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AP Calculus AB/BC · Mathematics · Lesson 9.1

9.1 Differentiating parametric equations (BC)

Level
Advanced Placement (AP)
Reading time
8 min reading
Errata

Abstract

When a curve is traced by x(t)x(t) and y(t)y(t), the slope is still rise over run — but both rise and run are measured against the parameter tt.

Concept preview

What you will be able to do

  1. 01Describe a curve given parametrically by x=x(t)x=x(t) and y=y(t)y=y(t), and explain why such a curve need not be the graph of a single function y=f(x)y=f(x).
  2. 02Compute the slope of a parametric curve with the chain-rule formula dydx=dy/dtdx/dt\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt} — dividing the vertical rate by the horizontal rate, not multiplying them.
  3. 03Locate horizontal tangents where dydt=0\dfrac{dy}{dt}=0 (and dxdt≠0\dfrac{dx}{dt}\neq 0) and vertical tangents where dxdt=0\dfrac{dx}{dt}=0 (and dydt≠0\dfrac{dy}{dt}\neq 0), and evaluate dydx\dfrac{dy}{dx} at a chosen value of tt.

A parametric curve is a pair of equations x=x(t)x=x(t) and y=y(t)y=y(t) sharing one parameter tt. Think of tt as time and the pair as the recipe for a point that moves through the plane. For example, x=3cos⁡tx=3\cos t and y=2sin⁡ty=2\sin t with tt running over [0, 2π][0,\,2\pi] traces an ellipse: at t=0t=0 the point sits at (3, 0)(3,\,0), at t=π2t=\tfrac{\pi}{2} it is at (0, 2)(0,\,2), at t=πt=\pi it is at (−3, 0)(-3,\,0), and it returns to the start at t=2πt=2\pi. No single y=f(x)y=f(x) describes this whole loop — for most xx there are two yy values — which is exactly why we need the parameter.

All-Access Plan

9.1 Differentiating parametric equations (BC)

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