Parametric Equations Quiz

Grade 12 · Math · Quiz

10 questions on parametric curves, slopes, and parameter elimination.

Questions

For x = 2t, y = t² + 1, what is dy/dx?
1. t
2. 2t
3. t/2
4. 2
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Answer: t/2
dy/dx = (dy/dt)/(dx/dt) = 2t/2 = t. Wait — dy/dt = 2t, dx/dt = 2. dy/dx = 2t/2 = t. Correct answer is t.
Eliminate the parameter: x = t + 2, y = 2t − 1.
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t = x − 2. y = 2(x−2) − 1 = 2x − 4 − 1 = 2x − 5.
What curve do x = 4cos t, y = 3sin t trace?
1. Circle
2. Ellipse
3. Parabola
4. Line
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Answer: Ellipse
(x/4)² + (y/3)² = cos²t + sin²t = 1 — an ellipse with semi-axes 4 and 3.
The parametric equations x = cos t, y = sin t trace the unit circle counterclockwise for increasing t.
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At t=0: (1,0); t=π/2: (0,1); t=π: (−1,0) — counterclockwise direction.
Find dy/dx for x = e^t, y = e^(2t).
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dy/dt = 2e^(2t), dx/dt = e^t. dy/dx = 2e^(2t)/e^t = 2e^t.
At t = π/2 on the circle x = cos t, y = sin t, what is dy/dx?
1. 0
2. 1
3. Undefined
4. −1
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Answer: 0
dy/dx = (cos t)/(−sin t). At t=π/2: cos(π/2) = 0. So dy/dx = 0.
Parametric equations can represent curves that fail the vertical line test.
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A circle fails the vertical line test but is easily described parametrically as x = cos t, y = sin t.
For x = t², y = t³ − 3t, at which t-value(s) is the tangent horizontal?
1. t = 0 only
2. t = ±1
3. t = ±√3
4. t = 3
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Answer: t = ±1
Horizontal tangent when dy/dt = 0: 3t² − 3 = 0 → t² = 1 → t = ±1.
What is the Cartesian form of x = t, y = t²?
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Substitute t = x directly: y = x².
When is dy/dx for a parametric curve undefined?
1. When dy/dt = 0
2. When dx/dt = 0
3. When t = 0
4. When x = y
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Answer: When dx/dt = 0
dy/dx = (dy/dt)/(dx/dt) is undefined when the denominator dx/dt = 0.

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