The table shows the position of a cyclist.

4/10/17, 6:21 PMSpring 2017 Math 1304-01 Homework #3

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1. –/1.25 pointsSCalcET8 2.1.501.XP.

The table shows the position of a cyclist.

t (seconds) 0 1 2 3 4 5

s (meters) 0 1.3 4.6 10.9 17.2 25.7

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

(i) [1, 3] m/s

(ii) [2, 3] m/s

(iii) [3, 5] m/s

(iv) [3, 4] m/s

(b) Estimate the instantaneous velocity when t = 3. m/s

Spring 2017 Math 1304-01 Homework #3 (Homework) Hamad Aldarweesh Math 1304, section 01, Spring 2017 Instructor: Chung-Hsing OuYang

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2. –/1 pointsSCalcET8 2.1.AE.003.

Time interval

Average velocity (m/s)

Video Example

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

SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by s(t) and measured 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 a single instant of time (t = 5), so no time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the 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 velocity over successively smaller time periods.

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

v = 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 49.0049 s(t) = 4.9t2.

average velocity =

=

=

= m/s.

change in position time elapsed

s(5.1) − s(5) 0.1

4.9 − 4.9

0.1

2

2

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3. –/1.25 pointsSCalcET8 2.1.005.

If a ball is thrown into the air with a velocity of 50 ft/s, its height in feet t seconds later is given by y = 50t − 16t2.

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

(i) 0.5 seconds ft/s

(ii) 0.1 seconds ft/s

(iii) 0.05 seconds ft/s

(iv) 0.01 seconds ft/s

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

4. –/1.5 pointsSCalcET8 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 soon as 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 the graph (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 that cuts (intersects) a curve more than once.]

We choose so that Then,

For instance, for the point Q(1.5, 20.25) 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, the closer mPQ is to . This suggests that the slope of the tangent line t should be m =

.

x mPQ x mPQ

2 60 0 4

1.5 32.5 .5 7.5

1.1 18.564 .9 13.756

y = 4x4

Q(x, 4x4)

x ≠ 1 Q ≠ P.

mPQ = . 4x4 − 4 x − 1

mPQ = = = . − 4 − 1 .5

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1.01 16.242 .99 15.762

1.001 16.024 .999 15.976

We say that the slope of the tangent line is the limit of the slopes of the secant 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 of the 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 rotate about P and approach the tangent line t.

5. –/3 pointsSCalcET8 2.1.502.XP.

The point P(9, 1) lies on the curve

(a) If Q is the point use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the

following values of x.

(i) 8.5

(ii) 8.9

(iii) 8.99

(iv) 8.999

(v) 9.5

mPQ = m and = .lim Q → P

lim x → 1

4x4 − 4 x − 1

y − = (x − 1) y = x − .

y = .x − 8

(x, ),x − 8

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(vi) 9.1

(vii) 9.01

(viii) 9.001

(b) Using the results of part (a), guess the value of the slope 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

(d) Sketch the curve, two of the secant lines, and the tangent line.

P(9, 1).

P(9, 1). y =

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