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Conics, Parametric Equations, and Polar Coordinates

10.3 Parametric Equations and Calculus

Focus on parametric derivatives, parametric area, tangent lines and second derivatives.

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Ideas in this lesson

01Parametric derivativesWorked steps · five questions02Parametric areaWorked steps · five questions03Tangent lines and second derivativesWorked steps · five questions
Idea 1 of 3

Parametric derivatives

Use dydx=y′tx′(t)\frac{dy}{dx} = y'\frac{t}{x'}(t) when x′(t)≠0x'(t) \ne 0. A second derivative requires dividing by x′x' again.

Worked example 1

Let x=2t,y=t3x = 2t, y = t^{3}. At t=1t = 1, find dydx\dfrac{dy}{dx}.
  1. dydx=3t22\dfrac{dy}{dx} = \dfrac{3t^{2}}{2}
  2. d2ydx2=12ddt(3t22)=6t4\dfrac{d^{2}y}{dx^{2}} = \dfrac{1}{2}\dfrac{d}{dt}(\dfrac{3t^{2}}{2}) = \dfrac{6t}{4}
  3. dydx\dfrac{dy}{dx} at t=1=32≈1.5t = 1 = \dfrac{3}{2} \approx 1.5

Answer: 32≈1.5\dfrac{3}{2} \approx 1.5

Worked example 2

Let x=3t,y=t3x = 3t, y = t^{3}. At t=2t = 2, find d2ydx2d^{2}\dfrac{y}{dx^{2}}.
  1. dydx=3t23\dfrac{dy}{dx} = \dfrac{3t^{2}}{3}
  2. d2ydx2=13ddt(3t23)=6t9\dfrac{d^{2}y}{dx^{2}} = \dfrac{1}{3}\dfrac{d}{dt}(\dfrac{3t^{2}}{3}) = \dfrac{6t}{9}
  3. d2ydx2d^{2}\dfrac{y}{dx^{2}} at t=2=43≈1.33t = 2 = \dfrac{4}{3} \approx 1.33

Answer: 43≈1.33\dfrac{4}{3} \approx 1.33

Idea 2 of 3

Parametric area

Integrate y(t)x′(t)dty(t)x'(t) dt for signed area, checking orientation and the traced interval.

Worked example 1

For x=2t,y=t2x = 2t, y = t^{2} with 0≤t≤10 \le t \le 1, find the area below the curve.
  1. x′(t)=2>0x'(t) = 2 > 0; y(t)=t2y(t) = t^{2}
  2. A=∫012t2 dtA = \int_{0}^{1} 2t^{2}\,dt
  3. A=2(1)33=23≈0.67A = \dfrac{2(1)^{3}}{3} = \dfrac{2}{3} \approx 0.67

Answer: 23≈0.67\dfrac{2}{3} \approx 0.67

Worked example 2

For x=3t,y=t2x = 3t, y = t^{2} with 0≤t≤20 \le t \le 2, find the area below the curve.
  1. x′(t)=3>0x'(t) = 3 > 0; y(t)=t2y(t) = t^{2}
  2. A=∫023t2 dtA = \int_{0}^{2} 3t^{2}\,dt
  3. A=3(2)33=8A = \dfrac{3(2)^{3}}{3} = 8

Answer: 88

Idea 3 of 3

Tangent lines and second derivatives

Find the point and slope at the same parameter value. The tangent’ss intercept follows from y−mxy - mx.

Worked example 1

Let x=2t,y=t3x = 2t, y = t^{3}. At t=1t = 1, find dydx\dfrac{dy}{dx}.
  1. dydx=3t22\dfrac{dy}{dx} = \dfrac{3t^{2}}{2}
  2. d2ydx2=12ddt(3t22)=6t4\dfrac{d^{2}y}{dx^{2}} = \dfrac{1}{2}\dfrac{d}{dt}(\dfrac{3t^{2}}{2}) = \dfrac{6t}{4}
  3. dydx\dfrac{dy}{dx} at t=1=32≈1.5t = 1 = \dfrac{3}{2} \approx 1.5

Answer: 32≈1.5\dfrac{3}{2} \approx 1.5

Worked example 2

Let x=3t,y=t3x = 3t, y = t^{3}. At t=2t = 2, find d2ydx2d^{2}\dfrac{y}{dx^{2}}.
  1. dydx=3t23\dfrac{dy}{dx} = \dfrac{3t^{2}}{3}
  2. d2ydx2=13ddt(3t23)=6t9\dfrac{d^{2}y}{dx^{2}} = \dfrac{1}{3}\dfrac{d}{dt}(\dfrac{3t^{2}}{3}) = \dfrac{6t}{9}
  3. d2ydx2d^{2}\dfrac{y}{dx^{2}} at t=2=43≈1.33t = 2 = \dfrac{4}{3} \approx 1.33

Answer: 43≈1.33\dfrac{4}{3} \approx 1.33

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §10.3,p10.3, p. 706706. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.