# 3. 0/5 points | Previous Answers

January 13, 2016

3. 0/5 points | Previous Answers
Lecture Question 3:
As in lecture 2, consider the distance function d(t)=16t 2 feet, where t is in units of seconds. The time interval [1,b] (with 1 < b)
on which the average velocity is 33 ft/sec is: (Select ALL correct answers; you will only be allowed two submissions on this
question)
4. 0/5 points | Previous Answers
Lecture Question 4:
As in lecture 2, consider the distance function d(t)=16t 2 feet, where t is in units of seconds. The slope of the tangent line at the
point P=(1,16) on the graph of the function is: (Select ALL correct answers; you will only be allowed two submissions on this
question)
5. 1/14 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).
[1,3]
[1,3/2]
[1,17/16]
[1,9/8]
[1,5/4]
The instantaneous velocity at time t=0 sec.
The average velocity on the time interval [1,2].
Negative.
The instantaneous velocity at time t=2 sec.
The instantaneous velocity at time t=1 sec.
Positive.
(a) The location of the top of the ladder will be given by parametric equations (0,y(t)). The formula for y(t)=
(b) The domain of t values for y(t) ranges from 0 to 24
(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] [2,4]
[6,8] [8,9]
(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=
(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”)
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
The table shows the position of a cyclist.
t (seconds) 0 1 2 3 4 5
s (meters) 0 1.1 5.4 10.5 17.5 26.8
(a) Find the average velocity for each time period.
(i) [1, 3]
4.7 m/s
(ii) [2, 3]
5.1 m/s
(iii) [3, 5]
8.15 m/s
(iv) [3, 4]
7 m/s
(b) Estimate the instantaneous velocity when t = 3.
6.6 m/s
Enhanced Feedback
Please try again. When calculating the average velocity, you are actually calculating the slope of the secant line between the two
given points. Use the table to calculate the difference in distance for the given 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 matching the trend for the given
time.
7. –/12 pointsSCalcET7 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 the
volume V of water remaining in the tank (in gallons) after t minutes.
t (min) 5 10 15 20 25 30
V (gal) 3470 2220 1225 580 140 0
(a) If P is the point (15, 1225) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph
Q slope
(5, 3470)
(10, 2220)
(20, 580)
(25, 140)
(30, 0)
(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.)
8. –/11 pointsSCalcET7 2.1.AE.001.
Video Example
EXAMPLE 1 Find an equation of the tangent line to the function
at the point P(1, 5).
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, 11.25) 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 15 0 5
y = 5×2
Q(x, 5×2)
x ? 1 Q ? P.
mPQ = . 5×2 – 5
x – 1
mPQ = = = . – 5
– 1 .5
1.5 12.5 .5 7.5
1.1 10.5 .9 9.500
1.01 10.050 .99 9.950
1.001 10.005 .999 9.995
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 pointslope
form of the equation of a line (see Appendix B) to write the equation of
the tangent line through (1, 5) 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.
lim mPQ = m and = . Q ? P
lim
x ? 1
5×2 – 5
x – 1
y – = (x – 1) y = x – .
9. –/10 pointsSCalcET7 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 six
decimal places) for the following values of x.
(i) 3.9
(ii) 3.99
(iii) 3.999
(iv) 3.9999
(v) 4.1
(vi) 4.01
(vii) 4.001
(viii) 4.0001
(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
P(4, -4) y = 4/(3 – x).
(x, 4/(3 – x)), mPQ
mPQ =
mPQ =
mPQ =
mPQ =
mPQ =
mPQ =
mPQ =
mPQ =
P(4, -4).
m =
P(4, -4).

Current Score : – / 66 Due : Tuesday, January 12 2016 11:00 PM PST
1. –/5 points
Lecture Question 3:
In lecture 3, the meaning of the notation
is carefully discussed and means:
“We can make the values of f(x) ARBITRARILY CLOSE to L by taking x SUFFICIENTLY CLOSE to a, BUT x is not equal to a.”
The lecture then looks at the example of f(x)=2-x+x2 and notes that . The lecture goes on to explain that if we
insist that the values of f(x) be within 1 unit of L=4, then we can determine how close the inputs x need to be to the number 2.
Suppose instead we want the values of f(x) to be within 1/2 unit of L=4, then this will happen if we take the inputs x in the
interval : (Select ALL correct answers; you will only be allowed one submission on this question)
(0,2).
( ).
(1,3).
( ).
(2,4).
2. –/5 points
Lecture Question 4:
Consider the multipart function f(x) which is equal to x2 for all non-zero x and has value 7 at x=0 : (Select ALL correct answers;
you will only be allowed one submission on this question)
does not exist
is positive.
is positive.
Homework Lesson 3 (Homework)
WEI TING LEE
MATH 124, section online(G), Winter 2016
Instructor: David Collingwood
WebAssign
lim f(x)=L
x ? a
2-x+x lim 2=4
x ? 2
, 1+ 5
2
v1+ 13
2
v
, 1+ 7
2
v1+ 11
2
v
lim f(x)
x ? 0
lim f(x)
x ? 1
lim f(x)
x ? -1
lim f(x) =0
x ? 0
lim f(x) =7
x ? 0
3. –/6 pointsSCalcET7 2.2.004.
Use the given graph of f to state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
(a)
(b)
(c)
(d)
(e)
(f)
lim f(x)
x ? 2-
lim f(x)
x ? 2+
lim f(x)
x ? 2
f(2)
lim f(x)
x ? 4
f(4)
4. –/8 pointsSCalcET7 2.2.007.
For the function g whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
lim g(t)
t ? 0-
lim g(t)
t ? 0+
lim g(t) t ? 0
lim g(t)
t ? 2-
lim g(t)
t ? 2+
lim g(t) t ? 2
g(2)
lim g(t) t ? 4
5. –/4 pointsSCalcET7 2.2.010.MI.
A patient receives a 150-mg injection of a drug every 4 hours. The graph shows the amount f(t) of the drug in the bloodstream
after t hours.
Find and
6. –/4 pointsSCalcET7 2.2.011.
Sketch the graph of the function.
lim f(t)
t ? 4-
lim f(t).
t ? 4+
lim f(t) = mg
t ? 4-
lim f(t) = mg
t ? 4+
f(x) =
4 + x if x < -1
x2 if -1 = x < 1
2 – x if x = 1
Use the graph to determine the values of a for which does not exist. (Enter your answers as a comma-separated list.)
7. –/4 pointsSCalcET7 2.2.025.
Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically.
8. –/4 pointsSCalcET7 2.2.029.
Determine the infinite limit.
8
-8
lim f(x)
x ? a
a =
lim
x ? 1
x8 – 1
x4 – 1
lim
x ? -4+
x + 5
x + 4
9. –/4 pointsSCalcET7 2.2.030.MI.
Determine the infinite limit.
8
-8
10.–/4 pointsSCalcET7 2.2.030.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 skipped part.
Tutorial Exercise
Determine the infinite limit.
11.–/4 pointsSCalcET7 2.2.031.
Determine the infinite limit.
8
-8
12.–/4 pointsSCalcET7 2.2.046.
In the theory of relativity, the mass of a particle with velocity v is
where m0 is the mass of the particle at rest and c is the speed of light. What happens as

lim
x ? -4-
x + 5
x + 4
lim
x ? -9-
x + 10
x + 9
lim
x ? 5
6 – x
(x – 5)2
m = ,
m0
1 – v2/c2
v ? c-?
m ? 8
m ? 0
m ? m0
m ? -8
13.–/5 pointsSCalcET7 2.2.503.XP.MI.
Use the given graph of f to state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
(a)
(b)
(c)
(d)
(e) f(7)
14.–/5 points
Consider . Using a table of values, the limiting value is
(Enter “DNE” if the limit does not exist.)
lim f(x)
x ? 3-
lim f(x)
x ? 3+
lim f(x)
x ? 3
lim f(x)
x ? 7
lim ( )
t ? 0+
-2 sin(2t)
sin(2t) + 2 t cos(2t)

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