Parametrizing Implicit Curves
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| parameter/pəˈræmɪtə/ | tham số |
| implicitly defined/ɪmˈplɪsɪtli dɪˈfaɪnd/ | xác định ngầm |
| parametric equations/ˌpærəˈmetrɪk ɪˈkweɪʒnz/ | phương trình tham số |
Making a static curve move
- An implicit equation like $x^2 + y^2 = 4$ pins down a shape but tells you nothing about motion.
- To animate it, or to plot it point by point, we want a rule that walks along the curve.
- That rule is a parametrization: express $x$ and $y$ in terms of a single parameter.
- It converts a frozen condition into a moving point that traces the curve.
From implicit to parametric
- Start with an implicitly defined 隐式定义 curve — an equation mixing $x$ and $y$.
- Find parametric equations 参数方程 $x(t)$, $y(t)$ whose points all satisfy it.
- As the parameter 参数 runs, the point sweeps out exactly that curve.
- Same curve, new description — now with direction and speed built in.
Parametrizing an implicit curve is useful because it…
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A parametrization gives a rule to walk along the curve, one $t$-value at a time — great for plotting and motion.
The circle, both ways
- Implicit: $x^2 + y^2 = 4$ says "distance $2$ from the origin".
- Parametric: $x = 2\cos t$, $y = 2\sin t$ walks around it as $t$ sweeps $0$ to $2\pi$.
- Check they match: $(2\cos t)^2 + (2\sin t)^2 = 4(\cos^2 t + \sin^2 t) = 4$. ✓
- The Pythagorean identity is exactly what makes the parametrization satisfy the equation.
Turn a circle equation into motion
The implicit circle x squared plus y squared equals 4 becomes the parametric x = 2 cos t, y = 2 sin t as the angle sweeps.
A parametrization of the implicit circle $x^2 + y^2 = 4$ is…
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Check: $(2\cos t)^2 + (2\sin t)^2 = 4(\cos^2 t + \sin^2 t) = 4$. ✓ It satisfies the implicit equation for every $t$.
Select all true statements about parametrizing implicit curves.
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Parametrizing does not change the shape — it describes the same curve. The other three are correct.
Verifying a parametrization
- To confirm $x(t), y(t)$ parametrizes an implicit curve, substitute them into the equation.
- If the equation holds for every $t$, the parametrization is valid.
- This is how you catch a wrong guess: plug in and see if it simplifies to a true statement.
- The trig identity $\cos^2 + \sin^2 = 1$ does the heavy lifting for circles and ellipses.
To verify a parametrization, substitute $x(t)$ and $y(t)$ into the ____ equation and check it holds.
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A valid parametrization must satisfy the implicit equation for every value of the parameter.
A single implicit curve can be parametrized in more than one way.
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Different directions, speeds, or starting points give different valid parametrizations of the same curve.
Many valid parametrizations
- A curve does not have a single "correct" parametrization.
- Reversing direction, changing speed, or shifting the start all give valid ones.
- $x = 2\cos t, y = -2\sin t$ traces the same circle clockwise instead.
- Choose the parametrization whose motion suits your problem.
A parametrization must satisfy the implicit equation for all parameter values, not just one. Verify by substituting and simplifying — a formula that works at a single $t$ but fails elsewhere is not a parametrization of the curve.
Parametrize the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$.
- Try $x = 3\cos t$, $y = 2\sin t$.
- Check: $\dfrac{9\cos^2 t}{9} + \dfrac{4\sin^2 t}{4} = \cos^2 t + \sin^2 t = 1$. ✓
- So $(3\cos t, 2\sin t)$ traces the whole ellipse as $t$ goes $0$ to $2\pi$.
Parametrizing turns an implicitly defined curve into parametric equations $x(t), y(t)$ that trace it as the parameter runs. Verify by substituting into the implicit equation and checking it holds for every $t$. A curve has many valid parametrizations, differing in direction and speed.
Write from a radius of 3
- x equals 3 cos t and y equals 3 sin t.
- Check that the path is the circle of radius 3.
The parametric equations x equals 3 cos t and y equals 3 sin t are meant to trace a circle. What is the radius of that circle?
9 is the square of the radius 3.
What is the radius?
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The identity reduces the path to radius 3.