Parametric Equations and Motion

Representing Curves with Parameters

A parametric curve describes a point by giving both coordinates as functions of a parameter: $x=x(t)$ and $y=y(t)$. As $t$ varies over an interval $a\le t\le b$, the point $(x(t),y(t))$ traces the curve. The interval and the direction of increasing $t$ matter because they determine which part of the curve is traced and in what order.

A useful example is the unit circle:

Using the identity $\cos^2t+\sin^2t=1$, the corresponding rectangular equation is $x^2+y^2=1$. At $t=0$, the point is $(1,0)$; as $t$ increases, the point moves counterclockwise and returns to its starting point at $t=2\pi$.

Connecting parametric and rectangular forms

Eliminating the parameter can reveal the shape of a curve. For example, if $x=t^2$ and $y=t+1$, then $t=y-1$, so substitution gives $x=(y-1)^2$. However, the rectangular equation does not by itself preserve the timing or direction encoded by $t$. It can also hide repeated traversal of a point.

Takeaway: Parametric equations describe both the geometry of a curve and the way the curve is traced.

Derivatives, Tangents, and Concavity

For a curve $x=x(t)$, $y=y(t)$, the slope of a parametric curve follows from the Chain Rule:

At $t=t_0$, the point is $(x(t_0),y(t_0))$, and the tangent-line equation is

For example, let $x=t^2+1$ and $y=t^3-2t$. At $t=1$, the point is $(2,-1)$, while $x'(t)=2t$ and $y'(t)=3t^2-2$. Thus the slope is $\frac{1}{2}$, and the tangent line is $y+1=\frac12(x-2)$.

A horizontal tangent occurs, at a regular point, when $y'(t)=0$ and $x'(t)\ne 0$. A vertical tangent occurs when $x'(t)=0$ and $y'(t)\ne 0$. If both derivatives are zero, the point is nonregular and may represent a cusp, a corner-like singularity, or another special behavior.

Concavity

Differentiate the first-derivative formula with respect to $t$, then convert the result to differentiation with respect to $x$:

For the example above, $\frac{dy}{dx}=\frac32t-\frac1t$, so $\frac{d}{dt}(dy/dx)=\frac32+\frac1{t^2}$. Since $dx/dt=2t$, the second derivative at $t=1$ is $\frac54>0$, indicating concavity up there.

Takeaway: Differentiate both coordinate functions first, then use their ratio for slope and the parameter-conversion rule for concavity.

Motion: Position, Velocity, and Acceleration

When the parameter represents time, the position vector is

Its derivative is Velocity:

The components describe horizontal and vertical rates of change. The corresponding Speed is the magnitude of velocity:

Speed is nonnegative, while velocity includes direction. The derivative of velocity is Acceleration:

Example: projectile motion

Suppose

Then

At $t=1$, velocity is $\langle 3,0.2\rangle$, so the projectile is still moving slightly upward. Its speed is

The highest point occurs when the vertical velocity is zero:

The geometric slope of the path is still $dy/dx=y'(t)/x'(t)$. This describes the path’s tangent, not the full motion of the particle. For instance, a particle may have a horizontal tangent when $y'(t)=0$ while still moving horizontally with $x'(t)\ne 0$.

Takeaway: Position, velocity, speed, and acceleration are successive layers of information about motion; slope describes the geometry of the path.

Distance, Arc Length, and Area

The differential distance along a parametric curve is

Therefore, Arc length over $a\le t\le b$ is

When $t$ is time, the same integral is distance traveled because it integrates speed. It measures total distance rather than displacement, so reversals of direction do not cancel.

For the unit circle $x=\cos t$, $y=\sin t$, $0\le t\le2\pi$, the derivatives are $x'=-\sin t$ and $y'=\cos t$. Thus

Parametric area

The usual area expression $A=\int y\,dx$ becomes a parametric integral after substituting $dx=x'(t)\,dt$. Signed area is therefore

For $x=t^2$, $y=t$, and $0\le t\le1$, we have $x'(t)=2t$, so

If the curve lies below the $x$-axis, the signed result is negative. For geometric area, split the interval where the curve crosses the axis or changes orientation, and use absolute values as appropriate. If $x'(t)$ changes sign, the parameter is not moving consistently from left to right, so interval splitting may also be necessary.

Takeaway: Arc length integrates speed, while parametric area integrates $y\,dx$; both formulas depend on how the parameter traces the curve.

A Unified Problem-Solving Strategy

A reliable solution process keeps the parameter, interval, derivatives, and interpretation connected.

1. Identify the parameter and interval. Decide whether $t$ represents time and determine the relevant values of $t$.

2. Differentiate both coordinates. Find $x'(t)$ and $y'(t)$; also find $x''(t)$ and $y''(t)$ when acceleration or concavity is needed.

3. Choose the matching formula.

   • Slope: $\dfrac{dy}{dx}=\dfrac{y'}{x'}$

   • Second derivative: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}(dy/dx)}{x'}$

   • Speed: $\sqrt{(x')^2+(y')^2}$

   • Arc length or distance: $\int_a^b\sqrt{(x')^2+(y')^2}\,dt$

   • Signed area: $\int_a^b y(t)x'(t)\,dt$

4. Check special parameter values. Test where $x'(t)=0$, where $y'(t)=0$, and where both derivatives vanish.

5. Interpret the answer. Include units in motion problems, distinguish speed from velocity, and distinguish signed area from geometric area.

The central relationships can be summarized as follows:

• Position: $\mathbf r(t)=\langle x(t),y(t)\rangle$

• Velocity: $\mathbf v(t)=\langle x'(t),y'(t)\rangle$

• Acceleration: $\mathbf a(t)=\langle x''(t),y''(t)\rangle$

• Path slope: $\dfrac{dy}{dx}=\dfrac{y'(t)}{x'(t)}$

• Speed: $\sqrt{[x'(t)]^2+[y'(t)]^2}$

Final takeaway: Parametric calculus uses the same coordinate derivatives to answer different questions: geometry through slope and concavity, motion through velocity and acceleration, distance through speed, and area through the conversion $dx=x'(t)\,dt$.