AP Precalculus Flashcards: Parametric Functions

Study Parametric Functions in AP Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Precalculus

Parametric Functions

0 mastered0 still learning

0% Complete

QUESTION
1/ 67

Convert x=3sin(t)x = 3 \sin(t), y=4cos(t)y = 4 \cos(t) to a Cartesian equation.

Tap card or press Space to flip

ANSWER

x29+y216=1\frac{x^2}{9} + \frac{y^2}{16} = 1. Ellipse using sine for xx and cosine for yy.

How well did you know it?

Card 1 / 67

What this deck covers

This deck focuses on Parametric Functions, 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.

All flashcards

Flashcard 1: Convert x=3sin(t)x = 3 \sin(t), y=4cos(t)y = 4 \cos(t) to a Cartesian equation.

Answer: x29+y216=1\frac{x^2}{9} + \frac{y^2}{16} = 1. Ellipse using sine for xx and cosine for yy.

Flashcard 2: Determine the Cartesian equation from x=4cos(t)x = 4 \cos(t), y=5sin(t)y = 5 \sin(t).

Answer: x216+y225=1\frac{x^2}{16} + \frac{y^2}{25} = 1. Ellipse with semi-axes 4 and 5.

Flashcard 3: What is the parameter in parametric equations?

Answer: A variable, often tt, that both xx and yy are functions of. The independent variable that controls both coordinates.

Flashcard 4: What is the parametric form for a circle with radius rr?

Answer: x=rcos(t)x = r \cos(t), y=rsin(t)y = r \sin(t). General circle equation with specified radius.

Flashcard 5: Convert x=1+2tx = 1 + 2t, y=3+4ty = 3 + 4t to a Cartesian equation.

Answer: y=2x+1y = 2x + 1. From x=1+2tx = 1 + 2t get t=x12t = \frac{x-1}{2}, substitute.

Flashcard 6: Find the Cartesian equation from x=1+tx = 1 + t, y=2t1y = 2t - 1.

Answer: y=2x3y = 2x - 3. From x=1+tx = 1 + t, get t=x1t = x - 1, substitute.

Flashcard 7: Convert x=cos(t)x = \cos(t), y=sin(t)y = \sin(t) to a Cartesian equation.

Answer: x2+y2=1x^2 + y^2 = 1. Unit circle using fundamental trigonometric identity.

Flashcard 8: What is the purpose of parametric equations in mathematics?

Answer: To describe geometric figures and motions. Enables modeling of complex paths and trajectories.

Flashcard 9: What is the role of tt in parametric equations?

Answer: It is the independent variable or parameter. Controls the position along the curve as it varies.

Flashcard 10: Convert x=5t2x = 5t - 2, y=3t+1y = 3t + 1 to a Cartesian equation.

Answer: y=35(x+2)1y = \frac{3}{5}(x + 2) - 1. From x=5t2x = 5t - 2 get t=x+25t = \frac{x+2}{5}, substitute.

Flashcard 11: What is a parametric equation?

Answer: An equation that expresses variables as functions of a parameter. Both xx and yy depend on the same parameter.

Flashcard 12: Identify parametric equations for a parabola y=x2y = x^2.

Answer: x=tx = t, y=t2y = t^2. Simplest parametrization using tt as xx-coordinate.

Flashcard 13: Identify parametric equations for a horizontal line y=cy = c.

Answer: x=tx = t, y=cy = c. Parameter varies while yy remains constant.

Flashcard 14: Convert x=4tx = 4t, y=9t2y = 9 - t^2 to a Cartesian equation.

Answer: y=9(x4)2y = 9 - \left(\frac{x}{4}\right)^2. Substitute t=x4t = \frac{x}{4} into the yy equation.

Flashcard 15: What is a parametric equation?

Answer: An equation that expresses variables as functions of a parameter. Both xx and yy depend on the same parameter.

Flashcard 16: Find the Cartesian equation from x=2cos(t)x = 2 \cos(t), y=3sin(t)y = 3 \sin(t).

Answer: x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1. Use identity cos2(t)+sin2(t)=1\cos^2(t) + \sin^2(t) = 1.

Flashcard 17: Define what a parameter is in terms of parametric equations.

Answer: An independent variable that defines a set of equations. The controlling variable in parametric representation.

Flashcard 18: What is the parameter in parametric equations?

Answer: A variable, often tt, that both xx and yy are functions of. The independent variable that controls both coordinates.

Flashcard 19: What is the parametric equation for a line with slope mm?

Answer: x=tx = t, y=mt+cy = mt + c. Standard form with slope mm and parameter tt.

Flashcard 20: Convert the parametric equations x=t+1x = t + 1, y=2ty = 2t to Cartesian form.

Answer: y=2(x1)y = 2(x - 1). Solve for tt from first equation, substitute into second.

Flashcard 21: Identify parametric equations for a horizontal line y=cy = c.

Answer: x=tx = t, y=cy = c. Parameter varies while yy remains constant.

Flashcard 22: What is the parametric equation for a vertical line x=cx = c?

Answer: x=cx = c, y=ty = t. Parameter varies while xx remains constant.

Flashcard 23: Which parametric equations describe the line segment from (1,2)(1,2) to (4,8)(4,8)?

Answer: x=1+3tx = 1 + 3t, y=2+6ty = 2 + 6t, 0t10 \leq t \leq 1. Direction vector (3,6)(3,6) with parameter range [0,1][0,1].

Flashcard 24: What is the parametric form for a parabola y=ax2y = ax^2?

Answer: x=tx = t, y=at2y = at^2. General parabola form with coefficient aa.

Flashcard 25: Convert x=4tx = 4t, y=9t2y = 9 - t^2 to a Cartesian equation.

Answer: y=9(x4)2y = 9 - \left(\frac{x}{4}\right)^2. Substitute t=x4t = \frac{x}{4} into the yy equation.

Flashcard 26: Identify the parametric form for the line y=3x+2y = 3x + 2.

Answer: x=tx = t, y=3t+2y = 3t + 2. Set parameter t=xt = x for simplest form.

Flashcard 27: What is the role of tt in parametric equations?

Answer: It is the independent variable or parameter. Controls the position along the curve as it varies.

Flashcard 28: Define what a parameter is in terms of parametric equations.

Answer: An independent variable that defines a set of equations. The controlling variable in parametric representation.

Flashcard 29: What are the parametric equations for a line parallel to y=2x+3y = 2x + 3?

Answer: x=tx = t, y=2t+cy = 2t + c. Same slope but different yy-intercept constant.

Flashcard 30: Identify the parametric form for the line y=3x+2y = 3x + 2.

Answer: x=tx = t, y=3t+2y = 3t + 2. Set parameter t=xt = x for simplest form.

Flashcard 31: What is the parametric form for a parabola y=ax2y = ax^2?

Answer: x=tx = t, y=at2y = at^2. General parabola form with coefficient aa.

Flashcard 32: What is the parametric equation for a line with slope mm?

Answer: x=tx = t, y=mt+cy = mt + c. Standard form with slope mm and parameter tt.

Flashcard 33: Convert x=cos(t)x = \cos(t), y=sin(t)y = \sin(t) to a Cartesian equation.

Answer: x2+y2=1x^2 + y^2 = 1. Unit circle using fundamental trigonometric identity.

Flashcard 34: State the parametric equations for a circle centered at the origin.

Answer: x=rcos(t)x = r \, \cos(t), y=rsin(t)y = r \, \sin(t). Standard form using trigonometric functions with radius rr.

Flashcard 35: Convert x=5t2x = 5t - 2, y=3t+1y = 3t + 1 to a Cartesian equation.

Answer: y=35(x+2)1y = \frac{3}{5}(x + 2) - 1. From x=5t2x = 5t - 2 get t=x+25t = \frac{x+2}{5}, substitute.

Flashcard 36: What is the parametric form of a line segment from (2,1)(2,1) to (5,4)(5,4)?

Answer: x=2+3tx = 2 + 3t, y=1+3ty = 1 + 3t, 0t10 \leq t \leq 1. Direction vector (3,3)(3,3) from start to end point.

Flashcard 37: Identify parametric equations for a parabola y=x2y = x^2.

Answer: x=tx = t, y=t2y = t^2. Simplest parametrization using tt as xx-coordinate.

Flashcard 38: Convert x=3tx = 3t, y=4ty = 4t to a Cartesian equation.

Answer: y=43xy = \frac{4}{3}x. Eliminate tt by solving x=3tx = 3t gives t=x3t = \frac{x}{3}.

Flashcard 39: Find the Cartesian equation from x=1+tx = 1 + t, y=2t1y = 2t - 1.

Answer: y=2x3y = 2x - 3. From x=1+tx = 1 + t, get t=x1t = x - 1, substitute.

Flashcard 40: Find the Cartesian equation from x=2cos(t)x = 2 \cos(t), y=3sin(t)y = 3 \sin(t).

Answer: x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1. Use identity cos2(t)+sin2(t)=1\cos^2(t) + \sin^2(t) = 1.

Flashcard 41: What is the parametric equation for a vertical line x=cx = c?

Answer: x=cx = c, y=ty = t. Parameter varies while xx remains constant.

Flashcard 42: What is the purpose of parametric equations in mathematics?

Answer: To describe geometric figures and motions. Enables modeling of complex paths and trajectories.

Flashcard 43: State the parametric equations for a circle centered at the origin.

Answer: x=rcos(t)x = r \, \cos(t), y=rsin(t)y = r \, \sin(t). Standard form using trigonometric functions with radius rr.

Flashcard 44: Convert x=3sin(t)x = 3 \sin(t), y=4cos(t)y = 4 \cos(t) to a Cartesian equation.

Answer: x29+y216=1\frac{x^2}{9} + \frac{y^2}{16} = 1. Ellipse using sine for xx and cosine for yy.

Flashcard 45: Convert x=3tx = 3t, y=4ty = 4t to a Cartesian equation.

Answer: y=43xy = \frac{4}{3}x. Eliminate tt by solving x=3tx = 3t gives t=x3t = \frac{x}{3}.

Flashcard 46: Convert x=1+2tx = 1 + 2t, y=3+4ty = 3 + 4t to a Cartesian equation.

Answer: y=2x+1y = 2x + 1. From x=1+2tx = 1 + 2t get t=x12t = \frac{x-1}{2}, substitute.

Flashcard 47: What is the general form of parametric equations for a line?

Answer: x=x0+atx = x_0 + at, y=y0+bty = y_0 + bt. Point (x0,y0)(x_0, y_0) with direction vector (a,b)(a,b).

Flashcard 48: What do aa and bb represent in the ellipse parametric equations x=acos(t)x = a \cos(t), y=bsin(t)y = b \sin(t)?

Answer: The semi-major and semi-minor axes. The lengths of the ellipse's major and minor axes.

Flashcard 49: What is an advantage of parametric equations?

Answer: They allow the description of curves not functions. Can represent curves that fail the vertical line test.

Flashcard 50: What do aa and bb represent in the ellipse parametric equations x=acos(t)x = a \cos(t), y=bsin(t)y = b \sin(t)?

Answer: The semi-major and semi-minor axes. The lengths of the ellipse's major and minor axes.

Flashcard 51: What is the parametric form of a line segment from (2,1)(2,1) to (5,4)(5,4)?

Answer: x=2+3tx = 2 + 3t, y=1+3ty = 1 + 3t, 0t10 \leq t \leq 1. Direction vector (3,3)(3,3) from start to end point.

Flashcard 52: What is an advantage of parametric equations?

Answer: They allow the description of curves not functions. Can represent curves that fail the vertical line test.

Flashcard 53: What are the parametric equations for a line parallel to y=2x+3y = 2x + 3?

Answer: x=tx = t, y=2t+cy = 2t + c. Same slope but different yy-intercept constant.

Flashcard 54: Convert x=2cos(t)x = 2 \cos(t), y=2sin(t)y = 2 \sin(t) to a Cartesian equation.

Answer: x2+y2=4x^2 + y^2 = 4. Circle with radius 2 centered at origin.

Flashcard 55: State the parametric equations for an ellipse centered at the origin.

Answer: x=acos(t)x = a \, \cos(t), y=bsin(t)y = b \, \sin(t). Uses different coefficients for horizontal and vertical radii.

Flashcard 56: Convert x=2cos(t)x = 2 \cos(t), y=2sin(t)y = 2 \sin(t) to a Cartesian equation.

Answer: x2+y2=4x^2 + y^2 = 4. Circle with radius 2 centered at origin.

Flashcard 57: Convert x=2tx = 2t, y=t2y = t^2 to a Cartesian equation.

Answer: y=x24y = \frac{x^2}{4}. Eliminate parameter: t=x2t = \frac{x}{2}, so y=(x2)2y = \left(\frac{x}{2}\right)^2.

Flashcard 58: Convert the parametric equations x=t+1x = t + 1, y=2ty = 2t to Cartesian form.

Answer: y=2(x1)y = 2(x - 1). Solve for tt from first equation, substitute into second.

Flashcard 59: What is the parametric form for a circle with radius rr?

Answer: x=rcos(t)x = r \cos(t), y=rsin(t)y = r \sin(t). General circle equation with specified radius.

Flashcard 60: What is the general form of parametric equations for a line?

Answer: x=x0+atx = x_0 + at, y=y0+bty = y_0 + bt. Point (x0,y0)(x_0, y_0) with direction vector (a,b)(a,b).

Flashcard 61: Which parametric equations describe the line segment from (1,2)(1,2) to (4,8)(4,8)?

Answer: x=1+3tx = 1 + 3t, y=2+6ty = 2 + 6t, 0t10 \leq t \leq 1. Direction vector (3,6)(3,6) with parameter range [0,1][0,1].

Flashcard 62: Convert x=5sin(t)x = 5 \sin(t), y=5cos(t)y = 5 \cos(t) to a Cartesian equation.

Answer: x2+y2=25x^2 + y^2 = 25. Apply Pythagorean identity to eliminate parameter.

Flashcard 63: Convert x=5sin(t)x = 5 \sin(t), y=5cos(t)y = 5 \cos(t) to a Cartesian equation.

Answer: x2+y2=25x^2 + y^2 = 25. Apply Pythagorean identity to eliminate parameter.

Flashcard 64: Convert x=2tx = 2t, y=t2y = t^2 to a Cartesian equation.

Answer: y=x24y = \frac{x^2}{4}. Eliminate parameter: t=x2t = \frac{x}{2}, so y=(x2)2y = \left(\frac{x}{2}\right)^2.

Flashcard 65: What is the parametric equation for a straight line through (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2)?

Answer: x=x1+(x2x1)tx = x_1 + (x_2 - x_1)t, y=y1+(y2y1)ty = y_1 + (y_2 - y_1)t. Linear interpolation between two given points.

Flashcard 66: What is the parametric equation for a straight line through (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2)?

Answer: x=x1+(x2x1)tx = x_1 + (x_2 - x_1)t, y=y1+(y2y1)ty = y_1 + (y_2 - y_1)t. Linear interpolation between two given points.

Flashcard 67: State the parametric equations for an ellipse centered at the origin.

Answer: x=acos(t)x = a \, \cos(t), y=bsin(t)y = b \, \sin(t). Uses different coefficients for horizontal and vertical radii.