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1.1 — Vectors & Dot Product
MATH 155 • Calculus II for Biological Sciences • SFU
1. Vectors in $\mathbb{R}^n$
A vector $\mathbf{v} \in \mathbb{R}^n$ is an ordered list of $n$ real numbers. In MATH 155 we mostly work in $\mathbb{R}^2$ and $\mathbb{R}^3$.
$$\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix}$$
| Operation | Definition | Result |
|---|---|---|
| Addition | $\mathbf{u}+\mathbf{v}$ | $(u_1+v_1, u_2+v_2)$ |
| Scalar multiply | $c\mathbf{v}$ | $(cv_1, cv_2)$ |
| Magnitude | $\|\mathbf{v}\|$ | $\sqrt{v_1^2+v_2^2}$ |
2. The Dot Product
For $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$:
$$\mathbf{u} \cdot \mathbf{v} = \sum_{i=1}^n u_i v_i$$
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Geometric meaning
$\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\|\mathbf{v}\|\cos\theta$. If $\mathbf{u}\cdot\mathbf{v}=0$ the vectors are orthogonal.
⚠
Exam trap
The dot product returns a scalar, not a vector. $\mathbf{u}\cdot\mathbf{v}=0$ means orthogonal, not that either vector is zero.
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Unit 1 — Matrices & Vectors
Q1
Compute $\mathbf{u}\cdot\mathbf{v}$ where $\mathbf{u}=(2,-3,1)$ and $\mathbf{v}=(4,1,-2)$.
EasyDot product
$\mathbf{u}\cdot\mathbf{v}=(2)(4)+(-3)(1)+(1)(-2)=8-3-2=\mathbf{3}$
Q2
Find $\|\mathbf{v}\|$ for $\mathbf{v}=(3,4)$.
EasyVectors
$\|\mathbf{v}\|=\sqrt{9+16}=5$
Q3
Find eigenvalues of $A=\begin{pmatrix}4&1\\2&3\end{pmatrix}$.
MediumEigenvalues
$(4-\lambda)(3-\lambda)-2=\lambda^2-7\lambda+10=(\lambda-5)(\lambda-2)=0$. So $\lambda_1=5, \lambda_2=2$.
Q4
Solve: $2x+y=5$, $4x-y=7$.
MediumLinear systems
Adding: $6x=12\Rightarrow x=2$. Then $y=1$.
Q5
Population model $\mathbf{x}_{t+1}=A\mathbf{x}_t$ with $A=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}$. Find eigenvalues and long-run behaviour.
HardPopulation models
$\lambda_1=1, \lambda_2=0.5$. Dominant eigenvalue is 1, so population approaches a steady state.
Unit 2 — Functions of Two Variables
Q6
Find $f_x$ and $f_y$ for $f(x,y)=3x^2y+\sin(xy)$.
EasyPartial derivatives
$f_x=6xy+y\cos(xy)$, $f_y=3x^2+x\cos(xy)$
Q7
Find $\nabla f$ for $f(x,y)=x^2+4y^2$ at $(1,1)$.
EasyGradient
$\nabla f=(2x,8y)$. At $(1,1)$: $\nabla f=(2,8)$.
Q8
Use Lagrange multipliers to find the max of $f(x,y)=xy$ subject to $x+2y=8$.
HardLagrange multipliers
$\nabla f=\lambda\nabla g$: $(y,x)=\lambda(1,2)\Rightarrow x=2y$. Substituting: $y=2, x=4$. Max $=f(4,2)=\mathbf{8}$.
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