9.1 Defining and Differentiating Parametric Equations

Learning objectives

—CHA-3.G—Calculate derivatives of parametric functions• CHA-3.G Calculate derivatives of parametric functions.• CHA-3.G.1 Methods for calculating derivatives of real-valued functions can be extended to parametric functions.• CHA-3.G.2 For a curve defined parametrically, the value of dy/dx at a point on the curve is the slope of the line tangent to the curve at that point. dy/dx, the slope of the line tangent to a curve defined using parametric equations, can be determined by dividing dy/dt by dx/dt, provided dx/dt does not equal zero.• Enduring understanding CHA-3: Derivatives allow us to solve real-world problems involving rates of change.

Parametric equations $x=x(t)$ and $y=y(t)$ assign both coordinates to the same parameter value. As $t$ changes, the point $(x(t),y(t))$ traces a curve. Because both coordinates depend on $t$ , their derivatives describe horizontal and vertical rates along that curve.

\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\quad\Longrightarrow\quad\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad \frac{dx}{dt}\ne0

Differentiate $x(t)$ and $y(t)$ separately with respect to $t$ .\n2. Divide $dy/dt$ by $dx/dt$ .\n3. Substitute the required parameter value.\n4. Use the same value in $x(t),y(t)$ to find the point.\n5. Combine point and slope if a tangent equation is needed.

Example: $x=t^2+1$ and $y=t^3-3t$ . Then $dx/dt=2t$ and $dy/dt=3t^2-3$ , so $dy/dx=(3t^2-3)/(2t)$ . At $t=2$ , the curve is at $(5,2)$ and the slope is $9/4$ . The tangent line is therefore $y-2=\frac94(x-5)$ .

Do not divide the coordinate functions themselves: $dy/dx$ is the ratio of their derivatives. The quotient formula requires $dx/dt\ne0$ at the parameter value; when $dx/dt=0$ , this calculation does not produce a finite tangent slope.

Next Topic

—9.2 Second Derivatives of Parametric EquationsRead