Copy of Team Homework 2_ Math 115
pdf
keyboard_arrow_up
School
University of Michigan, Dearborn *
*We aren’t endorsed by this school
Course
115
Subject
Mathematics
Date
Apr 3, 2024
Type
Pages
8
Uploaded by BrigadierSteel392
Math Team Homework Cover Sheet
Team Homework 2
1.
a.
Consider the function
i.
Explain why R(x) is a rational function. Be sure to refer to any relevant
definitions.
R(x) is a rational function because both the numerator and denominator
are polynomials.
ii.
Find any zeros, vertical asymptotes, and horizontal asymptotes of R(x). Be
sure to explain how you determined each of these things.
Vertical Asymptote: x =2, this is because at the value 2, the function is
undefined because the denominator of the function equals zero.
Horizontal Asymptote: y = 3, since both the denominator and the
numerator are at the same degree (1) we can look at the coefficient of x on
the numerator over the coefficient of x on the denominator (3/1 = 3)
Zeros: x = ⅓, this function has a zero at one third because at this value, the
function is equal to zero without being undefined (only the numerator
equals zero when x=⅓)
iii.
Find a polynomial p(x) such that R(x) ·p(x) has no vertical asymptotes. Be
sure to explain your process and how you know that the new function
doesn’t have any vertical asymptotes.
One possible polynomial for p(x) could be x-2. This would make the new
function (3x-1)(x-2)/(x-2). The term x-2, will then cancel out because it is
on both the numerator and denominator, and there will be no vertical
asymptote at this point, only a hole.
b.
Now consider the function
i.
Explain why tan(x) is not a rational function.
tan(x) is not a rational function because the numerator and denominator
are both trigonometric functions, not polynomials.
ii.
Does tan(x) have any horizontal asymptotes?
No tan(x) does not have any horizontal asymptotes because the graph of
tangent repeats every period as it is a trig function. Since the graph of
tan(x) has vertical asymptotes starting at pi/2 and repeats each half period,
tan(x) cannot have any horizontal asymptotes.
iii.
List or describe all zeros and vertical asymptotes of tan(x)
– on the interval [0,2],
– on the interval [0,5], and
– on the interval [0,∞).
Between the interval 0 ≤ x ≤ 2 there is a vertical asymptote at x= π/2. This
can be found by finding what makes sin(x) equal to zero on this interval,
without also equally what makes cos(x)=0.
Between the intervals 0 ≤ x ≤ 5 there is a vertical asymptote at x=
𝝅
/2 and
a zero at x=
𝝅
. These values can be found the same way as the first interval
just with different numerical values.
Finally, on the interval [0, ∞) there are vertical asymptotes at
𝝅
/2, 3
𝝅
/2,
5
𝝅
/2 …etc, and the zeros occur at 0,
𝝅
, 2
𝝅
, etc. Since this graph repeats
itself we can assume that there will also be repeating asymptotes and zeros
as x approaches infinity.
c.
As you found in (b), there is exactly one vertical asymptote of tan(x) on interval
[0,2]. Let us call this asymptote x = A.
i.
Find a number α such that the function
has no vertical asymptotes on [0,2]. It might be helpful to consider what
you did in part (a).
In order for there to be no vertical asymptote, we would need to multiply
the function by the polynomial (x -
𝝅
/2), so A =
𝝅
/2, because this is where
the vertical asymptote occurs on this interval. Therefore, when multiplied
by this polynomial, the numerator and denominator would cancel out, and
there would be a hole at this point instead.
ii.
Now that you have “eliminated” this vertical asymptote of tan(x), let’s
look closer at the function h(x) near x = A. Fill in the first two columns of
the table below by giving values for (x −α) and using a calculator to give
approximate values for tan(x) to several decimal places, or write DNE for
any value that Does Not Exist. Then use these columns to compute the
values of h(x) in the final column.
x
x-a
tan(x)
h(x)
A-1
-1
0.642
-0.642
A-0.1
-0.1
9.967
-0.997
A-0.01
-0.01
99.997
-1.000
A-0.001
-0.001
999.999
-1.000
A
0
DNE
DNE
A+0.001
0.001
-999.999
-1.000
A+0.01
0.01
-99.997
-1.000
A+0.1
0.1
-9.967
-0.997
A+1
1
-0.642
-0.642
iii.
Using your table, estimate
You are
welcome to graph h(x) in
Desmos to
check your answers.
As x approaches A from the right side, we can estimate that the limit = -1
since the values in the table depicting A + a number get closer to -1 the
smaller they are. Likewise, when subtracting from A (approaching from
the left side) we can also say that the limit = -1 since the values approach
-1 the closer they are to A. Finally, this means that the limit as x
approaches A is -1, since it is -1 from both the right and the left.
d.
Similarly, can you find a polynomial q(x) such that the product tan(x) ·q(x) has no
vertical asymptotes on the interval [0,5]?
We can use the same steps we used in part c, to cancel out the vertical asymptotes
and create holes instead. However, since this is a bigger interval, we used
multiplication of more polynomials to cancel out the additional asymptotes from
the interval. Our final polynomial was (x-
𝝅
/2)(x-3
𝝅
/2).
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Hint: you may want to use the function h(x) as a starting point.
e.
Can you describe a function k(x) such that the product tan(x) ·k(x) has no vertical
asymptotes on the interval [0,∞)? If so, is this function a polynomial?
Since it is impossible to cancel out every repeating polynomial to infinity by
using the same formula as the previous two parts, we decided to use 0x, where k
=0. This way the numerator of the function will always cancel out with the
denominator when it equals 0, and there will be no vertical asymptotes.
2.
You are at the Ann Arbor train station waiting for a train coming from Detroit, MI and
headed to Chicago, IL. Lambda, an app on your phone developed by your friends’
start-up, is showing you the graph of λ(t), the distance, in miles, from the Detroit station
to the train t minutes after noon. Also included is a magnified version of the graph for 0 ≤
t ≤9.
a.
You know that the train is scheduled to arrive at Ann Arbor train station at 12:45
pm. Given that the distance by rail between Detroit and Ann Arbor is 36 miles,
does this seem reasonable?
Yes this seems reasonable because looking at the graph, we are able to see that as
the time is at 12:45pm the train would indeed be 36 miles from Detroit, in Ann
Arbor.
b.
It’s currently 12:03 pm. Estimate the distance of the train from Detroit, and its
current velocity in miles per hour.
From the graph, 3 minutes looks to be 0.8 miles from Detroit. In order to find
velocity we used points that are close to 3 miles, so we took f(3.5)-f(2.5)/3.5-2.5
so plugging in our y values we get 1-0.5/1 giving us an answer of 0.5 miles per
minute.
c.
Estimate the values in the following table.
t
mins
7
9
13
15
20
25
,
λ(𝑡)
mi/hr
60
66
70
0
0
0
Well in order to convert miles to minutes and this into miles per hour, we must
multiply by 60 (mph) in which the minutes cancel each other out so the units
come out to miles / hour. So to get the correct value for Lambda(t) we have to
multiply the approx. derivative at t minutes by 60.
For
t
= 7 the tangent line slope is 1 so the derivative in mph is about 60.
For
t
= 9 the tangent line slope is about 1.1 resulting in 66 mph.
For
t
= 13 the tangent line slope is about 1.15 resulting in about 70 mph.
For
t
= 15 the graph is constant so there is zero slope for the tangent line
For
t
= 20 the graph is constant so there is zero slope for the tangent line
For
t
= 25 the graph is constant so there is zero slope for the tangent line
What can you say about the motion of the train between 12:07 and 12:14?
between 12:14 and 12:25?
At 12:07-12:14 the graph looks linear so the train was moving at a constant speed.
Whereas at 12:14-12:25 the graph is constant meaning that the train did not move
at all and was at rest.
d.
You want to suggest to your friends that it would be helpful to also provide a
graph of ω(h), the distance, in miles, of the train from the Ann Arbor station h
hours after it leaves the Detroit station. Write a formula for ω(h) in terms of λ.
w(h) = 36 - λ(60h)
The total distance from Detroit to Ann Arbor is 36 miles so that is why we
subtract
λ(60h)
from 36 as
λ(60h)
represents the distance of the train from Detroit
in hours. We use
λ(60h)
instead of
λ(t)
because
w(h)
is in hours and to convert
from minutes to hours we have to multiply by 60 thus instead of
λ(t)
in minutes
we use
λ(60h)
in hours.
e.
After all this careful study of the graph of λ(t), you are now unsure that it is right,
and want to find the errors and provide feedback so your friends can make the app
better.
i.
Estimate the velocity of the train at a time just before t = 45 minutes. What
does the graph imply is happening with the motion of the train as it
approaches 45? Is this reasonable?
The velocity of the train at a time just before t = 45 mins is about 240
mins. This can be found as we can approximate the value of λ’(44).
Using Δλ/Δt we can approximate the speed as the train approaches Ann
Arbor.
Thus:
Δλ/Δt → [λ’(44) - λ’(43)] / (44 - 43)
≈ 4 mpm * 60 mins / hr = 240 mph
According to the graph it implies that this is not the limit of the speed of
the train as the graph still looks to be exponentially growing at t = 45.
This velocity is not reasonable as the graph shows that the train is going to
keep speeding up without bounds. As the train nears Ann Arbor you
would expect it to slow down and stop, not continue to speed up.
ii.
After doing some more estimates, you find that the train’s velocity just
before t = 37 minutes is about 23 miles per minute, and it is about −1
miles per minute right after t = 37 minutes. Explain why these values don’t
make sense from a physics standpoint.
Speaking in a physics perspective the change in velocity from positive 23
miles per minute to negative 1 mile per minute would mean that in an
instant the train transitioned from moving forwards at 23 miles per minute
to backwards at 1 mile per minute. This does not make sense physically as
the train can’t just instantaneously switch from moving forward to moving
backwards without a significant amount of time to slow down and speed
up in the reverse direction.
iii.
Similarly, explain what is wrong with the graph around t = 14 minutes.
Use your results from part (c) of this problem to support your claim(s).
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
At t = 14 minutes the train switches from having a positive speed to no
speed at all. Physically speaking the train would instantaneously stop
when the travel time hits 14 minutes. This does not make sense as this
cannot be physically achieved unless a much larger force is applied on the
train to stop it. This force would have to be so great that it has to at the
very least equal the entirety of the force of the train moving at the speed it
was moving at, if not more. The appearance of such a force at such a time
to stop the train at t = 14 minutes is near impossible thus this event just
like the one before is not reasonable in the viewpoint of physics.
f.
One of the current passengers on the train, Ilia, boarded in Detroit, and
continually sips his tea during the trip to Ann Arbor. Define the function T(t) to be
the amount of tea, in ounces, remaining in Ilia’s mug t minutes after noon.
Assume that T(t) is invertible.
T: min to ounces
t^-1 : ounces to min
For each of the following, write an equation involving T and/or T−1 representing
the given sentence.
i.
At 12:05pm, Ilia has 14.3 ounces of tea remaining in his mug.
T is in minutes, and our output gives ounces of tea that is remaining.
Giving us an equation of T(5) = 14.3
ii.
Ten minutes after noon, the amount of tea in Ilia’s mug is half of what it
was at 12:00.
Ten minutes after noon so 12:10pm Ilia has half of her mug filled with tea
so we can get T(10) = 1/2
Write a sentence giving a practical interpretation of each of the following
equations, or explain why the equation does not make sense in context.
iii.
λ(T^−1(4.3)) = 20.7
T^-1(4.3) will be how many ounces to minutes, and 20.7 is our output in
miles. . We are able to interpret this equation to, “Ilia has about 4.3 ounces
of tea left in her mug when the train is approximately 20.7 miles from
Detroit.
iv.
T(λ(11)) = 12.9
Input gives distance in miles, and T also has an input of miles, so this
equation does not make sense.