010. `query 10
Question:
`qquery
the family of linear functions, Problem 2.
Describe the graphs of y = A f(x) for A = -.3 and A = 1.3 and
compare; explain the comparison.
Your solution:
Confidence Assessment:
Given Solution:
** For the basic linear function f(x) = x the A = -.3 graph
is obtained by vertically stretching the y = x function by factor -.3, resulting
in a straight line thru the origin with slope -.3, basic points (0,0) and (1,
-.3), and
the A = 1.3 graph is obtained by vertically stretching the y
= x function by factor 1.3, resulting in is a straight line thru the origin with
slope 1.3, basic points (0,0) and (1, 1.3). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qdescribe the graphs of y = f(x) + c for c = .3 and c = -2.7 and
compare; explain the comparison.
Your solution:
Confidence Assessment:
Given Solution:
** The graphs will have slopes identical to that of the
original function, but their y intercepts will vary from -2.7 to .3. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 4. linear function y = f(x) = -1.77 x - 3.87
What are your symbolic expressions, using x1 and x2, for the
corresponding y coordinates y1 and y2.
Your solution:
Confidence Assessment:
Given Solution:
** y1 = f(x1) = -1.77 x1 - 3.87
y2 = f(x2) = -1.77 x2 - 3.87.
`dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77
x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1).
Thus slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77.
This is the slope of the straight line, showing that these
symbolic calculations are consistent. **
STUDENT QUESTION
My question is how did you take -1.77 x2 +
1.77 x1 and get -1.77(x2 – x1)? I understand the x2-x1 but what happened to the
1.77?
INSTRUCTOR RESPONSE
This may be clearer if we work backwards:
-1.77 * (x2 - x1) = -1.77 * x2 - (-1.77 * x1) = -1.77 x2 + 1.77 x1, which is the
same thing as 1.77 x1 - 1.77 x2.
-1.77 * (x2 - x1) was chosen as the form for the numerator, so we could easily
divide it by x2 - x1.
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery problem 5. graphs of families for y = mx + b.
Describe your graph of the family: m = 2, b varies from -3 to 3 by
step 1.
Your solution:
Confidence Assessment:
Given Solution:
** The graphs will all have slope 2 and will pass thru the y
axis between y = -3 and y = 3.
The family will consist of all such graphs. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 6. three basic points graph of y = .5 x + 1
what are your three basic points?
Your solution:
Confidence Assessment:
Given Solution:
** This is of the form y = mx + b with b= 1. So the y intercept is (0, 1).
The point 1 unit to the right is (1, 1.5).
The x-intercept occurs when y = 0, which implies .5 x + 1 = 0
with solution x = -2, so the x-intercept is (-2, 0). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 6. three basic points graph of y = .5 x + 1
What are your three basic points?
Your solution:
Confidence Assessment:
Given Solution:
** The y intercept occurs when x = 0, at which point we have
y = .5 * 0 + 1 = 1. So one basic
point is (0, 1).
The point 1 unit to the right of the y axis occurs at x = 1,
where we get y = .5 * 1 + 1 = 1.5 to give us the second basic point (1, 1.5)
}The third point, which is not really necessary, is the x
intercept, which occurs when y = 0.
This gives us the equation 0 = .5 x + 1, with solution x = -2. So the third basic point is (-2, 0). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 7. simple pendulum force vs. displacement
What are your two points and what line do you get from the
two points?
Self-critique (if necessary):
Self-critique Rating:
Question:
`qSTUDENT RESPONSE:
The two points are (1.1, .21) and (2.0, .54).
These points give us the two simultaneous equations
.21- m(1.1) + b
.54= m(2.0) +b.
If we solve for m and b we will get our y = mx + b form.
INSTRUCTOR COMMENT:
I believe those are data points. I doubt if the best-fit line goes exactly
through two data points.
In the future you should use points on your sketched line,
not data points. However, we'll see
how the rest of your solution goes based on these points. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qwhat equation do you get from the slope and y-intercept?
Your solution:
Confidence Assessment:
Given Solution:
STUDENT RESPONSE:
b= .21
m=.19
INSTRUCTOR COMMENT:
** b would be the y intercept, which is not .21 since y = .21
when x = 1.1 and the slope is nonzero.
If you solve the two equations above for m and b you obtain m
= .367 and b = -.193.
This gives you equation y = mx + b or y = .367 x - .193. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qwhat is your linear regression model?
Your solution:
Confidence Assessment:
Given Solution:
** Your linear regression model would be obtained using a
graphing calculator or DERIVE. As a
distance student you are not required to use these tools but you should be aware
that they exist and you may need to use them in other courses. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat force would be required to hold the pendulum 47 centimeters
from its equilibrium position? what equation did you solve to obtain this
result?
Your solution:
Confidence Assessment:
Given Solution:
** If your model is y = .367 x - .193, with y = force and x=
number of cm from equilibrium, then we have x = 47 and we get
force = y = .367 * 47 - .193 = 17 approx. The force would be 17 force units. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhy would it not make sense to ask what force would be necessary to
hold the pendulum 80 meters from its equilibrium position? what equation did you
solve to obtain this result?
Your solution:
Confidence Assessment:
Given Solution:
STUDENT RESPONSE:
I used the equation f= .10*47+.21
and got the answer 15.41 which would be to much force to push
or pull
INSTRUCTOR COMMENT:
** The problem is that you can't hold a pendulum further at a
distance greater than its length from its equilibrium point--the string isn't
long enough. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qHow far could you hold the pendulum from its equilibrium position
using a string with a breaking strength of 25 pounds? what equation did you
solve to obtain this result?
Your solution:
Confidence Assessment:
Given Solution:
** Using the model y = .367 * x - .193 with y = force = 25
lbs we get the equation
25 = .367 x - .193, which we solve to obtain
x = 69 (approx.).
Note that this displacement is also unrealistic for this
pendulum. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the average rate of change associated with this model?
Explain this average rate in common-sense terms.
Your solution:
Confidence Assessment:
Given Solution:
** Using the model y = .367 * x - .193 with y = force and x =
displacement from equilibrium we can use any two (x, y) points to get the rate
of change. In all cases we will get
rate of change = change in y / change in x = .367.
The change in y is the change in the force, while the change
in x is the change in position. The
rate of change therefore tells us how much the force changes per unit of change
in position (e.g., the force increases by 15 pounds for every inch of
displacement). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the average slope associated with this model? Explain this
average slope in common-sense terms.
Your solution:
Confidence Assessment:
Given Solution:
** Using the model y = .367 * x - .193 with y = force and x =
displacement from equilibrium the average slope is .367. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qAs you gradually pull the pendulum from a point 30 centimeters from
its equilibrium position to a point 80 centimeters from its equilibrium
position, what average force must you exert?
Your solution:
Confidence Assessment:
Given Solution:
** if it was possible to pull the pendulum back this far and
if the model applies you will get
Force at 30 cm:
y = .367 * 30 - .193 = 10.8
approx. and
Force at 80 cm:
y = .367 * 80 - .193 = 29
approx. so that
ave force between 30 cm and 80 cm is therefore
(10.8 + 29) / 2 = 20 approx.. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 8. flow range
What is the linear function range(time)?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT RESPONSE:
I obtained model one by drawing a line through the data points and picking two points on the line and
finding the slope between them. I then substituted this value for m and used one
of my data points on my line for the x and y value and solved for b. the line I
got was range(t) = -.95t + 112.38.
y = -16/15x + 98
INSTRUCTOR COMMENT:
This looks like a good model.
According to the instructions it should however be expressed
as range(time) = -16/15 * time + 98.
**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the significance of the average rate of change? Explain
this average rate in common-sense terms.
Your solution:
Confidence Assessment:
Given Solution:
** the average rate of change is change in range / change in
clock time. The average rate of
change indicates the average rate at which range in cm is changing with respect
to clock time in sec, i.e., the average number of cm / sec at which the range
changes. Thus the average rate tells
us how fast, on the average, the range changes. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the average slope associated with this model? Explain this
average slope in common-sense terms.
Your solution:
Confidence Assessment:
Given Solution:
** it's the average rate at which the range of the flow
changes--the average rate at which the position of the end of the stream
changes. It's the speed with which
the point where the stream reaches the ground moves across the ground. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 9. If your total wealth at clock time t = 0 hours is $3956, and
you earn $8/hour for the next 10 hours, then what is your total wealth function
totalWealth( t ), where t is time in hours?
Your solution:
Confidence Assessment:
Given Solution:
** Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would have
y intercept 3956, since that is the t = 0 value, and slope 8, since slope
represents change in total wealth / change in t, i.e., the number of dollars per
hour.
A graph with y-intercept b and slope m has equation y = m t +
b. Thus we have
totalWealth(t) = 8 * t + 3956 . **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qAt what clock time will your total wealth reach $4000? what
equation did you solve to obtain this result?
Your solution:
Confidence Assessment:
Given Solution:
STUDENT RESPONSE:
To find the clock time when my total wealth will reach 4000 I
solved the equation totalWealth(t) = 4000. The value I got when I solved for t
was t = 5.5 hours.
4.4 hours needed to reach 4000 4000 = 10x + 3956
INSTRUCTOR COMMENT:
Almost right.
You should solve 4000 = 8 x + 3956, obtaining 5.5 hours. This is equivalent to solving
totalWealth(t) = 4000 = 8 t + 3956, which is the more meaningful form of the
relationship. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the meaning of the slope of your graph?
Your solution:
Confidence Assessment:
Given Solution:
GOOD STUDENT RESPONSE: The slope of the graph shows the steady
rate at which money is earned on an hourly basis. It shows a steady increase in
wealth.
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 10. Experience shows that when a certain widget is sold for $30,
a certain store can expect to sell 200 widgets per week, while a selling price
of $28 increases the number sold to 300.
What linear function numberSold(price) describes this
situation?
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 10. Experience shows that when a certain widget is sold for $30,
a certain store can expect to sell 200 widgets per week, while a selling price
of $28 increases the number sold to 300.
What linear function numberSold(price) describes this
situation?
Self-critique (if necessary):
Self-critique Rating:
Question:
`qIf you make a graph of y = numberSold vs. x = price you have graph
points (30, 200) and (28, 300).
You need the equation of the straight line through these points.
You plug these coordinates into the form y = m x + b and
solve for m and b. Or you can use
another method. Whatever method you
use you get y = -50 x + 1700.
Then to put this into the notation of the problem you write
numberSold(price) instead of y and price instead of x.
You end up with the equation
numberSold(price) = -50 * price + 1700. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qIf the store must meet a quota by selling 220 units per week, what
price should they set? what equation did you solve to obtain this result?
Your solution:
Confidence Assessment:
Given Solution:
** If the variables are y and x, you know y so you can solve
for x.
For the function numberSold(price) = -50 * price + 1700 you
substitute 220 for numbersold(price) and solve for price.
You get the equation
220 = -50 * price + 1700
which you can solve to get
price = 30, approx.
**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qIf each widget costs the store $25, then how much total profit will
be expected from selling prices of $28, $29 and $30? what equation did you solve
to obtain this result?
Your solution:
Confidence Assessment:
Given Solution:
STUDENT RESPONSE:
If each widget costs the store $25, then they should expect
to earn a profit of 300 dollars from a selling price of $28, 250 dollars from a price of $29 and 200
dollars from a price of $30. To find
this I solved the equations numberSold(28); numberSold(29), and numberSold(30).
Solving for y after putting the price values in for p.
They will sell 300, 250 and 200 widgets, respectively (found
by solving the given equation).
To get the total profit you have to multiply the number of
widgets by the profit per widget.
At $28 the profit per widgit is $3 and the total profit is $3 * 300 =
$900; at $30 the profit per widgit is $5 and 200 are sold for profit $1000; at
$29 what happens? **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 11. quadratic function depth(t) = .01 t^2 - 2t + 100 representing
water depth vs.
What is the equation of the straight line connecting the t =
20 point of the graph to the t = 60 point?
Your solution:
Confidence Assessment:
Given Solution:
** The t = 20 point is (20,64) and the t = 60 point is (60,
16), so the slope is (-48 / 20) = -1.2.
This can be plugged into the form y = m t + b to get y = -1.2
t + b.
Then plugging in the x and y coordinates of either point you
get b = 88.
y = -1.2 t + 88 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 13. quadratic depth function y = depth(t) = .01 t^2 - 2t + 100.
What is `dy / `dt based on the two time values t = 30 sec and
t = 40 sec.
Your solution:
Confidence Assessment:
Given Solution:
** For t = 30 we have y = 49 and for t = 40 we have y = 36.
The slope between (30, 49) and (40,36) is (36 - 49) / (40 -
30) = -1.3.
This tells you that the depth is changing at an average rate
of -1.3 cm / sec. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qwhat is `dy / `dt based on t = 30 sec and t = 31 sec.
Your solution:
Confidence Assessment:
Given Solution:
** Based on t =
30 and t = 31 the value for `dy /
`dt is -1.39, following the same steps as before **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qwhat is `dy / `dt based on t = 30 sec and t = 30.1 sec.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT RESPONSE:
The value for 'dy / `dt based on t = 30 sec and t = 30.1 sec is -1.4
INSTRUCTOR COMMENT:
** Right if you round off the answer. However the answer shouldn't be rounded
off. Since you are looking at a
progression of numbers (-1.3, -1.39, and this one) and the differences in these
numbers get smaller and smaller, you have to use a precision that will always
show you the difference. Exact
values are feasible here and shoud be used. I believe that this one comes out to
-1.399. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat do you think you would get for `dy / `dt if you continued this
process?
Your solution:
Confidence Assessment:
Given Solution:
STUDENT RESPONSE:
An even more and more accurate slope value. I don't think it would have
to continue to decrease.
INSTRUCTOR COMMENT
**If you look at the sequence -1.3, -1.39, -1.399, ..., what
do you think happens?
It should be apparent
that the limiting value is -1.4 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat does the linear function tell you?
Your solution:
Confidence Assessment:
Given Solution:
** The function tells you that at any clock time t the rate
of depth change is given by the function .02 t - 2.
For t = 30, for example, this gives us .02 * 30 - 2 = -1.4,
which is the rate we obtained from the sequence of calculations above. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 14. linear function y = f(x) = .37 x + 8.09
.
Self-critique (if necessary):
Self-critique Rating:
Question:
`qwhat are the first five terms of the basic sequence {f(n), n = 1,
2, 3, ...} for this function.
Your solution:
Confidence Assessment:
Given Solution:
** The first five terms are 8.46, 8.83, 9.2, 9.57, and 9.94
**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the pattern of these numbers?
Your solution:
Confidence Assessment:
Given Solution:
** These numbers increase by .37 at each interval. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qIf you didn't know the equation for the function, how would you go
about finding the 100th member of the sequence? How can you tell your method is
valid?
Your solution:
Confidence Assessment:
Given Solution:
** You could find the 100th member by noting that you have 99
‘jumps’ between the first number and the 100 th, each ‘jump’ being .37.
Multiplying 99 times .37 and then adding the result to the
'starting value' (8.46). STUDENT RESPONSE:
simply put 100 as the x in the formula .37x +8.09
INSTRUCTOR COMMENT:
That's what you do if you have the equation.
Given just the numbers you could find the 100th member by
multiplying 99 times .37 and then adding the result to the first value 8.46. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qfor quadratic function y = g(x) = .01 x^2 - 2x + 100 what are the
first five terms of the basic sequence {g(n), n = 1, 2, 3, ...}?
Your solution:
Confidence Assessment:
Given Solution:
** We have
g(1) = .01 * 1^2 - 2 * 1 + 100 = 98.01
g(2) = .01 * 2^2 - 2 * 2 + 100 = 96.04,
etc.
The first 5 terms are therefore {98.01, 96.04, 94.09, 92.16,
90.25}
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the pattern of these numbers?
Your solution:
Confidence Assessment:
Given Solution:
** The changes in these numbers are -1.97, -1.95, -1.93,
-1.91. With each interval of x, the
change in y is .02 greater than for the previous interval. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qIf you didn't know the equation for the function, how would you go
about finding the next three members of the sequence?
Your solution:
Confidence Assessment:
Given Solution:
** According to the pattern established above, the next three
changes are -1.89, -1.87, -1.85.
This gives us
g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85.
**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qHow can you verify that your method is valid?
Your solution:
Confidence Assessment:
Given Solution:
** You can verify the result using the original formula; if
you evaluate it at 5, 6 and 7 it should confirm your results.
That's the best answer that can be given at this point.
You should understand, though that even if you verified it
for the first million terms, that wouldn't really prove it (who knows what might
happen at the ten millionth term, or whatever). It turns out that to prove it would
require calculus or the equivalent. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 15. The difference equation a(n+1) = a(n) + .4, a(1) = 5
If you substitute n = 1 into a(n+1) = a(n) + .4, how do you
determine a(2)?
Your solution:
Confidence Assessment:
Given Solution:
** You get a(1+1) = a(1) + .4, or
a(2) = a(1) + .4.
Knowing a(1) = 5 you get a(2) = 5.4. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q If you substitute n = 2 into a(n+1) = a(n) + .4 how do you
determine a(3)?
Your solution:
Confidence Assessment:
Given Solution:
** You have to do the substitution.
You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) +
.4
Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qIf you substitute n = 3 into a(n+1) = a(n) + .4, how do you
determine a(4)?
Your solution:
Confidence Assessment:
Given Solution:
** We get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is a(100)?
Your solution:
Confidence Assessment:
Given Solution:
** a(100) would be equal to a(1) plus 99 jumps of .4, or 5 +
99*.4 = 44.6. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 17. difference equation a(n+1) = a(n) + 2 n, a(1) = 4.
What is the pattern of the sequence?
Your solution:
Confidence Assessment:
Given Solution:
** For n = 1 we have n+1 = 2 so that the equation
a(n+1) = a(n) + 2 n
becomes
a(2) = a(1) + 2 * 1
Since a(1) = 4 (this was given) we have
a(2) = a(1) + 2 * 1 = 4 + 2 = 6.
Reasoning similarly, n = 2 gives us
a(3) = a(2) + 2 * 2 = 6 + 4 = 10.
n = 3 gives us
a(4) = a(3) + 2 * 3 = 10 + 6 = 16; etc.
The sequence is 4, 6, 10, 16, 24, 34, ... . **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat kind of function do you think a(n) is (e.g., linear,
quadratic, exponential, etc.)?
Your solution:
Confidence Assessment:
Given Solution:
** The differences of the sequence are 4,
6, 8, 10, 12, . . ..
The difference change by the same amount each time, which is
a property of quadratic functions. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery the slope =
slope equation
Explain the logic of the slope = slope equation (your may
take a little time on this one)
Your solution:
Confidence Assessment:
Given Solution:
** The slope = slope equation sets the slope between two
given points equal to the slope between one of those points and the variable
point (x, y).
Since all three points lie on the same straight line, the
slope between any two of the three points must be equal to the slope between any
other pair. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 7. streamRange(t), 50 centimeters at t = 20 seconds, range
changes by -10 centimeters over 5 seconds.
what is your function?
Your solution:
Confidence Assessment:
Given Solution:
** The rate at which streamRange changes is change in
streamRange / change in t = -10 cm / (5 sec) = -2 cm/s. This will be the slope m of the graph.
Since streamRange is 50 cm when t = 20 sec the point (20, 50)
lies on the graph. So the graph
passes through (20, 50) and has slope -2.
The function is therefore of the form y = m t + b with m =
-2, and b such that 50 = -2 * 20 + b.
Thus b = 90.
The function is therefore y = -2 t + 90, or using the
meaningful name of the function
steamRange(t) = -2t + 90
You need to use function notation. y = f(x) = -2x + 90 would be OK, or just f(x) =
-2x + 90. The point is that you need
to give the funcion a name.
Another idea here is that we can use the 'word' streamRange
to stand for the function. If you
had 50 different functions and, for example, called them f1, f2, f3, ..., f50
you wouldn't remember which one was which so none of the function names would
mean anything. If you call the
function streamRange it has a meaning.
Of course shorter words are sometimes preferable; just understand that
function don't have to be confined to single letters and sometimes it's not a
bad idea to make the names easily recognizable.
STUDENT RESPONSE:
y = -2x + 50
INSTRUCTOR COMMENT:
** At t = 20 sec this would give you y = -2 * 20 + 50 = 10.
But y = 50 cm when t = 20 sec.
Slope is -10 cm / (5 sec) = -2 cm/s, so you have y = -2 t +
b.
Plug in y = 50 cm and t = 20 sec and solve for b.
You get b = 90 cm.
The equation is y = -2 t + 90, or
streamRange(t) = -2t + 90. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qwhat is the clock time
at which the stream range first falls to 12 centimeters?
Your solution:
Confidence Assessment:
Given Solution:
** Using the correct equation streamRange(t) = -2t + 90, you
would set streamRange(t) = 12 and solve 12 = -2t + 90, obtaining t = 39 sec. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qquery
problem 9. equation of the straight line through t = 5 sec and the t = 7
sec points of the quadratic function depth(t) = .01 t^2 - 2t + 100
What is the slope and what does it tell you about the depth
function?
Your solution:
Confidence Assessment:
Given Solution:
** You have to get the whole equation. y = m t + b is now y = -1.88 t + b. You have to solve for b. Plug in the coordinates of the t = 7
point and find b.
You get 90.9 = -1.88 * 7 + b so b = 104, approximately. Find the correct value.
The equation will end up something like y = depth(t) = -1.88
t + 104. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qThe slope of the linear function is -1.88. This tells me that the
depth is decreasing as the time is increasing at a rate of 1.88 cm per sec.
Self-critique (if necessary):
Self-critique Rating:
Question:
`qHow closely does the linear function approximate the quadratic
function at each of the given times?
Your solution:
Confidence Assessment:
Given Solution:
** The deviations are for t = 3, 4, 5, 6, 7, & 8 as follows:
.08, .03. 0. -.01, 0, .03.
**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qat what t value do we obtain the closest values?
Your solution:
Confidence Assessment:
Given Solution:
** Not counting t= 5 and t = 7, which are 0, the next closest
t value is t = 6, the deviation for this is -.01. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qOn which side of the t = 5 and t = 7 points is the linear
approximation closer to the quadratic function? On which side does the quadratic
function 'curve away' from the linear most rapidly?
Your solution:
Confidence Assessment:
Given Solution:
** On the t = 4 side the approximation is closer. The
quadratic function curves away on the positive x side. **
Query Add comments on any surprises or
insights you experienced as a result of this assignment.
The slope = slope helped me out a lot. Learning that I can
solve a linear in different ways was helpful.
Self-critique (if necessary):
Self-critique Rating:
STUDENT COMMENT
I found the difference equation to be a
bit challenge to comprehend (it seems it can get pretty complicated) but very
exciting as well. I'm still not entirely sure what uses it will have in the
future, but it seems like an important concept to have
for future reference.
INSTRUCTOR RESPONSE
The difference equation is a way of specifying how a quantity changes, step by step.
There are numerous situations in which all was know is the initial value of a quantity and the rules for how it changes.
For example when water flows from a hole in the bottom of a uniform cylinder, it is the depth of water that determines how fast it comes out.
All we know, then, is the initial depth of the water and the rule for how quickly the depth changes.
It turns out that we can approximate the behavior of the depth function using the difference equation y(n+1) = y(n) - k * sqrt(y(n)), where k is a constant number determined by the diameter of the cylinder.
If you continue your study of mathematics you will eventually get to the
fourth semester of the standard calculus sequence, a course entitled
'Introduction to Ordinary Differential Equations'. Most second-semester calculus
courses also include a briefer introduction to the subject.
Your exposure to difference equations in this course will be usefu helpful to
you when you reach that point.