3 Parametric Equations

Representing and tracing curves

A parametrically defined point has coordinates determined by a parameter $t$:

As $t$ changes, the point $(x(t),y(t))$ traces a path. The parameter may represent time, but it can also be an auxiliary variable chosen because it describes the curve conveniently. This approach is especially useful for paths that cannot be represented conveniently as one function $y=f(x)$.

A Parametric curve contains more information than its rectangular equation alone. The parameter can record the starting point, direction, and speed of traversal. For example, consider

Solving the first equation for the parameter gives $t=x-1$, so the rectangular equation is

As $t$ increases, $x=t+1$ increases, so this parametrization traces the parabola from left to right. Another parametrization could trace the same parabola in the opposite direction or at a different speed.

Useful standard forms include a line through $(x_0,y_0)$ with direction vector $\langle a,b\rangle$:

A circle centered at $(h,k)$ with radius $r$ can be written as

For the unit circle, $x=\cos t$ and $y=\sin t$; increasing $t$ from $0$ to $2\pi$ traces the circle counterclockwise.

Takeaway: Always identify both the geometric path and how the parameter moves along it.

Slopes and tangent lines

The slope of a parametrically defined curve is found by differentiating both coordinate functions. The chain rule gives

Therefore, when $x'(t)\ne 0$,

To find a Tangent line at $t=t_0$, first find the point $(x(t_0),y(t_0))$, then evaluate the slope, and finally use point-slope form:

For the circle

we have

so

At $t=\frac{\pi}{4}$, the point is $\left(\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2}\right)$, and the slope is $-1$. Thus the tangent line is

which simplifies to

Special tangent behavior follows directly from the coordinate derivatives:

• A horizontal tangent occurs 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 usual slope formula is inconclusive. The point may be a cusp, a corner, or another singular point and requires additional analysis.

Takeaway: Find the point and both coordinate derivatives before deciding whether the tangent is ordinary, horizontal, vertical, or singular.

Second derivatives and concavity

To study concavity, differentiate the first-derivative formula with respect to the parameter and then divide by $dx/dt$:

An equivalent formula is

provided $x'(t)\ne 0$. A positive value indicates concavity up, while a negative value indicates concavity down.

For

we obtain, for $t\ne 0$,

Differentiating with respect to $t$ gives

and therefore

Thus the curve is concave down for $t<0$ and concave up for $t>0$, wherever the derivative is defined.

Takeaway: Concavity depends on how the tangent slope changes with respect to $x$, so the final division by $dx/dt$ is essential.

Distance and area

For a smooth curve defined by $x=x(t)$ and $y=y(t)$ on $a\le t\le b$, Arc length is

The expression under the integral is the rate at which distance accumulates. For a circle of radius $r$, use

Then

so

The area under a parametric curve is based on $A=\int y\,dx$. Since $dx=x'(t)\,dt$, the corresponding formula is

This produces Signed area. Check whether the curve lies below the horizontal axis or whether $x(t)$ decreases. If either occurs, the integral may be negative, so split the interval and use absolute values or add geometric areas separately when ordinary area is required.

For

we have $x'(t)=2t$, and therefore

Here $x(t)$ increases and $y(t)$ is nonnegative, so the signed integral equals the ordinary geometric area.

Takeaway: Differentiate the coordinate functions first, then use the appropriate integral and inspect its sign.

Parametric motion

Parametric equations also describe motion. If the position is

then the Velocity vector is

Its magnitude is the Speed:

The Acceleration vector is

Because speed is the rate at which arc length accumulates, total distance traveled over an interval is

For the projectile model

where position is measured in meters and time in seconds, the velocity is

and the speed is

The acceleration is constant:

The tangent slope is

When $30-9.8t=0$, the vertical component of velocity is zero, so the trajectory has a horizontal tangent and the object reaches its maximum height.

Takeaway: Coordinate derivatives have both geometric and physical meanings: their ratio gives tangent slope, their vector gives velocity, their magnitude gives speed, and their second derivatives give acceleration.

A unified problem-solving workflow

A reliable workflow keeps the parameter interval, geometry, derivatives, and signs connected:

1. Identify the parameter interval and determine the direction of travel as $t$ increases.

2. For a specified parameter value, calculate the point $(x(t),y(t))$.

3. Differentiate both coordinate functions to obtain $x'(t)$ and $y'(t)$.

4. Compute the tangent slope with

   $$\frac{dy}{dx}=\frac{y'(t)}{x'(t)},$$

   provided $x'(t)\ne 0$.

5. Check separately for horizontal tangents, vertical tangents, and values where both derivatives vanish.

6. For concavity, use

   $$\frac{d^2y}{dx^2}=\frac{d(dy/dx)/dt}{dx/dt}.$$

7. For distance, integrate the speed:

   $$L=\int_a^b\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt.$$

8. For area, use

   $$A=\int_a^b y(t)x'(t)\,dt,$$

   then inspect whether the result is signed or geometric area.

9. For motion, interpret $\langle x'(t),y'(t)\rangle$ as velocity, $\langle x''(t),y''(t)\rangle$ as acceleration, and the velocity magnitude as speed.

The central idea is that a parametrization describes both where a curve is and how it is traversed. Derivatives reveal local direction and changing slope, while integrals accumulate distance or signed area.