Copy of Team Homework 3_ Math 115
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University of Michigan, Dearborn *
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Course
115
Subject
Mathematics
Date
Apr 3, 2024
Type
Pages
11
Uploaded by BrigadierSteel392
Math Team Homework Cover Sheet
Team Homework 3
1.
Consider the function
.
?(?) = ?
?
· (??𝑠(?)
− 𝑠𝑖?(?))
a)
First, answer the following by hand, without using a calculator (including
Desmos):
i)
What is
? What is
?
?(0)
?(𝜋/4)
f(0) = e
(0)
* (cos(0) - sin(0))
f(0) = 1 * (1- 0)
f(0) = 1
f(
) =
e
𝜋
/4
* (cos(
𝜋
/4)
- sin(
/4
))
𝜋/4
𝜋
f(
𝜋
/4)
= e
𝜋
/4
* (
2
/2 −
2 /2)
f(
/4) =
e
𝜋
/4
* 0
𝜋
f(
𝜋
/4)
= 0
ii)
Find
.
?'(?)
f’(x) = e
x
(cosx - sinx) + (-sinx - cosx)e
x
f’(x) = e
x
cosx - e
x
sinx - e
x
sinx - e
x
cosx
f’(x) = -2e
x
(sinx)
iii)
Use your work above to describe the values of
on the interval [0,
?(?)
π/4]. For example, are the values increasing? Decreasing?
The derivative shows that on the interval between [0, π/4] the function
values were decreasing. As the derivative will give a negative value due
to the -2 and always positive e^x and sinx (because of the interval, sin is
always positive in the first quarter of the unit circle) we know that the
original function can only be decreasing during this interval.
b)
Next, let’s work on a way to approximate many values of
near x = 0 at once.
?(?)
i)
First, explain what it means for a function to be “an approximation” of
. When and why might you use an approximating function instead of
?(?)
just using
itself?
?(?)
An “approximation” function is used when you want to find the values
near a point where the actual function is very complicated and hard to
solve for. Thus an “approximation” function is used and oftentimes it's the
tangent line approximation which gives a simple linear function which is
much easier to work with than the complex original function. Also since
the linear function is the tangent line approximation then the values close
to the point where they intersect/touch are actually very close such that 3
digit accuracy can be obtained and in the real world that is what is usually
used (three digit accuracy). Below is an image to show the wording above
as a drawing:
ii)
Compute the linear approximation
for
near x = 0.
𝑇(?)
?(?)
y = y
1
+ m(x - x
1
) → T(x) = f(x
1
) + f’(x
1
)(x - x
1
)
T(x) = f(0) + f’(0)(x - 0)
At x = 0 → f(0) = 1
T(x) = 1 + 0(x-0)
T(x) = 1
iii)
Would you consider
to be a good approximation of
near x = 0?
𝑇(?)
?(?)
Why or why not?
No T(x) is not a good approximation because f’(0) is 0 which results in
T(x) = 1. This doesn’t give us a function to plug in x values near x = 0 to
approximate the y-values which is the whole point of the linear function
T(x).
iv)
If you were given another function approximating
near x = 0, how
?(?)
would you decide which approximation is better?
As long as the other function has variables within the approximation while
also being a tangent line approximation then it would be better than T(x)
above. This is because you can actually input values of x to try and
approximate y-values near the point where the two functions
intersect/touch, thus achieving the whole point of using a linear
approximation.
c)
Let’s explore whether another linear function can do a better job approximating
near x = 0.
?(?)
i)
Plot the function
, and the linear function
given below, including
?(?)
𝐿(?)
a slider for a, in Desmos.
?(?) = ?
?
· (??𝑠(?)
− 𝑠𝑖?(?))
𝐿(?)
=
1
+
? · ? Why is the function
an approximation for
near 0? In particular,
𝐿(?)
?(?)
there is a property that both functions share at x = 0, regardless of the
value of a. What is it?
L(x) is an approximation for f(x) near 0 because it is a tangent line
approximation to f(x) at x = 0. Regardless of the value of
a
, L(x) will
always pass through (0,1) which is the same point that f(x) passes through.
ii)
With just these two graphs, it is not necessarily clear which value of a
gives us a better or worse approximation. We can use an error function to
investigate this more carefully:
E
1
(?) = |?(?) − 𝐿(?)|
Add this error function
E
1
to your Desmos graph, then briefly explain
(?)
what the values of this function mean in terms of determining which
values of
give better or worse approximations.
?
The values of the error function provide the numerical difference between
the original function f(x) and the approximation function L(x). As it can
be seen on the graph below, the error function (purple) is equal to zero at
x = 0 as that is where the points on the approximation function (blue) and
original function (red) “touch” and as you move farther away from zero
the values of the error function increase as the approximation values and
the actual values differ.
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iii)
Usually, we are not interested in values of the error function
E
1
(?)
everywhere. Instead, we only expected that
would give an
𝐿(?)
approximation to
near x = 0. Use Desmos to fill in the following
?(?)
table of values.
? =− 0. 5
? =− 0. 05
? = 0
? = 0. 05
? = 0. 5
E
1
for
a
= -0.3
(?)
0.3269
0.0174
0
0.0124
0.1936
E
1
for
a
= 0
(?)
0.1769
0.0024
0
0.0026
0.3436
E
1
for
a
= 0.3
(?)
0.0269
0.0126
0
0.0176
0.4936
iv)
Using the slider for
a
and your table from the previous part, explain which
value of a gives you the best, most uniform approximation of
near x
?(?)
= 0.
The value of
a
that gives the best approximation is
a
= 0. This value has
the smallest value from the error function for 3 out of the five values
checked when compared to either
a
= 0.3 and
a
= -0.3. This can be seen
when looking at the table of values and comparing the values listed in
each column by each other and the result is that the middle row of values
has the smallest values on average compared to the other two rows.
v)
With this particular value of a, there are now two properties that the
functions
and
share at x = 0. What are they? Was
from
?(?)
𝐿(?)
𝑇(?)
part (b) the best linear function approximating
after all?
?(?)
The two properties that they share is that the actual value of both functions
at x = 0 is equal to 1 and that the derivative of both functions at x = 0 is
also the shared value of 0.
d)
Now that we have considered linear approximation, let’s try using a quadratic
function as an approximation.
i)
Add to your Desmos graph the following function
, adding the slider
𝑄(?)
for
b
:
𝑄(?)
=
𝑇(?)
+
? · ?
2
where
is the linear approximation you found in part (b). Similarly to
𝑇(?)
part (c), we want to find the optimal value for b, so also graph the error
function
E
2
analogous to
E
1
:
(?)
(?)
E
1
(?) = |?(?) − 𝑄(?)|
Using logic analogous to that in part (c), find the optimal value of
b
.
x = -0.5
x = -0.05
x = 0
x = 0.05
x = 0.5
E
2
for
b
= -0.85
(?)
0.0356
0.0003
0
0.0005
0.1311
E
2
for
b
= -1
(?)
0.0731
0.0001
0
0.0001
0.0936
E
2
for
b
= -1.15
(?)
0.1106
0.0005
0
0.0003
0.0561
The optimal value for
b
is
b
= -1. Using the same way as in part (b), using
a table of values it is clear that the value b = -1 has the smallest error
values thus making it the most optimal value of
b
.
ii)
Compute
for the optimal value of
b
you found in the previous part.
𝑄′′(0)
Also compute
. How do the two values compare? Can you guess the
? ′′(0)
general formula for
(the best)
quadratic approximation of a general
function
near x = 0?
?(?)
Q(x) = T(x) + (-1)x
2
Q’(x) = T’(x) + (-1)2x
Q’’(x) = T’’(x)+ (-1)2
T(x) = 1 → T’(x) = 0 → T’’(x) = 0
Q’’(x) = -2
Q’’(0) = -2
f(x) = e
x
· (cos(x) - sin(x))
f’(x) = -2e
x
(sin(x))
f’’(x) = -2[e
x
cos(x)+sin(x)e
x
]
f’’(0) = -2[e
0
cos(0)+sin(0)e
0
]
f’’(0) = -2[1 · 1 + 0 · 1]
f’’(0) = -2(1)
f’’(0) = -2
When comparing the two second derivative values we find that the values
are actually exactly the same! Thus, it can be concluded that the general
formula for the best quadratic approximation of a general function g(x)
near x = 0 would be where their second derivatives are equal. This can be
seen from the work above where I found the second derivative functions
of each respective function (f(x) and Q(x)) then solved for when x = 0 and
the result was that both values came out to be -2.
iii)
Finally, compare your best quadratic approximation to your best linear one
by looking at the graph of
E
2
compared to
E
1
(with the optimal
(?)
(?)
values of
a
and
b
you found). As you can see, one of these functions is
much smaller around x = 0. Discuss:
●
What does this imply about the accuracy of the linear vs quadratic
approximations? Can you explain graphically why this is the case?
When looking at the accuracy of the linear vs quadratic
approximations it is very clear that the quadratic approximation is
much more accurate. This can be easily explained using the graph
of f(x). The graph of f(x) near x = 0 looks to be quadratic, thus a
quadratic approximation would be much more accurate compared
to a linear approximation. This can be seen in the image below:
Key: f(x) is red, T(x) is green, Q(x) is orange
●
Why might someone choose to use a less accurate approximation,
here and in general?
In general, real life applications only care for about 3 digit (after
decimal point) accuracy. In this case while the quadratic
approximation would be much more accurate, the linear
approximation would still be usable as it is still accurate within a
certain range. Comparing the error function graphs of T(x) and
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Q(x), we actually see that they are quite similar. The use for linear
approximations is to test values close to the point that the tangent
line approximation shares with the original graph, such error is
acceptable and negligible especially as you move closer to the
point where the tangent line passes through along with the original
function.
●
Is there a best approximation for a function at a given point?
There will never be a best approximation for a function as the best
approximation is the actual value which can be obtained from the
original function. Thus the best approximation for a function at a
given point would be the original function itself.
2)
You and your friends from the
Lambda
app are on a hiking trip. The path you will follow
can be modeled by the curve
?
3
+ ?
3
= 3??
Where the positive
x
-axis is directed to the East, and the positive
y
-axis is directed to the
North
Note that the direction you are traveling at any point can be determined from the picture
above.
a)
Use calculus to find all points
on your path where you will move in one of
(?, ?)
the cardinal directions, that is, directly North, South, East, or West. Be sure to
include all necessary algebraic computations and for any point you find, specify
the direction of travel. Your work will at some point involve substituting an
equation relating
and
into the equation for the curve.
Hint:
There is no reason
?
?
to involve a square root in this step…
x
3
+
y
3
= 3
xy
3x
2
+ 3y
2
dy/dx = 3y + 3xdy/dx
3y
2
dy/dx - 3xdy/dx = 3y - 3x
2
dy/dx(3y
2
- 3x) = 3y - 3x
2
dy/dx = (3y - 3x
2
)/(3y
2
- 3x)
3y - 3x
2
= 0, y = x
2
, → x
3
+ (x
2
)
3
= 3x(x
2
)
x
3
+ x
6
= 3x
3
x
6
- 2x
3
= 0
x
3
- 2 = 0
x = cube root of 2 (2
1/3
) and 0
Therefore: (2
1/3
)
3
+ y
3
= 3(2
1/3
)y → 2 = y
3
+ 3(2
1/3
)y, y = {0.58, 1.58, -2.17}
y must be 1.58 because the point (2
1/3
, 1.58) is the only one that appears to have a
slope of 0
0 + y
3
= 0, y = 0, (0,0)
3y
2
- 3x = 0, x = y
2
→ (y
2
)
3
+ y
3
= 3y
2
(y)
y
6
+ y
3
= 3y
3
y
3
(y
3
+ 1 = 3)
y
3
= 2
Y = 2
1/3
and 0
x
3
+ (2
1/3
)
3
= 3x(2
1/3
)
Same process as above except switching x and y. = {.58, 1.58, -2.17} = 1.58
(according to graph)
x
3
+ 0 = 0
x = 0, point = (0,0)
Therefore points:
North = (1.58, 2
1/3
)
West = (2
1/3
, 1.58)
East = (0,0)
South = (0,0)
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b)
Now find all points on your route where you will move directly North-East or
South-West. This time, you will need the help of Desmos: in addition to the
equation of the curve given above, find one more equation that must be satisfied
by the coordinates of such points, and then plot both equations in Desmos to find
their approximate points of intersection. Be sure to explain why these points of
intersection are relevant to your solution.
The points of intersection are: (1.2, 0.53) northeast, (0.53, 1.2) south west. In
order to find these points we set the derivative equal to one, since this would be
the slope if a person were traveling exactly northeast and southwest (45 degrees
upward and downward). We used desmos to graph the derivative at 1 and the
original function, and that points of intersection were (1.2, 0.53) and (0.53, 1.2).
c)
Using the same general steps as in part (b), now find any points on your route
where you will move directly North-West or South-East.
We used the same process but this time set the derivative equal to negative 1 in
desmos since we would be traveling in the opposite direction. The only point on
the curve where these two lines intersect is (1.5, 1.5) and this is when we are
traveling northwest.
d)
In part (c), one of the curves you graphed can be considered an asymptote to the
curve pictured above. What does this imply in the context of your hike?
The asymptote implies that you never walk southeast because the slope -1 never
intersects the (curve other than at the point 1.5, 1.5).
e)
As you may have already noticed, the point (0, 0) is an unusual point on your
hiking route. Why? What can you say about dy/dx at this point? How does this
relate to your work in each of the parts (a)–(c) above?
It is unusual because it is intersected multiple times from both of the two lines
graphed in part c and b. This implies that throughout the hike we pass by the same
point twice, from both the northeast and southeast. At both times the point is
passed, we are facing a different direction.