**Unformatted text preview: **6/5/2019 hw02S2.1
FEVEN WELDEYES YARED
Math 124, section A, Spring 2019
Instructor: Jarod Alper TA WebAssign
hw02S2.1 (Homework)
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View Key 1. 10/10 points | Previous AnswersSCalcET8 2.1.501.XP. The table shows the position of a cyclist.
t (seconds) 0
s (meters) 1 2 3 4 5 0 1.3 4.7 10.6 17.5 26.9 (a) Find the average velocity for each time period.
(i) [1, 3] 4.65 m/s (ii) [2, 3] 5.9 m/s (iii) [3, 5] 8.15 m/s (iv) [3, 4] 6.9 m/s (b) Estimate the instantaneous velocity when t = 3. 6.4 m/s 1/10 6/5/2019 hw02S2.1 2. 12/12 points | Previous AnswersSCalcET8 2.1.001.MI. A tank holds 3000 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) 2091 1290 765 327 75 0 (a) If P is the point (15, 765) on the graph of V, find the slopes of the secant lines PQ when Q is the point
on the graph with the following values. (Round your answers to one decimal place.)
Q slope (5, 2091) 132.6 (10, 1290) 105 (20, 327) 87.6 (25, 75) 69 (30, 0) 51 (b) Estimate the slope of the tangent line at P by averaging the slopes of two adjacent secant lines. (Round
your answer to one decimal place.) 96.3 2/10 6/5/2019 hw02S2.1 3. 11/11 points | Previous AnswersSCalcET8 2.1.AE.001. EXAMPLE 1 Find an equation of the tangent line to the function y = 4x4 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 Q(x, 4x4) 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
Video Example (intersects) a curve more than once.] We choose x ≠ 1 so that Q ≠ P. Then,
4 mPQ = 4x − 4 .
x − 1
For instance, for the point Q(1.5, 20.25) we have
20.25 − 4
16.25 = = 32.5 .
1.5 − 1
.5
The tables below show the values of mPQ for several values of x close
mPQ = 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 16 . This suggests that the slope of the tangent line t should be m = 16 x mPQ x mPQ 2 60 0 4 1.5 32.5 .5 7.5 1.1 18.564 .9 13.756 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
4
lim mPQ = m and lim 4x − 4 = 16 .
Q →P x → 1 x − 1
Assuming that this is indeed the slope of the tangent line, we use the pointslope form of the equation of a line (see Appendix B) to write
the equation of the tangent line through (1, 4) as
y − 4 y = 16 = 16 x − 12 (x − 1) or . The graphs below illustrate the limiting process that occurs in this
example. As Q approaches P along the graph, the corresponding 3/10 6/5/2019 hw02S2.1 secant lines rotate about P and approach the tangent line t. 4/10 6/5/2019 hw02S2.1 4. 10/10 points | Previous AnswersSCalcET8 2.1.003. The point P(7, −4) lies on the curve y = 4/(6 − x).
(a) If Q is the point (x, 4/(6 − x)), use your calculator to find the slope mPQ of the secant line PQ (correct
to six decimal places) for the following values of x.
(i) 6.9 mPQ = 4.444444 (ii) 6.99 mPQ = 4.040404 (iii) 6.999 mPQ = 4.004004 (iv) 6.9999 mPQ = 4.000400 (v) 7.1 mPQ = 3.636364 (vi) 7.01 mPQ = 3.960396 (vii) 7.001 mPQ = 3.996004 (viii) 7.0001 mPQ = 3.999600 (b) Using the results of part (a), guess the value of the slope m of the tangent line to the curve at P(7, −4). m = 4 (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(7, −4). $$y=4x−32 5/10 6/5/2019 hw02S2.1 5. 14/14 points | Previous Answers A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall.
The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The
location of the foot of the ladder at time t seconds is given by the parametric equations (7+2t,0). (a) The location of the top of the ladder will be given by parametric equations (0,y(t)). The formula for
y(t)=
$$√576−28t−4t2 . (Put your cursor in the box, click and a palette will come up to help you enter your symbolic answer.) (b) The domain of t values for y(t) ranges from 0 to 9 (c) Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three
decimal places):
time interval ave velocity time interval ave velocity [0,2] .775 [2,4] 1.225 [6,8] 3.225 [8,9] 9.798 (d) Find a time interval [a,9] so that the average velocity of the top of the ladder on this time interval is 20
ft/sec i.e. a= 8.7525 (e) Using your work above and this picture of the graph of the function y(t) given below, answer these
true/false questions: (Type in the word "True" or "False") 6/10 6/5/2019 hw02S2.1 The top of the ladder is moving down the wall at a constant rate
T
F The foot of the ladder is moving along the ground at a constant rate
T
F There is a time at which the average velocity of the top of the ladder on the time interval
[a,9] is 1 ft/sec
T
F There is a time at which the average velocity of the top of the ladder on the time interval
[a,9] is 0 ft/sec
T
F There is a time at which the average velocity of the top of the ladder on the time interval
[a,9] is 100 ft/sec
T
F There is a time at which the average velocity of the top of the ladder on the time interval
[a,9] is less than 100 ft/sec
T
F 7/10 6/5/2019 hw02S2.1 6. 4/4 points | Previous Answers The cup on the 9th hole of a golf course is located dead center in the middle of a circular green which is 40 feet in
radius. Your ball is located as in the picture below. The ball follows a straight line path and exits the green at the
rightmost edge. Assume the ball travels 10 ft/sec. Introduce coordinates so that the cup is the origin of an xy
coordinate system. Provide numerical answers below with two decimal places of accuracy. (a) The xcoordinate of the position where the ball enters the green will be 17.53 . (b) The ball will exit the green exactly 9.43 seconds after it is hit. (c) Suppose that L is a line tangent to the boundary of the golf green and parallel to the path of the ball. Let
Q be the point where the line is tangent to the circle. Notice that there are two possible positions for Q. Find
the possible xcoordinates of Q: smallest xcoordinate = 21.2 largest xcoordinate = 21.2 8/10 6/5/2019 hw02S2.1 7. 4/4 points | Previous Answers A Ferris wheel of radius 100 feet is rotating at a constant angular speed ω rad/sec counterclockwise. Using a
stopwatch, the rider finds it takes 4 seconds to go from the lowest point on the ride to a point Q, which is level with
the top of a 44 ft pole. Assume the lowest point of the ride is 3 feet above ground level. Let Q(t)=(x(t),y(t)) be the coordinates of the rider at time t seconds; i.e., the parametric equations. Assuming the
rider begins at the lowest point on the wheel, then the parametric equations will have the form: x(t)=rcos(ωtπ/2)
and y(t)=rsin(ωt π/2), where r,ω can be determined from the information given. Provide answers below accurate
to 3 decimal places. (Note: We have imposed a coordinate system so that the center of the ferris wheel is the
origin. There are other ways to impose coordinates, leading to different parametric equations.)
(a) r = 100 feet (b) ω = .235 rad/sec (c) During the first revolution of the wheel, find the times when the rider's height above the ground is 80
feet. first time = 5.7 second time= 21 sec sec 9/10 6/5/2019 hw02S2.1 8. 3/3 points | Previous AnswersSCalcET8 10.1.033. Find parametric equations for the path of a particle that moves along the circle x2 + (y − 3)2 = 16 in the manner
described. (Enter your answer as a commaseparated list of equations. Let x and y be in terms of t.)
(a) Once around clockwise, starting at (4, 3). 0 ≤ t ≤ 2π. $$x=4cos(t), y=3−4sin(t) (b) Four times around counterclockwise, starting at (4, 3). 0 ≤ t ≤ 8π. $$x=4cos(t), y=4sin(t)+3 (c) Halfway around counterclockwise, starting at (0, 7). 0 ≤ t ≤ π. $$x=−4sin(t), y=4cos(t)+3 10/10 ...

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