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

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

Flashcard 3: 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 4: 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 5: What is the trajectory of $x = t$, $y = 3t + 2$?

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

Flashcard 6: 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 7: Convert $x = t$, $y = 2t + 3$ to Cartesian form.

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

Flashcard 8: 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 9: What shape does $x = t$, $y = t^2$ describe?

Answer: A parabola. Standard parabola opening upward.

Flashcard 10: 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 11: 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 12: State the parametric form for a line parallel to $x$-axis.

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

Flashcard 13: 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 14: 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 15: 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 16: What is the path of $x = t$, $y = t^3$?

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

Flashcard 17: 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 18: 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 19: Find $x$ at $t = 2$ for $x = 4t + 3$.

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

Flashcard 20: 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 21: State parametric equations for the horizontal line $y = c$.

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

Flashcard 22: What motion does $x = 3 \cos(t)$, $y = 3 \sin(t)$ represent?

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

Flashcard 23: 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 24: 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 25: What is the meaning of $t$ in parametric equations?

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

Flashcard 26: Find $y$ at $t = 2$ for $y = 3t - 5$.

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

Flashcard 27: Find the coordinates at $t = 0$ for $x = t^2$, $y = 2t$.

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

Flashcard 28: 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 29: 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 30: 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 31: Find $y$ at $t = 3$ for $y = 2t - 1$.

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

Flashcard 32: Find the initial point of $x = 2t + 1$, $y = 3t$ at $t = 0$.

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

Flashcard 33: 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 34: 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 35: Find $x$ at $t = 4$ for $x = 5t - 3$.

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

Flashcard 36: 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.