AP Precalculus Flashcards: Parametrically Defined Circles And Lines

Study Parametrically Defined Circles And Lines 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

Parametrically Defined Circles And Lines

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QUESTION
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Identify the tt value for the point (r,0)(-r, 0) on x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t).

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ANSWER

t=πt = \pi. At t=πt = \pi, cos(π)=1\cos(\pi) = -1 and sin(π)=0\sin(\pi) = 0.

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This deck focuses on Parametrically Defined Circles And Lines, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.

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Flashcard 1: Identify the tt value for the point (r,0)(-r, 0) on x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t).

Answer: t=πt = \pi. At t=πt = \pi, cos(π)=1\cos(\pi) = -1 and sin(π)=0\sin(\pi) = 0.

Flashcard 2: Convert x=2t,y=3+tx = 2t, \, y = 3 + t to its Cartesian form.

Answer: y=3+x2y = 3 + \frac{x}{2}. Eliminate tt: t=x2t = \frac{x}{2}, substitute into yy.

Flashcard 3: Find the Cartesian equation of x=5cos(t),y=5sin(t)x = 5 \, \cos(t), \, y = 5 \, \sin(t).

Answer: x2+y2=25x^2 + y^2 = 25. Use identity cos2(t)+sin2(t)=1\cos^2(t) + \sin^2(t) = 1 with radius 5.

Flashcard 4: What is the parametric form for a line parallel to y=3x+2y = 3x + 2 through (4,1)(4, 1)?

Answer: x=4+t,y=1+3tx = 4 + t, \, y = 1 + 3t. Parallel lines have same slope; direction vector (1,3)(1, 3).

Flashcard 5: Convert x=1+t,y=2tx = 1 + t, \, y = 2t to its Cartesian form.

Answer: y=2(x1)y = 2(x - 1). Eliminate tt: t=x1t = x - 1, substitute into yy.

Flashcard 6: What is the parametric form of a vertical line x=cx = c?

Answer: x=c,y=tx = c, \, y = t. Parameter tt varies vertically, xx remains constant.

Flashcard 7: Determine the yy-coordinate at t=0t = 0 for x=3cos(t),y=3sin(t)x = 3 \, \cos(t), \, y = 3 \, \sin(t).

Answer: y=0y = 0. At t=0t = 0, sin(0)=0\sin(0) = 0 and cos(0)=1\cos(0) = 1.

Flashcard 8: Find the xx-coordinate when t=θ2t = \frac{\theta}{2} for x=5cos(t),y=5sin(t)x = 5 \, \cos(t), \, y = 5 \, \sin(t).

Answer: x=5cos(θ2)x = 5 \, \cos\left(\frac{\theta}{2}\right). Substitute t=θ2t = \frac{\theta}{2} into the xx equation.

Flashcard 9: Convert x=3t,y=4tx = 3t, \, y = 4 - t to its Cartesian form.

Answer: y=413xy = 4 - \frac{1}{3}x. Eliminate tt: t=x3t = \frac{x}{3}, substitute into yy.

Flashcard 10: Identify the tt value for the point (0,r)(0, r) on x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t).

Answer: t=π2t = \frac{\pi}{2}. At t=π2t = \frac{\pi}{2}, cos(t)=0\cos(t) = 0 and sin(t)=1\sin(t) = 1.

Flashcard 11: Find xx when t=0t = 0 for x=4+2t,y=3tx = 4 + 2t, \, y = 3 - t.

Answer: x=4x = 4. Substitute t=0t = 0 into the xx equation.

Flashcard 12: Determine the tt value for the point (r,0)(-r, 0) on x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t).

Answer: t=πt = \pi. At t=πt = \pi, cos(π)=1\cos(\pi) = -1 and sin(π)=0\sin(\pi) = 0.

Flashcard 13: Find the xx-coordinate when t=πt = \pi for x=2+3t,y=4tx = 2 + 3t, \, y = 4 - t.

Answer: x=2+3πx = 2 + 3\pi. Substitute t=πt = \pi into the xx equation.

Flashcard 14: Identify the tt value for the point (0,r)(0, -r) on x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t).

Answer: t=3π2t = \frac{3\pi}{2}. At t=3π2t = \frac{3\pi}{2}, cos(t)=0\cos(t) = 0 and sin(t)=1\sin(t) = -1.

Flashcard 15: Determine the tt value for (0,r)(0, r) on x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t).

Answer: t=π2t = \frac{\pi}{2}. At t=π2t = \frac{\pi}{2}, cos(t)=0\cos(t) = 0 and sin(t)=1\sin(t) = 1.

Flashcard 16: What is the parametric form of a line through (a,b)(a, b) parallel to y=mx+cy = mx + c?

Answer: x=a+t,y=b+mtx = a + t, \, y = b + mt. Direction vector from slope mm of parallel line.

Flashcard 17: Find the Cartesian equation of x=1+2t,y=3tx = 1 + 2t, \, y = 3 - t.

Answer: y=312(x1)y = 3 - \frac{1}{2}(x - 1). Eliminate tt: t=x12t = \frac{x-1}{2}, substitute into yy.

Flashcard 18: Convert x=2+3t,y=4+5tx = 2 + 3t, \, y = 4 + 5t to its Cartesian form.

Answer: y=4+53(x2)y = 4 + \frac{5}{3}(x - 2). Eliminate parameter: t=x23t = \frac{x-2}{3}, substitute into yy.

Flashcard 19: Convert x=t,y=2t+1x = t, \, y = 2t + 1 to its Cartesian form.

Answer: y=2x+1y = 2x + 1. Direct substitution since x=tx = t.

Flashcard 20: Find the yy-coordinate when t=π4t = \frac{\pi}{4} for x=4cos(t),y=4sin(t)x = 4 \, \cos(t), \, y = 4 \, \sin(t).

Answer: y=4sin(π4)y = 4 \, \sin\left(\frac{\pi}{4}\right). Substitute t=π4t = \frac{\pi}{4} into the yy equation.

Flashcard 21: Find xx when t=2t = 2 for x=1+3t,y=2tx = 1 + 3t, \, y = 2t.

Answer: x=7x = 7. Substitute t=2t = 2 into the xx equation.

Flashcard 22: What is the parametric form of a circle centered at the origin with radius rr?

Answer: x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t). Standard circular parametrization using trigonometric functions.

Flashcard 23: What is the parametric form of a circle with center (0,0)(0, 0) and radius rr?

Answer: x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t). Standard circular parametrization centered at origin.

Flashcard 24: Convert x=2+4t,y=3+2tx = 2 + 4t, \, y = 3 + 2t to its Cartesian form.

Answer: y=3+12(x2)y = 3 + \frac{1}{2}(x - 2). Eliminate tt: t=x24t = \frac{x-2}{4}, substitute into yy.

Flashcard 25: Identify the parameter tt value at the topmost point of x=rcos(t),y=rsin(t)x = r \, \cos(t), y = r \, \sin(t).

Answer: t=π2t = \frac{\pi}{2}. At t=π2t = \frac{\pi}{2}, cos(t)=0\cos(t) = 0 and sin(t)=1\sin(t) = 1.

Flashcard 26: What is the parametric form of a circle with center (h,k)(h, k) and radius rr?

Answer: x=h+rcos(t),y=k+rsin(t)x = h + r \, \cos(t), \, y = k + r \, \sin(t). Translation of circle center from origin to (h,k)(h, k).

Flashcard 27: Convert x=1+2t,y=3tx = 1 + 2t, \, y = 3t to its Cartesian form.

Answer: y=32(x1)y = \frac{3}{2}(x - 1). Eliminate tt: t=x12t = \frac{x-1}{2}, substitute into yy.

Flashcard 28: What is the parametric form of a line with slope mm passing through (0,0)(0, 0)?

Answer: x=t,y=mtx = t, \, y = mt. Line through origin with direction vector (1,m)(1, m).

Flashcard 29: What are the parametric equations for a line through (x1,y1)(x_1, y_1) with slope mm?

Answer: x=x1+t,y=y1+mtx = x_1 + t, \, y = y_1 + mt. Direction vector (1,m)(1, m) from point-slope form.

Flashcard 30: Identify the tt value for the point (r,0)(r, 0) on x=rcos(t),y=rsin(t)x = r \, \cos(t), \, y = r \, \sin(t).

Answer: t=0t = 0. At t=0t = 0, cos(0)=1\cos(0) = 1 and sin(0)=0\sin(0) = 0.

Flashcard 31: Find yy when t=1t = 1 for x=3+2t,y=4tx = 3 + 2t, \, y = 4 - t.

Answer: y=3y = 3. Substitute t=1t = 1 into the yy equation.

Flashcard 32: What is the parametric form of a horizontal line y=cy = c?

Answer: x=t,y=cx = t, \, y = c. Parameter tt varies along the line, yy remains constant.

Flashcard 33: What tt value corresponds to the point (h+r,k)(h + r, k) on x=h+rcos(t),y=k+rsin(t)x = h + r \, \cos(t), y = k + r \, \sin(t)?

Answer: t=0t = 0. At t=0t = 0, cos(0)=1\cos(0) = 1 gives rightmost point.

Flashcard 34: What are the parametric equations for a line through (0,0)(0, 0) with slope mm?

Answer: x=t,y=mtx = t, \, y = mt. Line through origin with direction vector (1,m)(1, m).

Flashcard 35: What are the parametric equations for a line through (x0,y0)(x_0, y_0) with direction vector (a,b)(a, b)?

Answer: x=x0+at,y=y0+btx = x_0 + at, \, y = y_0 + bt. Direction vector (a,b)(a, b) from point (x0,y0)(x_0, y_0).

Flashcard 36: Convert x=4cos(t),y=4sin(t)x = 4 \, \cos(t), \, y = 4 \, \sin(t) to its Cartesian form.

Answer: x2+y2=16x^2 + y^2 = 16. Use identity cos2(t)+sin2(t)=1\cos^2(t) + \sin^2(t) = 1 with radius 4.

Flashcard 37: Determine the yy-coordinate when t=πt = \pi for x=2+3t,y=4tx = 2 + 3t, \, y = 4 - t.

Answer: y=4πy = 4 - \pi. Substitute t=πt = \pi into the yy equation.