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

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

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ANSWER

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

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

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Flashcard 1: 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 2: 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 3: 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 4: 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 5: What is the purpose of parametric equations in mathematics?

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

Flashcard 6: 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 7: 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 8: 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 9: Identify parametric equations for a horizontal line y=cy = c.

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

Flashcard 10: 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 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: 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 27: 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 28: 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 29: 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 30: 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 31: 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 32: 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 33: 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 34: 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.