Unit 4
Functions Involving Parameters, Vectors, and Matrices
Notes
Parametric functions
A parametric function gives $x$ and $y$ separately in terms of a parameter $t$: $(x(t), y(t))$. Think of $t$ as time and the point as a moving particle.
- To graph, make a table of $t$, $x$ and $y$. Arrows show the direction of motion as $t$ increases.
- Eliminate the parameter by solving one equation for $t$ and substituting. Example: $x = t + 1$, $y = t^2$ becomes $y = (x - 1)^2$.
- A circle: $x = r\cos t$, $y = r\sin t$.
- Average rate of change of $x(t)$ on $[a, b]$ is $\displaystyle \frac{x(b) - x(a)}{b - a}$ (horizontal speed on average).
Conic sections
- Parabola: $y = a(x - h)^2 + k$.
- Ellipse: $\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.
- Hyperbola: $\displaystyle \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$.
Vectors
A vector has magnitude and direction. $\vec v = \langle a, b \rangle$.
- Magnitude $\displaystyle |\vec v| = \sqrt{a^2 + b^2}$. Unit vector: $\displaystyle \frac{\vec v}{|\vec v|}$.
- Add componentwise. Scalar multiples stretch the vector.
- Dot product: $\langle a, b\rangle \cdot \langle c, d\rangle = ac + bd = |\vec u||\vec v|\cos\theta$. A dot product of 0 means the vectors are perpendicular.
Matrices
- Multiply an $m \times n$ matrix by an $n \times p$ matrix to get $m \times p$. Order matters.
- For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\det A = ad - bc$ and $\displaystyle A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ when $\det A \ne 0$.
- A matrix can represent a linear transformation of the plane. The absolute value of the determinant is the area scale factor.
- Transition matrices model states changing over time: multiply the current state vector by the matrix to get the next state.
Topics in this unit
Each topic has its own short notes and practice.
- 4.1Parametric Functions
- 4.2Parametric Functions Modeling Planar Motion
- 4.3Parametric Functions and Rates of Change
- 4.4Parametrically Defined Circles and Lines
- 4.5Implicitly Defined Functions
- 4.6Conic Sections
- 4.7Parametrization of Implicitly Defined Functions
- 4.8Vectors
- 4.9Vector-Valued Functions
- 4.10Matrices
- 4.11The Inverse and Determinant of a Matrix
- 4.12Linear Transformations and Matrices
- 4.13Matrices as Functions
- 4.14Matrices Modeling Contexts
Practice
$\displaystyle t = \frac{x}{2}$, so $\displaystyle y = \left(\frac{x}{2}\right)^2 - 1$.
$\displaystyle \sqrt{25 + 144} = \sqrt{169} = 13$.
$2(4) + 3(-1) = 8 - 3 = 5$.
$3(4) + (-2)(6) = 0$, so the dot product is 0.
$3 \cdot 4 - 2 \cdot 1 = 10$.
Area scales by $|\det| = 3$, so $5 \cdot 3 = 15$.
$2 \cdot 3 - 6 \cdot 1 = 0$, so the determinant is 0.
$x^2 + y^2 = 9\cos^2 t + 9\sin^2 t = 9$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A particle moves so that $x(t) = t^2 - 4t$ and $y(t) = 2t + 1$ for $0 \le t \le 5$.
(a) Find the particle's position at $t = 1$ and at $t = 3$.
Show the answer and scoring guide
At $t = 1$: $(-3, 3)$. At $t = 3$: $(-3, 7)$.
Check off the points you earned:
(b) Find the average rate of change of $x(t)$ on $[1, 3]$ and explain what it means.
Show the answer and scoring guide
$\displaystyle \frac{x(3) - x(1)}{3 - 1} = \frac{-3 - (-3)}{2} = 0$. On average, the particle's horizontal position didn't change over that time; it moved left and then came back right.
Check off the points you earned:
(c) Write $\vec v$, the displacement vector from $t = 0$ to $t = 5$, and find its magnitude.
Show the answer and scoring guide
$t = 0$: $(0, 1)$; $t = 5$: $(5, 11)$. $\vec v = \langle 5, 10 \rangle$, $\displaystyle |\vec v| = \sqrt{125} = 5\sqrt5 \approx 11.18$.
Check off the points you earned: