Instructor: Prof. Mahesh Sunkula
Office: Mathematical Sciences Building (MATH), Room 842
Email: msunkula@purdue.edu
Course Webpage: https://www.math.purdue.edu/MA261 and Brightspace
Office Hours: Monday and Friday 9:45 AM – 11:00 AM, Thursday 11:00 AM – 12:00 PM, or by appointment
MA261 extends what you already know from Calculus 1 and 2 to more variables. The course covers multivariable functions, partial derivatives, multiple integrals, and vector fields. These topics have applications in optimization, physics, and machine learning.
The course uses MyLab Math for online homework and includes weekly quizzes in recitation sessions. There will be 37 online assignments due almost every Tuesday and Thursday at 11:59 PM. No late homework is accepted, but the 3 lowest homework scores are dropped.
Quizzes are administered in almost every recitation session, covering material from lessons whose homework was due the previous week. Each quiz is no more than 15 minutes long. No make-up quizzes will be given, but the 2 lowest quiz scores are dropped.
Location and more details will be provided before each exam.
| Component | Percentage |
|---|---|
| Homework | 15% |
| Quizzes | 15% |
| Midterms (20% each) | 40% |
| Final | 30% |
Letter grade cutoffs: A+ (97%), A (93%), A- (90%), B+ (87%), B (83%), B- (80%), C+ (77%), C (73%), C- (70%), D+ (67%), D (63%), D- (60%). At the end of the semester, if the distribution of letter grades is significantly different from the traditional distribution, these cutoff points may be lowered but will never be raised.
Important: Students with accommodations should schedule exams directly with the DRC, contact their recitation TA for quiz accommodations, and contact the instructor for any other needs.
Definition: A vector is a quantity that has both magnitude and direction.
Consider two points \(P(2, -1)\) and \(Q(3, 5)\). The vector from \(P\) to \(Q\), denoted \(\overrightarrow{PQ}\), is calculated by subtracting the coordinates of \(P\) from the coordinates of \(Q\):
\[\overrightarrow{PQ} = \langle 3-2, 5-(-1) \rangle = \langle 1, 6 \rangle\]Similarly, the vector from \(Q\) to \(P\) is:
\[\overrightarrow{QP} = \langle 2-3, -1-5 \rangle = \langle -1, -6 \rangle\]Notice that \(\overrightarrow{QP} = -\overrightarrow{PQ}\).
Diagram Description: A coordinate plane showing two points P and Q. Two vectors are drawn: one blue vector pointing from P to Q (pointing upward and to the right), and one green vector pointing from Q to P (pointing downward and to the left). The vectors are parallel but point in opposite directions, illustrating that they are negatives of each other.
For a vector \(\vec{u} = \langle 1, 6 \rangle\), the length (or magnitude) is calculated using the Pythagorean theorem:
\[|\overrightarrow{PQ}| = \sqrt{1^2 + 6^2} = \sqrt{37}\]The length of \(\overrightarrow{QP}\) is the same:
\[|\overrightarrow{QP}| = |\vec{u}| = \sqrt{37}\]Definition: In general, for a vector \(\vec{u} = \langle x, y, z \rangle\) in three dimensions, the length is:
\[|\vec{u}| = \sqrt{x^2 + y^2 + z^2}\]Definition: A unit vector is a vector of length 1. The unit vector in the direction of \(\vec{u}\) is denoted \(\hat{u}\) and calculated as:
\[\hat{u} = \frac{\vec{u}}{|\vec{u}|}\]Diagram Description: A coordinate system showing a vector drawn from the origin. A dashed line from the tip of the vector perpendicular to the x-axis illustrates the relationship between the vector's length and its components.
Let \(\vec{u} = \langle 1, 2, -1 \rangle\). Find the unit vector in the direction of \(\vec{u}\), and find a vector of length 7 in the opposite direction.
Solution:
First, find the length of \(\vec{u}\):
\[|\vec{u}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\]The unit vector in the direction of \(\vec{u}\) is:
\[\hat{u} = \frac{\vec{u}}{|\vec{u}|} = \frac{\langle 1, 2, -1 \rangle}{\sqrt{6}} = \left\langle \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}} \right\rangle\]For a vector of length 7 in the opposite direction to \(\vec{u}\), we use:
\[-7\hat{u} = -7 \cdot \frac{\vec{u}}{|\vec{u}|} = \frac{-7}{\sqrt{6}} \langle 1, 2, -1 \rangle = \frac{7}{\sqrt{6}} \langle -1, -2, 1 \rangle\]Alternatively:
\[-7\hat{u} = \left\langle \frac{-7}{\sqrt{6}}, \frac{-14}{\sqrt{6}}, \frac{7}{\sqrt{6}} \right\rangle\]Vectors can be added and subtracted both algebraically and geometrically.
Given \(\vec{u} = \langle 1, 2, -1 \rangle\) and \(\vec{v} = \langle 2, 3, 4 \rangle\):
\[\vec{u} + \vec{v} = \langle 1+2, 2+3, -1+4 \rangle = \langle 3, 5, 3 \rangle\] \[\vec{u} - \vec{v} = \langle 1-2, 2-3, -1-4 \rangle = \langle -1, -1, -5 \rangle\]To add vectors geometrically, place the tail of the second vector at the head of the first vector. The sum is the vector from the tail of the first to the head of the second.
Diagram Description: A parallelogram showing vector addition. Two vectors, \(\vec{u}\) and \(\vec{v}\), are drawn from a common origin. A third vector \(\vec{u} + \vec{v}\) is shown as the diagonal of the parallelogram formed by \(\vec{u}\) and \(\vec{v}\). Additionally, the vector \(\vec{u} - \vec{v}\) is shown, illustrating subtraction as addition of the negative vector.
There are two ways to multiply vectors: the dot product and the cross product.
Definition: The dot product of two vectors \(\vec{u} = \langle u_1, u_2, u_3 \rangle\) and \(\vec{v} = \langle v_1, v_2, v_3 \rangle\) is calculated by multiplying corresponding components and adding the results:
\[\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3\]Let \(\vec{u} = \langle 1, 2, -1 \rangle\) and \(\vec{v} = \langle 2, 3, 4 \rangle\). Find \(\vec{u} \cdot \vec{v}\).
Solution:
\[\vec{u} \cdot \vec{v} = (1)(2) + (2)(3) + (-1)(4) = 2 + 6 - 4 = 4\]Geometric Interpretation: The dot product can also be expressed in terms of the angle \(\theta\) between the two vectors:
\[\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos\theta\]Diagram Description: Two vectors \(\vec{u}\) and \(\vec{v}\) emanating from a common origin, with the angle \(\theta\) between them marked.
Definition: The cross product of two vectors \(\vec{u}\) and \(\vec{v}\), denoted \(\vec{u} \times \vec{v}\), is a vector perpendicular to both \(\vec{u}\) and \(\vec{v}\).
The magnitude of the cross product is:
\[|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin\theta\]where \(\theta\) is the angle between \(\vec{u}\) and \(\vec{v}\).
The direction of \(\vec{u} \times \vec{v}\) is determined by the right-hand rule: point your fingers in the direction of \(\vec{u}\), curl them toward \(\vec{v}\), and your thumb points in the direction of \(\vec{u} \times \vec{v}\). Note that \(\vec{v} \times \vec{u} = -(\vec{u} \times \vec{v})\).
Diagram Description: A three-dimensional coordinate system showing two vectors \(\vec{u}\) and \(\vec{v}\) in a plane, with the cross product vector \(\vec{u} \times \vec{v}\) pointing perpendicular to that plane. The angle \(\theta\) between \(\vec{u}\) and \(\vec{v}\) is marked.
For vectors \(\vec{u} = \langle u_1, u_2, u_3 \rangle\) and \(\vec{v} = \langle v_1, v_2, v_3 \rangle\), the cross product is computed using the determinant:
\[\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}\]Expanding this determinant gives:
\[\vec{u} \times \vec{v} = \langle u_2 v_3 - u_3 v_2, -(u_1 v_3 - u_3 v_1), u_1 v_2 - u_2 v_1 \rangle\]Let \(\vec{u} = \langle 1, 2, -1 \rangle\) and \(\vec{v} = \langle 2, 3, 4 \rangle\). Find \(\vec{u} \times \vec{v}\).
Solution:
We compute the cross product using the determinant formula:
\[\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & -1 \\ 2 & 3 & 4 \end{vmatrix}\]Expanding along the first row:
\[\vec{u} \times \vec{v} = \vec{i}(2 \cdot 4 - (-1) \cdot 3) - \vec{j}(1 \cdot 4 - (-1) \cdot 2) + \vec{k}(1 \cdot 3 - 2 \cdot 2)\] \[= \vec{i}(8 + 3) - \vec{j}(4 + 2) + \vec{k}(3 - 4)\] \[= 11\vec{i} - 6\vec{j} - 1\vec{k}\] \[= \langle 11, -6, -1 \rangle\]Diagram Description: A 3×3 matrix showing the determinant calculation for the cross product. The first row contains the unit vectors \(\vec{i}\), \(\vec{j}\), \(\vec{k}\). The second row contains the components of \(\vec{u}\): 1, 2, -1. The third row contains the components of \(\vec{v}\): 2, 3, 4. Curved arrows indicate the calculation pattern for each component, with specific pairs of numbers circled to show which terms are multiplied.
Key Properties: