Parametric derivatives and tangent lines

Understand the idea

A parametric curve gives both coordinates as functions of a third variable, often time. The derivative $\displaystyle dy/dx$ compares the rates at which the two coordinates change. It is a slope in the coordinate plane, not the vertical coordinate's rate of change with respect to time.

Use the parameter to locate the point before writing a tangent line. Differentiate $\displaystyle x(t)$ and $\displaystyle y(t)$ separately, then take their ratio wherever $\displaystyle x'(t)\ne0$. Eliminating the parameter can be a useful check, but it may hide restrictions or the direction in which the curve is traced.

A horizontal tangent is supported by $\displaystyle y'(t)=0$ with $\displaystyle x'(t)\ne0$. A vertical tangent is supported by $\displaystyle x'(t)=0$ with $\displaystyle y'(t)\ne0$. If both derivatives vanish, those quick tests are inconclusive; examine nearby behavior rather than treating $\displaystyle 0/0$ as a slope.

Worked example

For $\displaystyle x=t^2+1$ and $\displaystyle y=t^3-3t$, find the tangent line at $\displaystyle t=2$.

1. The point is $\displaystyle (x(2),y(2))=(5,2)$.
2. The coordinate derivatives are $\displaystyle x'=2t$ and $\displaystyle y'=3t^2-3$.
3. At $\displaystyle t=2$, the slope is $\displaystyle 9/4$ because $\displaystyle x'=4\ne0$.