Defining and Differentiating Parametric Equations
Introduced| English | Meaning |
|---|---|
| parametric equations/ˌpærəˈmetrɪk ɪˈkweɪʒnz/ | Equations expressing coordinates or variables in terms of one or more common parameters. |
| parameter/pəˈræmɪtə/ | A value that specifies a particular member of a family of functions, models or systems. |
Track a moving point with two coordinates
- A drone's position needs both a horizontal coordinate and a vertical coordinate at each time. Parametric equations 参数方程 give these together: $x=f(t)$ and $y=g(t)$.
- The parameter 参数 $t$ controls both coordinates. It often represents time, but a mathematical parameter need not be time. The ordered pair $(x(t),y(t))$ traces the path.
In $x=f(t)$, $y=g(t)$, the variable $t$ is called the ____.
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The parameter drives both coordinates.
Evaluate a pair, then follow its order
- For $x=t^2$, $y=t^3$, the values $t=-1,0,1,2$ give $(1,-1),(0,0),(1,1),(4,8)$. Substitute the same parameter value into both equations.
- As $t$ increases, the point follows that order. At $t=-1$ and $t=1$, the same $x$ has different $y$ values, so the whole path is not one function $y=f(x)$.
For $x=t^2$, $y=t^3$, what is the y-coordinate at $t=-1$?
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$(-1)^3=-1$; the point is $(1,-1)$.
The whole curve $x=t^2$, $y=t^3$ for $-2\le t\le2$ is one function $y=f(x)$.
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At $x=1$, it includes both $y=1$ and $y=-1$.
Keep the interval and direction
- For $x=\cos t$, $y=\sin t$, eliminating $t$ gives $x^2+y^2=1$. With $0\le t\le\pi$, only the upper semicircle is traced, from $(1,0)$ through $(0,1)$ to $(-1,0)$.
- With $0\le t\le2\pi$, the full circle is traced counterclockwise. Replacing $y=\sin t$ by $y=-\sin t$ reverses that direction. A Cartesian equation alone loses the parameter interval and travel order.
Which parametric equations trace the unit circle?
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$\cos^2 t+\sin^2 t=1$ traces the unit circle.
For $x=\cos t$, $y=\sin t$, $0\le t\le\pi$, which path is traced?
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The y-coordinate is nonnegative and the endpoints are $(1,0)$ and $(-1,0)$.
Derive the slope and check its denominator
- Where $dx/dt\ne0$, the chain rule gives $dy/dt=(dy/dx)(dx/dt)$, hence $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. Both derivatives are evaluated at the same $t$.
- This is a ratio of coordinate derivatives. It is not the derivative of $y/x$, not their product, and not the inverted ratio. When $dx/dt=0$, this division cannot decide the tangent.
For a parametric curve, $\dfrac{dy}{dx}=$
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Top over bottom: $\frac{dy/dt}{dx/dt}$.
The formula $(dy/dt)/(dx/dt)$ can be evaluated directly even when $dx/dt=0$.
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Division requires $dx/dt\ne0$. A zero denominator needs a separate tangent check.
Work a slope and a tangent line
- For $x=t^2$, $y=t^3$, $dx/dt=2t$ and $dy/dt=3t^2$. At $t=2$, the point is $(4,8)$ and the slope is $12/4=3$.
- Therefore the tangent line is $y-8=3(x-4)$. For $t\ne0$, the slope formula simplifies to $3t/2$; cancelling $t$ does not make the original ratio valid at $t=0$.
Trace the parameter and inspect the tangent
For x=t² and y=t³ at t=2, locate (4,8), compare the two coordinate derivatives and see slope 3. Compare regular horizontal/vertical tangents with a both-zero derivative ratio.
For $x=t^2$, $y=t^3$, $\dfrac{dy}{dx}=\dfrac{3t}{2}$. Find the slope at $t=2$.
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$\frac{3(2)}{2}=3$.
For $x=t^2$, $y=t^3$ at $t=2$, what is the x-coordinate used in the tangent-line equation?
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The point is $(4,8)$, not $(2,8)$: substitute into both coordinate functions.
The tangent $y-8=3(x-4)$ has what y-intercept?
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$y=3x-4$, so the intercept is $-4$.
A regular horizontal tangent
- If $dy/dt=0$ and $dx/dt\ne0$, the tangent is horizontal. The vertical coordinate has zero rate while the horizontal coordinate has a nonzero rate.
- For $x=t$, $y=t^2$ at $t=0$, the derivatives are $1$ and $0$. The point is $(0,0)$, the slope is $0$, and the tangent line is $y=0$.
A parametric curve has a horizontal tangent where...
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The regular horizontal test needs $dy/dt=0$ and $dx/dt\ne0$, so their ratio is 0.
At $t=0$ for $x=t$, $y=t^2$, which tangent occurs?
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The derivatives are 1 and 0, giving slope 0 and line $y=0$.
A regular vertical tangent
- If $dx/dt=0$ and $dy/dt\ne0$, the tangent is vertical. Its equation is $x=x(t_0)$ at the parameter value $t_0$; it has no finite slope.
- For $x=t^2$, $y=t$ at $t=0$, the derivatives are $0$ and $1$. The point is $(0,0)$ and the tangent line is $x=0$. An undefined slope by itself is not enough to identify a vertical tangent.
A vertical tangent occurs where $\dfrac{dx}{dt}=$ ____ (and $\tfrac{dy}{dt}\neq0$).
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Here $dx/dt=0$ and the stated condition $dy/dt\ne0$ gives a regular vertical tangent.
At $t=0$ for $x=t^2+3$, $y=2t$, what is the tangent?
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The point is $(3,0)$ and the derivatives are 0 and 2, so the vertical tangent is $x=3$.
When both derivatives are zero
- If $dx/dt=dy/dt=0$, the ratio is $0/0$ and the regular horizontal/vertical tests do not apply. Do not automatically label this point vertical or say no tangent exists.
- For $x=t^3$, $y=t^3$ at $t=0$, both derivatives vanish, but the path is the line $y=x$, with tangent slope $1$. For $x=t^2$, $y=t^3$, the secant slope from the origin is $y/x=t\to0$, giving the tangent line $y=0$ there.
If $dx/dt=dy/dt=0$, what follows from the derivative ratio alone?
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The ratio is $0/0$. It gives neither a regular tangent classification nor proof that no tangent exists.
For $x=t^3$, $y=t^3$, the path is $y=x$. What is its tangent slope at the origin?
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The geometric path is a straight line of slope 1 even though both coordinate derivatives vanish at $t=0$.
Direction and tangent slope are different
- The line $x=t$, $y=2t$ is travelled toward increasing $x$ as $t$ increases. The parametrization $x=-u$, $y=-2u$ follows the same line in the opposite direction.
- The slopes are $2/1=2$ and $(-2)/(-1)=2$. Reversing travel does not reverse the tangent line's slope at the same point: both coordinate rates change sign together.
For $x=-u$, $y=-2u$, find the tangent slope.
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$(-2)/(-1)=2$. Reversing the travel direction changes both rates, preserving this slope.
Find candidates, then verify them
- For $x=t^2-1$, $y=t^3-3t$, horizontal candidates satisfy $3t^2-3=0$, so $t=\pm1$. At both, $2t\ne0$; the tangent lines are $y=-2$ and $y=2$.
- The vertical candidate is $t=0$, where $2t=0$ but $3t^2-3=-3\ne0$. Its point is $(-1,0)$ and its tangent is $x=-1$. Check the allowed parameter interval before accepting any candidate.
For $x=t^2-1$, $y=t^3-3t$, what positive t gives a regular horizontal tangent?
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$3t^2-3=0$ gives $t=\pm1$; $dx/dt=2t\ne0$ at both.
For that same curve, the vertical tangent is $x=$ what value?
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At $t=0$, the point is $(-1,0)$ and $dy/dt=-3\ne0$, so $x=-1$.
For $x=t^2-1$, $y=t^3-3t$ restricted to $0\le t\le2$, $t=-1$ is an allowed horizontal-tangent parameter.
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Although it solves the derivative equation, $t=-1$ is outside the allowed interval.
Try a fresh curve
- For $x=t^2+2$, $y=t^3-12t$ on $-3\le t\le3$, find all regular horizontal and vertical tangents. Give parameter values, points and line equations.
- For $x=2t+1$, $y=t^2$, find the tangent at $t=-1$. Then explain why $x=t^3$, $y=2t^3$ at $t=0$ needs more than the regular derivative-ratio test.
For $x=t^2+2$, $y=t^3-12t$, $-3\le t\le3$, find all regular horizontal and vertical tangents. Give t, point and line.
Compare your derivative checks, point and tangent conclusion with the model.
For $x=2t+1$, $y=t^2$, find the tangent line at $t=-1$.
Compare your derivative checks, point and tangent conclusion with the model.
A reliable parametric method
- Find the point, differentiate both coordinates, and divide only when $dx/dt\ne0$. A finite slope $m$ at $(x_0,y_0)$ gives $y-y_0=m(x-x_0)$.
- Check both derivatives for regular horizontal or vertical tangents; investigate $0/0$ separately. Keep the parameter interval and direction, which can disappear when you eliminate the parameter.
What can be lost when the parameter is eliminated?
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A Cartesian relation may show the whole geometric curve while omitting which part is traced and its direction.
For $x=t^3$, $y=2t^3$, explain what the regular ratio test says at $t=0$, then find the actual tangent.
Compare your derivative checks, point and tangent conclusion with the model.