5), so no time interval is involved. However, wecan approximate the desired quantity by computing the average velocity overthe brief time interval of a tenth of a second from t = 5 to t = 5.1:

The table shows the results of similar calculations of the average velocityover successively smaller time periods.

It appears that as we shorten the time period, the average velocity isbecoming closer to 49 m/s (rounded to one decimal place). Theinstantaneous velocity when t = 5 is defined to be the limiting value ofthese average velocities over shorter and shorter time periods that start at t= 5. Thus the (instantaneous) velocity after 5 s is the following. (Round youranswer to one decimal place.)

v = 49 m/s

5 ≤ t ≤ 6 53.9

5 ≤ t ≤ 5.1 49.49

5 ≤ t ≤ 5.05 49.245

5 ≤ t ≤ 5.01 49.049

5 ≤ t ≤ 5.001 49.0049s(t) = 4.9t2.

average velocity =

=

=

= 49.49 m/s.

change in positiontime elapsed

s(5.1) − s(5)0.1

4.9 5.1 − 4.9 5

0.1

2

2


9. 2/2 points | Previous Answers SCalcET7 2.1.JIT.003.MI.

Find the slope-intercept form of the equation of the line passing through the given points.

y =

Sketch the line.

Master-ItFind the slope-intercept form of the equation of the line passing through the given points.

Part 1 of 3

(1, 1), (5, − )13

Submission Data

(2, 2), (4, − )32

Use the two-point form of the linear equation. Fill in the missing blanks using (2, 2) for (x1, y1). You will need both points to

determine the slope, .y2 − y1x2 − x1

y − 2 = (x − (No Response) )− − (No Response)

(No Response) − (No Response)

32


10.1/1 points | Previous Answers SCalcET7 2.1.JIT.004.MI.

Find an equation of the line that satisfies the given conditions.

Through (2, 8); parallel to the line passing through

(3, 6) and (−1, 2)


section 2.1 quiz

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