M408M Learning Module Pages
Main page ## Chapter 10: Parametric Equations and Polar Coordinates## Learning module LM 10.1: Parametrized Curves:## Learning module LM 10.2: Calculus with Parametrized Curves:Slope and areaArc length and surface area Summary and simplification Higher Derivatives ## Learning module LM 10.3: Polar Coordinates:## Learning module LM 10.4: Areas and Lengths of Polar Curves:## Learning module LM 10.5: Conic Sections:## Learning module LM 10.6: Conic Sections in Polar Coordinates:## Chapter 12: Vectors and the Geometry of Space## Chapter 13: Vector Functions## Chapter 14: Partial Derivatives## Chapter 15: Multiple Integrals |
## Higher derivativesTo determine whether a parametrized curve is concave up or concave down, we need to see whether $d^2 y/dx^2$ is positive or negative. To compute these higher derivatives, we start with the formula for $dy/dx$ and apply the chain rule: $$\frac{d\hbox{(anything)}}{dt} = \frac{d\hbox{(anything)}}{dx}\frac{dx}{dt}.$$Dividing both sides by $dx/dt$ gives$$\frac{d\hbox{(anything)}}{dx} = \frac{1}{dx/dt} \frac{d\hbox{(anything)}}{dt}.$$ In particular, we have already seen that $$\frac{dy}{dx} = \frac{1}{dx/dt} \frac{dy}{dt} = \frac{dy/dt}{dx/dt}.$$ In addition, we have $$\frac{d^2y}{dx^2} = \frac{d}{dx}\frac{dy}{dx} = \frac{1}{dx/dt} \frac{d (dy/dx)}{dt}.$$ $$\frac{d^3y}{dx^3} = \frac{d}{dx}\frac{d^2y}{dx^2} = \frac{1}{dx/dt} \frac{d (d^2y/dx^2)}{dt}.$$ |