Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
Parametric Equations and Parametric Curves
What a parametric equation is (and why you use it)
A parametric representation of a curve describes the coordinates separately in terms of a third variable called a parameter (often time). Instead of thinking “y is a function of x,” you think “both coordinates are functions of the parameter.” This is especially natural for describing the position of a moving object or the shape traced out over time.
A parametric curve is typically written as:
x=f(t)
y=g(t)
Equivalently, you may see:
x=x(t)
y=y(t)
The key dependency idea is that x and y are dependent variables, and time/parameter t is the independent variable.
This matters because many real situations naturally produce coordinates as functions of time: a moving particle, a projectile, a point on a rolling wheel, or a curve traced by a robot arm. Some curves also fail the vertical line test (so they are not functions y=f(x)), but they still can be described cleanly with parameters.
A key idea: the same geometric curve can have many parametric descriptions. The parameter controls how the curve is traced (its orientation and speed) without changing the set of points (unless the parameter interval changes).
Reading a parametric curve: position, orientation, and repeats
When you have x=x(t) and y=y(t) over an interval a \le t \le b, you should always think about which points appear (the path), in what direction the curve is traced as t increases, and whether points repeat (the curve may loop and cross itself).
A common misconception is to treat parametric equations like two independent graphs. But the curve is not “the graph of x(t) plus the graph of y(t).” It is the set of points \big(x(t),y(t)\big) at matching parameter values.
Eliminating the parameter (when it helps and when it misleads)
Sometimes you can solve for t from one equation and substitute into the other to get a rectangular equation relating x and y. This can help you recognize the shape.
However, eliminating the parameter usually throws away orientation information and can also hide restrictions coming from the allowed t-interval. Two different parameter intervals can give the same rectangular equation but trace different portions of the curve.
Calculus with parametric equations: slope of the tangent line
To find the slope of the tangent line to a parametric curve, you want the derivative of y with respect to x. Since both depend on t, you use the chain rule:
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
This formula works when dx/dt \ne 0. Geometrically:
- If dx/dt=0 and dy/dt \ne 0, then the tangent line is vertical.
- If dy/dt=0 and dx/dt \ne 0, then the tangent line is horizontal.
A frequent error is to compute dy/dt and treat it as the slope. It is not; it is the rate of change of y with respect to the parameter, not with respect to x.
Example 1: slope and tangent line
Let
x=t^2+1
y=t^3-3t
Find the slope at t=1 and the tangent line.
Compute derivatives:
\frac{dx}{dt}=2t
\frac{dy}{dt}=3t^2-3
So
\frac{dy}{dx}=\frac{3t^2-3}{2t}
At t=1:
\frac{dy}{dx}=\frac{3-3}{2}=0
Now find the point on the curve at t=1:
x(1)=2
y(1)=-2
A slope of zero means a horizontal tangent line through \big(2,-2\big):
y=-2
Second derivative and concavity for parametric curves
Concavity is about how the slope changes as you move along the curve with respect to x. The second derivative can be computed by differentiating dy/dx with respect to t and dividing by dx/dt again:
\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}
This formula is easy to misuse if you forget that the outer derivative must be taken with respect to the parameter first.
A conceptual way to stay grounded: you are converting “change in slope per change in parameter” into “change in slope per change in x.”
Example 2: concavity at a parameter value
Using the same curve:
\frac{dy}{dx}=\frac{3t^2-3}{2t}=\frac{3}{2}\left(t-\frac{1}{t}\right)
Differentiate with respect to t:
\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{3}{2}\left(1+\frac{1}{t^2}\right)
Then divide by dx/dt=2t:
\frac{d^2y}{dx^2}=\frac{\frac{3}{2}\left(1+\frac{1}{t^2}\right)}{2t}=\frac{3\left(t^2+1\right)}{4t^3}
At t=1:
\frac{d^2y}{dx^2}=\frac{3(2)}{4}=\frac{3}{2}
Positive second derivative means the curve is concave up at that point (as a function of x locally).
Accumulated change along a parametric curve: area under the curve
If a parametric curve moves left-to-right without doubling back too much, you can compute “area under the curve” using the idea that small changes in area look like y times a small change in x.
Since dx=(dx/dt)dt, the signed area from t=a to t=b is:
A=\int_a^b y(t)\,\frac{dx}{dt}\,dt
This gives signed area: if the curve travels right-to-left (so dx/dt