notes / updates
November 18, 2024 | Resource: WolframAlpha
WolframAlpha is a reliable resource that you can use to confirm various mathematical calculations. To get started, visit wolframalpha.com and type a query using natural language.
For example: derivative of cos(3x) at x equals pi/6 . The following url gives the results:
https://www.wolframalpha.com/input?i=derivative+of+cos%283x%29+at+x+equals+pi%2F6
November 18, 2024 | Quiz notes.
P1.
\( \vec{r}(t) = < 0, 4 \cos t, 3 \sin t > \)
P2.
The SageMath code for problem 2 is below.
t = var('t')
# Problem 2 - part a
r1 = vector((t, e^t,-t*e^t))
r1.integrate(t,0,2)
# Problem 2 - part b)
f(t) = sqrt(t^2+t^4)
f.integrate(t,0,3)
November 15, 2024 | Quiz feedback
P1.
Write out the normal vector of the plane: \( \vec{n} = < \sqrt{3},1, 0 > \)
Write out the derivative of the given curve: \( \vec{r'(t)} = <-2\sin t, 2\cos t, e^t> \)
The derivative of the curve at some point has to be perpendicular to the normal vector to the plane in order for the tangent line to parallel, so we set the dot product equal to zero:
\( \vec{r'(t^*)} \cdot \vec{n} = 0 \rightarrow -2\sqrt{3}\sin t + 2\cos t = 0 \rightarrow \tan t = \frac{\sqrt{3}}{3} \rightarrow t=\frac{\pi}{6} \)
The corresponding point is \( (\sqrt{3}, 1, e^{\pi/6}) \)
P2.
To parametrize the intersection, set \( x(t) = t \). From the first surface we get \( y=\sqrt{25-t^2}\).
From the second surface we get: \( z = \sqrt{20-(25-t^2)} = \sqrt{t^2-5}\)
Therefore, a parametric equation of the intersection between the surfaces is \( \vec{r(t)} = < t,\sqrt{25-t^2} , \sqrt{t^2-5}>\) At the given point, we know that \( x = t = 3 \), so we find the direction vector of the tangent line by evaluating:
\( \vec{r'(t)} = < 1, \frac{-t}{\sqrt{25-t^2}},\frac{t}{\sqrt{t^2-5}} > \) at \( t = 3\).
The direction vector is \( \vec{r'(3)} = < 1, \frac{-3}{4}, \frac{3}{2} >\) or we can also use its multiple: \( <4, -3, 6> \)
The vector equation of the tangent line is: \( <3,4,2> + t*<4,-3,6>\).
Feel free to plot in SageMath, by the way, the original curve and the tangent line. You could even include the original surfaces.
P3.
a) The curve is an ellipse on the yz plane, centered at the origin and with equation: \( \frac{y^2}{4} + \frac{z^2}{9} = 1\). The major axis is 6 and the minor axis is 4 units.
b) Find \( | \vec{r'}(t) | = \sqrt{4\sin^2t+9\cos^2t} \). The least value occurs when cosine is zero, which means sine is 1 or -1, therefore the minimum is 2.
The largest value occurs when sine is 0, which means cosine is 1 or -1, so the largest value is 3.
Similar reasoning for the length of the second derivative leads to the same conclusion: the minimum value is 2 and the maximum is 3.
October 24, 2024 | Notes on Problem 48 | Section 11.7
If we let \( u = x^2, v = y^2 \), we get: \( 2u + 2v + (u+v)^2 = 3 \).
Rewrite as: \( (u + v )^2 + 2 (u+v) + 1 = 4 \)
Rewrite as: \( (u+v+1)^2 = 4 \), which is the same as \( u + v = 1\) or \( x^2 + y^2 = 1\).
Hence the shape is a circle.
October 3, 2024 | Cross Product with the TI-NSPIRE
tutorial here
September 30, 2024 | Problem 47 in Section 11.3
Hi all, desmos partially agrees with the author's answer to problem 47. The angle itself is the same as the author's (around 40 degrees) but the x-coordinate where this angle occurs is different.
September 27, 2024 | Quiz Notes
Below find some quiz notes. Graded work will be returned in class. If there are mistakes or typos below, feel free to email me.
P1
Let the points be: A(1, 2), B(3,1), and C(8,4). Form three vectors:
\( \vec{AB} = < 2, -1 >, \vec{BC} = < 5, 3 >, \vec{AC} = < 7, 2> \)
Find the three possibilities by adding any vector to the point that wasn't used to form that vector. For example, to the vector \( \vec{OC} \) we add the vector \( \vec{AB} \). The missing points are: (10, 3), (-4, -1), and (6, 5), as shown below:
There were no issues with problem 2, except be sure to use vectors:
If the points of trisection are M and N and the vector connecting the two given points is \( \vec{AB} = < 6, 3 > \), then:
\( \vec{OM} = \vec{OA} + \frac{1}{3}\vec{AB} = <3, 3 > \)
Similarly: \( \vec{ON} = \vec{OA} + \frac{2}{3}\vec{AB} = <5, 4 > \)
Therefore the points are: (3,3) and (5,4).
September 23, 2024 | Quiz Notes
Below find some quiz notes. Graded work will be returned in class. If there are mistakes or typos below, feel free to email me.
P1
Form three vectors using the points: A(4, -2, 7), B(-2,0,3), C(7, -3, 9).
\( \vec{AB} = < -6, 2, -4 >, \vec{AC} = < 3, -1, 2 >, \vec{BC} = < 9, -3, 6 > \)
Note that \( \vec{AB} = -2* \vec{AC} \text{ and } \vec{BC} = (-3/2)* \vec{AB} \)
The three vectors are therefore multiples of each other, so the points are collinear.
Note that the problem asked us to use vectors.
P2
A(2, 9, 1), B(3,11,4), C(0, 10, 12), D(1, 12, 5)
Same idea here, except the four resulting vectors should be two equal pairs:
\( \vec{AB} = < 1, 2, 3 > = \vec{CD} \text{ and } \vec{AC} = < -2, 1, 1 > = \vec{BD} \)
Note that the problem asked us to use vectors.
Here is the SageMath code:
#Problem 2
A=(2,9,1)
B=(3,11,4)
C=(0,10,2)
D=(1,12,5)
#form vectors
AB = vector([B[0]-A[0],B[1]-A[1],B[2]-A[2]])
BC = vector([C[0]-B[0],C[1]-B[1],C[2]-B[2]])
CD = vector([D[0]-C[0],D[1]-C[1],D[2]-C[2]])
AD = vector([D[0]-A[0],D[1]-A[1],D[2]-A[2]])
AC = vector([C[0]-A[0],C[1]-A[1],C[2]-A[2]])
BD = vector([D[0]-B[0],D[1]-B[1],D[2]-B[2]])
print("AB:", AB)
print("BC:", BC)
print("CD:", CD)
print("AD:", AD)
print("AC:", AC)
print("BD:", BD)
plot(AC, start=A) + plot(CD, start=C) + plot(-BD, start=D) + plot(-AB, start=B)
P3 \( t\vec{v} \) consists of all points on the line in the direction of vector \( \vec{v} \), passing through the origin.
\( \vec{u} + t\vec{v} \) describes all points on the lines that passes through the tip of \( \vec{u} \) in the direction of \( \vec{v} \). Here we assume that vector \( \vec{u} \) starts at the origin.
\( s\vec{u} + t\vec{v} \) describes all points on the plane as long as \( \vec{u}, \vec{v} \) are not parallel to each other. If they are parallel, the resulting set is the line passing through the origin in the direction of either vector.
September 22, 2024 | Weekly updates & Current grades
Hi everyone. I write to clarify a couple of things regarding grades and grading.
First, on myCushing under your progress tab you should be able to see your current grade for this course, which is a weighted average of the assessments we have written so far. While individual quiz scores will not be published on myCushing, the current grade will be updated approximately once week.
Second, when you read your weekly update, be sure to check the comments as well as the grade. Sometimes I add individual feedback based on recent performances or in-class observations.
As always, let me know if you have any questions.
September 10, 2024
If you haven't yet, be sure to install a recent version of Geogebra: https://www.geogebra.org/download
September 1, 2024
Hi everyone. In this area of the course page we will post notes, solutions, reminders, and links to other resources throughout the year. Entries will be dated with the most recent posted at the top of the page.