9.2 Second Derivatives of Parametric Equations

Learning objectives

—CHA-3.G—Calculate derivatives of parametric functions—Topic 9.2• CHA-3.G Calculate derivatives of parametric functions.• CHA-3.G.3 d²y/dx² can be calculated by dividing (d/dt)(dy/dx) by dx/dt.• Enduring understanding CHA-3: Derivatives allow us to solve real-world problems involving rates of change.

For a parametric curve, $dy/dx$ is usually still written in terms of $t$ . Differentiating it with respect to $t$ gives change in slope per unit $t$ , not per unit $x$ . Divide by $dx/dt$ once more to convert that rate into the second derivative with respect to $x$ .

\frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt},\qquad \frac{dx}{dt}\ne0

Find $dy/dx=(dy/dt)/(dx/dt)$ .\n2. Simplify that slope as a function of $t$ .\n3. Differentiate the slope with respect to $t$ .\n4. Divide the result by $dx/dt$ .\n5. Substitute the requested parameter value and interpret the sign where the expression is defined.

Example: $x=t^2+1$ and $y=t^3-3t$ . Then $dy/dx=(3t^2-3)/(2t)=\frac32(t-1/t)$ . Hence $d(dy/dx)/dt=\frac32(1+1/t^2)$ , and $d^2y/dx^2=\frac{3(t^2+1)}{4t^3}$ . At $t=2$ , $d^2y/dx^2=15/32>0$ , so the curve is concave up there.

Stopping after $d(dy/dx)/dt$ leaves a rate with respect to $t$ ; it is not $d^2y/dx^2$ . The conversion formula requires $dx/dt\ne0$ , and any parameter values excluded while simplifying must remain excluded from the final expression.

Next Topic

—9.3 Finding Arc Lengths of Curves Given by Parametric EquationsRead