AP Precalculus Flashcards: Parametric Functions Modeling Planar Motion

What this deck covers

This deck focuses on Parametric Functions Modeling Planar Motion, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Find $y$ at $t = 3$ for $y = 2t - 1$.

Answer: $y = 5$. Substitute $t = 3$: $y = 2(3) - 1 = 5$.

Flashcard 2: What type of curve is $x = a \cos(t)$, $y = a \sin(t)$?

Answer: A circle. A circle centered at origin with radius $a$.

Flashcard 3: Find $x$ at $t = 0$ for $x = 7t - 2$.

Answer: $x = -2$. Substitute $t = 0$: $x = 7(0) - 2 = -2$.

Flashcard 4: Find $x$ at $t = 2$ for $x = 4t + 3$.

Answer: $x = 11$. Substitute $t = 2$: $x = 4(2) + 3 = 11$.

Flashcard 5: Convert $x = t^2 - 1$, $y = 2t$ to a Cartesian equation.

Answer: $y^2 = 4(x + 1)$. From $y = 2t$, get $t = \frac{y}{2}$, substitute into $x$.

Flashcard 6: What is the path of $x = 2 \cos(t)$, $y = 3 \sin(t)$?

Answer: An ellipse. Standard form of an ellipse with semi-axes 2 and 3.

Flashcard 7: What shape does $x = t$, $y = t^2$ describe?

Answer: A parabola. Standard parabola opening upward.

Flashcard 8: Find the coordinates at $t = 0$ for $x = t^2$, $y = 2t$.

Answer: $(0, 0)$. Both coordinates are zero when $t = 0$.

Flashcard 9: Convert $x = 6t$, $y = 2t + 3$ to a Cartesian equation.

Answer: $y = \frac{1}{3}x + 3$. From $x = 6t$, get $t = \frac{x}{6}$, substitute into $y$.

Flashcard 10: What is the path of $x = 2 \cos(t)$, $y = 3 \sin(t)$?

Answer: An ellipse. Standard form of an ellipse with semi-axes 2 and 3.

Flashcard 11: What is the trajectory of $x = t$, $y = 3t + 2$?

Answer: A line. Linear relationship between $x$ and $y$ with slope 3.

Flashcard 12: State the parametric equations for a circle with radius $r$.

Answer: $x = r \, \cos(t)$, $y = r \, \sin(t)$. Standard form using cosine for $x$ and sine for $y$ components.

Flashcard 13: Convert $x = t$, $y = 2t + 3$ to Cartesian form.

Answer: $y = 2x + 3$. Since $x = t$, substitute directly into $y = 2t + 3$.

Flashcard 14: Find the range of $y = \sin(t)$ for $0 \leq t \leq 2\pi$.

Answer: $-1 \leq y \leq 1$. Standard range of the sine function.

Flashcard 15: Find the slope of the line for $x = 2t + 1$, $y = 3t - 4$.

Answer: Slope is $\frac{dy}{dx} = \frac{3}{2}$. Slope equals $\frac{dy/dt}{dx/dt} = \frac{3}{2}$.

Flashcard 16: Find the slope of the line for $x = 2t + 1$, $y = 3t - 4$.

Answer: Slope is $\frac{dy}{dx} = \frac{3}{2}$. Slope equals $\frac{dy/dt}{dx/dt} = \frac{3}{2}$.

Flashcard 17: What shape does $x = t$, $y = t^2$ describe?

Answer: A parabola. Standard parabola opening upward.

Flashcard 18: Convert $x = 4t$, $y = 5t + 1$ to a Cartesian equation.

Answer: $y = \frac{5}{4}x + 1$. From $x = 4t$, get $t = \frac{x}{4}$, substitute into $y$.

Flashcard 19: Convert $x = 3t + 1$, $y = 2t + 4$ to Cartesian equation.

Answer: $y = \frac{2}{3}x + \frac{10}{3}$. From $x = 3t + 1$, get $t = \frac{x-1}{3}$, substitute into $y$.

Flashcard 20: Convert $x = t^2$, $y = 2t$ to a Cartesian equation.

Answer: $y^2 = 4x$. From $x = t^2$, get $t = \pm\sqrt{x}$, substitute into $y = 2t$.

Flashcard 21: What is the path of $x = t$, $y = t^3$?

Answer: A cubic curve. Third-degree polynomial relationship.

Flashcard 22: State the parametric form for a line parallel to $x$-axis.

Answer: $x = t$, $y = c$. Horizontal line where $y$ remains constant.

Flashcard 23: Identify the parameter in the equations $x = 3t$, $y = 2t + 1$.

Answer: The parameter is $t$. The independent variable that both $x$ and $y$ depend on.

Flashcard 24: Find $x$ at $t = 4$ for $x = 5t - 3$.

Answer: $x = 17$. Substitute $t = 4$: $x = 5(4) - 3 = 17$.

Flashcard 25: State the parametric equations for a circle with radius $r$.

Answer: $x = r \, \cos(t)$, $y = r \, \sin(t)$. Standard form using cosine for $x$ and sine for $y$ components.

Flashcard 26: Convert $x = t + 1$, $y = t^2$ to a Cartesian equation.

Answer: $y = (x-1)^2$. From $x = t + 1$, get $t = x - 1$, substitute into $y = t^2$.

Flashcard 27: State the parametric form of a line through $(x_0, y_0)$ with slope $m$.

Answer: $x = x_0 + t$, $y = y_0 + mt$. General form where $t$ acts as the parameter for direction.

Flashcard 28: What is the trajectory of $x = t$, $y = 3t + 2$?

Answer: A line. Linear relationship between $x$ and $y$ with slope 3.

Flashcard 29: What is the path of $x = t$, $y = t^3$?

Answer: A cubic curve. Third-degree polynomial relationship.

Flashcard 30: Convert $x = t^2$, $y = 2t$ to a Cartesian equation.

Answer: $y^2 = 4x$. From $x = t^2$, get $t = \pm\sqrt{x}$, substitute into $y = 2t$.

Flashcard 31: Convert $x = t^2 - 1$, $y = 2t$ to a Cartesian equation.

Answer: $y^2 = 4(x + 1)$. From $y = 2t$, get $t = \frac{y}{2}$, substitute into $x$.

Flashcard 32: What is the range of $y = 2 \sin(t)$ for $0 \leq t \leq 2\pi$?

Answer: $-2 \leq y \leq 2$. Sine function oscillates between -1 and 1, scaled by factor 2.

Flashcard 33: Convert $x = t$, $y = 2t + 3$ to Cartesian form.

Answer: $y = 2x + 3$. Since $x = t$, substitute directly into $y = 2t + 3$.

Flashcard 34: State the parametric form of a line through $(x_0, y_0)$ with slope $m$.

Answer: $x = x_0 + t$, $y = y_0 + mt$. General form where $t$ acts as the parameter for direction.

Flashcard 35: Convert $x = 6t$, $y = 2t + 3$ to a Cartesian equation.

Answer: $y = \frac{1}{3}x + 3$. From $x = 6t$, get $t = \frac{x}{6}$, substitute into $y$.

Flashcard 36: Convert $x = 3t + 1$, $y = 2t + 4$ to Cartesian equation.

Answer: $y = \frac{2}{3}x + \frac{10}{3}$. From $x = 3t + 1$, get $t = \frac{x-1}{3}$, substitute into $y$.

Flashcard 37: Find $x$ at $t = 0$ for $x = 7t - 2$.

Answer: $x = -2$. Substitute $t = 0$: $x = 7(0) - 2 = -2$.

Flashcard 38: What is the result of $x = 2t$, $y = 3t$?

Answer: A line through the origin. Linear relationship with slope $\frac{3}{2}$ passing through origin.

Flashcard 39: Find $x$ at $t = 2$ for $x = 4t + 3$.

Answer: $x = 11$. Substitute $t = 2$: $x = 4(2) + 3 = 11$.

Flashcard 40: Convert $x = 4t$, $y = 5t + 1$ to a Cartesian equation.

Answer: $y = \frac{5}{4}x + 1$. From $x = 4t$, get $t = \frac{x}{4}$, substitute into $y$.

Flashcard 41: State the parametric equations for a line with slope $m$.

Answer: $x = x_0 + at$, $y = y_0 + mt$. General form with direction vector $(a, m)$ and slope $\frac{m}{a}$.

Flashcard 42: Identify the parameter in the equations $x = 3t$, $y = 2t + 1$.

Answer: The parameter is $t$. The independent variable that both $x$ and $y$ depend on.

Flashcard 43: What is the range of $y = 2 \sin(t)$ for $0 \leq t \leq 2\pi$?

Answer: $-2 \leq y \leq 2$. Sine function oscillates between -1 and 1, scaled by factor 2.

Flashcard 44: State parametric equations for the horizontal line $y = c$.

Answer: $x = t$, $y = c$. Let $t$ vary while keeping $y$ constant at $c$.

Flashcard 45: State the parametric form of a vertical line $x = c$.

Answer: $x = c$, $y = t$. Let $t$ vary while keeping $x$ constant at $c$.

Flashcard 46: What is the meaning of $t$ in parametric equations?

Answer: A parameter, often representing time. Usually represents time or another independent variable.

Flashcard 47: What motion does $x = 3 \cos(t)$, $y = 3 \sin(t)$ represent?

Answer: Circular motion. Parametric equations for a circle with radius 3.

Flashcard 48: What motion does $x = 3 \cos(t)$, $y = 3 \sin(t)$ represent?

Answer: Circular motion. Parametric equations for a circle with radius 3.

Flashcard 49: Find the range of $y = \sin(t)$ for $0 \leq t \leq 2\pi$.

Answer: $-1 \leq y \leq 1$. Standard range of the sine function.

Flashcard 50: What type of curve is $x = a \cos(t)$, $y = a \sin(t)$?

Answer: A circle. A circle centered at origin with radius $a$.

Flashcard 51: Convert $x = t + 1$, $y = t^2$ to a Cartesian equation.

Answer: $y = (x-1)^2$. From $x = t + 1$, get $t = x - 1$, substitute into $y = t^2$.

Flashcard 52: Convert $x = \cos(t)$, $y = \sin(t)$ to a Cartesian equation.

Answer: $x^2 + y^2 = 1$. Uses the Pythagorean identity $\cos^2(t) + \sin^2(t) = 1$.

Flashcard 53: State the parametric form for a line parallel to $x$-axis.

Answer: $x = t$, $y = c$. Horizontal line where $y$ remains constant.

Flashcard 54: What is the meaning of $t$ in parametric equations?

Answer: A parameter, often representing time. Usually represents time or another independent variable.

Flashcard 55: Find $y$ at $t = 2$ for $y = 3t - 5$.

Answer: $y = 1$. Substitute $t = 2$: $y = 3(2) - 5 = 1$.

Flashcard 56: Find $y$ at $t = 2$ for $y = 3t - 5$.

Answer: $y = 1$. Substitute $t = 2$: $y = 3(2) - 5 = 1$.

Flashcard 57: Find the coordinates at $t = 0$ for $x = t^2$, $y = 2t$.

Answer: $(0, 0)$. Both coordinates are zero when $t = 0$.

Flashcard 58: State parametric equations for the horizontal line $y = c$.

Answer: $x = t$, $y = c$. Let $t$ vary while keeping $y$ constant at $c$.

Flashcard 59: Convert $x = 2t + 3$, $y = 4t - 1$ to Cartesian form.

Answer: $y = 2x - 7$. From $x = 2t + 3$, get $t = \frac{x-3}{2}$, substitute into $y$.

Flashcard 60: State the parametric equations for a line with slope $m$.

Answer: $x = x_0 + at$, $y = y_0 + mt$. General form with direction vector $(a, m)$ and slope $\frac{m}{a}$.

Flashcard 61: Find the point at $t = 1$ for $x = 3t$, $y = t^2 + 1$.

Answer: $(3, 2)$. Substitute $t = 1$: $x = 3$, $y = 1 + 1 = 2$.

Flashcard 62: Find $y$ at $t = 3$ for $y = 2t - 1$.

Answer: $y = 5$. Substitute $t = 3$: $y = 2(3) - 1 = 5$.

Flashcard 63: Find the initial point of $x = 2t + 1$, $y = 3t$ at $t = 0$.

Answer: $(1, 0)$. Substitute $t = 0$: $x = 1$, $y = 0$.

Flashcard 64: Find the point at $t = 1$ for $x = 3t$, $y = t^2 + 1$.

Answer: $(3, 2)$. Substitute $t = 1$: $x = 3$, $y = 1 + 1 = 2$.

Flashcard 65: Find the initial point of $x = 2t + 1$, $y = 3t$ at $t = 0$.

Answer: $(1, 0)$. Substitute $t = 0$: $x = 1$, $y = 0$.

Flashcard 66: State the parametric form of a vertical line $x = c$.

Answer: $x = c$, $y = t$. Let $t$ vary while keeping $x$ constant at $c$.

Flashcard 67: What is the result of $x = 2t$, $y = 3t$?

Answer: A line through the origin. Linear relationship with slope $\frac{3}{2}$ passing through origin.

Flashcard 68: Find $x$ at $t = 4$ for $x = 5t - 3$.

Answer: $x = 17$. Substitute $t = 4$: $x = 5(4) - 3 = 17$.

Flashcard 69: What is a parametric equation?

Answer: An equation expressing coordinates as functions of a parameter. Both $x$ and $y$ are expressed in terms of an independent variable.

Flashcard 70: Convert $x = 2t + 3$, $y = 4t - 1$ to Cartesian form.

Answer: $y = 2x - 7$. From $x = 2t + 3$, get $t = \frac{x-3}{2}$, substitute into $y$.