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Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Syllabus
2020
Section
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Level
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Exam analysis

9.1 Defining and Differentiating Parametric Equations

Syllabus
2020
Topic
9.1
Level
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Differentiate a Curve Through Its Shared Parameter

Parametric equations x=x(t)x=x(t) and y=y(t)y=y(t) assign both coordinates to the same parameter value. As tt changes, the point (x(t),y(t))(x(t),y(t)) traces a curve. Because both coordinates depend on tt, 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

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

Example: x=t2+1x=t^2+1 and y=t3−3ty=t^3-3t. Then dx/dt=2tdx/dt=2t and dy/dt=3t2−3dy/dt=3t^2-3, so dy/dx=(3t2−3)/(2t)dy/dx=(3t^2-3)/(2t). At t=2t=2, the curve is at (5,2)(5,2) and the slope is 9/49/4. The tangent line is therefore y−2=94(x−5)y-2=\frac94(x-5).

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

9.2 Second Derivatives of Parametric Equations

Syllabus
2020
Topic
9.2
Level
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Differentiate Parametric Slope with Respect to x

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

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

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

Example: x=t2+1x=t^2+1 and y=t3−3ty=t^3-3t. Then dy/dx=(3t2−3)/(2t)=32(t−1/t)dy/dx=(3t^2-3)/(2t)=\frac32(t-1/t). Hence d(dy/dx)/dt=32(1+1/t2)d(dy/dx)/dt=\frac32(1+1/t^2), and d2y/dx2=3(t2+1)4t3d^2y/dx^2=\frac{3(t^2+1)}{4t^3}. At t=2t=2, d2y/dx2=15/32>0d^2y/dx^2=15/32>0, so the curve is concave up there.

Stopping after d(dy/dx)/dtd(dy/dx)/dt leaves a rate with respect to tt; it is not d2y/dx2d^2y/dx^2. The conversion formula requires dx/dt≠0dx/dt\ne0, and any parameter values excluded while simplifying must remain excluded from the final expression.

9.3 Finding Arc Lengths of Curves Given by Parametric Equations

Syllabus
2020
Topic
9.3
Level
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Integrate Parametric Speed to Find Curve Length

For x=x(t)x=x(t) and y=y(t)y=y(t), a tiny parameter change produces horizontal change (dx/dt)dt(dx/dt)dt and vertical change (dy/dt)dt(dy/dt)dt. The Pythagorean magnitude of these components is the nonnegative rate at which length is traced.

L=\int_\alpha^\beta\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt

  1. Differentiate x(t)x(t) and y(t)y(t).\n2. Square and add the two component derivatives.\n3. Take the nonnegative square root.\n4. Integrate over the stated tt-interval.\n5. Check the interval describes the intended part of the curve.

Example: x=3cos⁡tx=3\cos t and y=3sin⁡ty=3\sin t for 0≤t≤π/20\le t\le\pi/2 trace one quarter of a circle of radius 3. Here dx/dt=−3sin⁡tdx/dt=-3\sin t and dy/dt=3cos⁡tdy/dt=3\cos t, so the speed is 9sin⁡2t+9cos⁡2t=3\sqrt{9\sin^2t+9\cos^2t}=3. Thus L=∫0π/23 dt=3π/2L=\int_0^{\pi/2}3\,dt=3\pi/2 units.

Use the parameter bounds, not automatically the curve's xx-coordinates. The square root is the magnitude of both component rates, so neither sign can make length negative. If the parameterization retraces a segment, this integral counts the retraced distance again.

9.4 Defining and Differentiating Vector-Valued Functions

Syllabus
2020
Topic
9.4
Level
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Differentiate a Vector One Component at a Time

A planar vector-valued function r(t)=⟨x(t),y(t)⟩\mathbf r(t)=\langle x(t),y(t)\rangle gives both coordinates of a moving point at the same parameter value. Its derivative records the instantaneous horizontal and vertical rates together, so it points tangent to the path when it is nonzero.

\mathbf r'(t)=\left\langle x'(t),y'(t)\right\rangle

  1. Keep the component order fixed.\n2. Differentiate each component with respect to the same parameter.\n3. Reassemble the derivatives inside vector brackets.\n4. Substitute a parameter value componentwise when an instantaneous vector is required.

Example: r(t)=⟨t2−1,t3+2t⟩\mathbf r(t)=\langle t^2-1,t^3+2t\rangle. Then r′(t)=⟨2t,3t2+2⟩\mathbf r'(t)=\langle2t,3t^2+2\rangle. At t=1t=1, the point is r(1)=⟨0,3⟩\mathbf r(1)=\langle0,3\rangle and the derivative is r′(1)=⟨2,5⟩\mathbf r'(1)=\langle2,5\rangle, a tangent direction and, for a position function, the instantaneous velocity vector.

The derivative is a vector, not the sum or magnitude of the component derivatives. Its magnitude ∥r′(t)∥=[x′(t)]2+[y′(t)]2\|\mathbf r'(t)\|=\sqrt{[x'(t)]^2+[y'(t)]^2} is a separate scalar quantity; for motion, that magnitude is speed.

9.5 Integrating Vector-Valued Functions

Syllabus
2020
Topic
9.5
Level
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Integrate Each Rate Component, Then Anchor the Path

If r′(t)=v(t)=⟨vx(t),vy(t)⟩\mathbf r'(t)=\mathbf v(t)=\langle v_x(t),v_y(t)\rangle, then the position vector is found by integrating the horizontal and vertical rates separately. An initial position fixes both constants and selects the one particular solution that describes the motion.

\mathbf r(t)=\mathbf r(t_0)+\int_{t_0}^{t}\mathbf v(u),du=\mathbf r(t_0)+\left\langle\int_{t_0}^{t}v_x(u),du,\int_{t_0}^{t}v_y(u),du\right\rangle

  1. Identify the given rate vector and the time t0t_0 of the initial position.
  2. Integrate each component over the same interval from t0t_0 to tt.
  3. Add the corresponding components of r(t0)\mathbf r(t_0).
  4. Differentiate the result and substitute t=t0t=t_0 to check both conditions.

Example: suppose v(t)=⟨2t,3⟩\mathbf v(t)=\langle2t,3\rangle and r(1)=⟨4,−2⟩\mathbf r(1)=\langle4,-2\rangle. Then
r(t)=⟨4,−2⟩+⟨∫1t2u du,∫1t3 du⟩=⟨4,−2⟩+⟨t2−1,3t−3⟩=⟨t2+3,3t−5⟩.\mathbf r(t)=\langle4,-2\rangle+\left\langle\int_1^t2u\,du,\int_1^t3\,du\right\rangle=\langle4,-2\rangle+\langle t^2-1,3t-3\rangle=\langle t^2+3,3t-5\rangle.
Indeed, r′(t)=⟨2t,3⟩\mathbf r'(t)=\langle2t,3\rangle and r(1)=⟨4,−2⟩\mathbf r(1)=\langle4,-2\rangle, so both the rate vector and initial condition are satisfied.

Do not integrate the speed ∥v(t)∥\|\mathbf v(t)\| when the question asks for position: speed is scalar and its integral gives distance traveled. Integrating the velocity vector gives vector displacement. With indefinite integrals, remember that the two components can have different constants before the initial position determines them.

9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

Syllabus
2020
Topic
9.6
Level
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Match Each Motion Quantity to the Right Operation

For planar motion, r(t)=⟨x(t),y(t)⟩\mathbf r(t)=\langle x(t),y(t)\rangle is position. Differentiate once for velocity and twice for acceleration; take the magnitude of velocity for speed. Integrate velocity to obtain displacement, or integrate speed to obtain total distance traveled.

\mathbf v(t)=\mathbf r'(t)=\langle x'(t),y'(t)\rangle,\qquad \text{speed}=|\mathbf v(t)|=\sqrt{[x'(t)]^2+[y'(t)]^2},\qquad \mathbf a(t)=\mathbf v'(t)

\text{displacement on }[a,b]=\int_a^b\mathbf v(t),dt=\mathbf r(b)-\mathbf r(a),\qquad \text{distance}=\int_a^b|\mathbf v(t)|,dt

Example: a particle's position in meters is r(t)=⟨t2+1,t3−3t⟩\mathbf r(t)=\langle t^2+1,t^3-3t\rangle, with tt in seconds. Then v(t)=⟨2t,3t2−3⟩\mathbf v(t)=\langle2t,3t^2-3\rangle m/s and a(t)=⟨2,6t⟩\mathbf a(t)=\langle2,6t\rangle m/s2^2. At t=1t=1, its position is ⟨2,−2⟩\langle2,-2\rangle m, velocity is ⟨2,0⟩\langle2,0\rangle m/s, speed is 22+02=2\sqrt{2^2+0^2}=2 m/s, and acceleration is ⟨2,6⟩\langle2,6\rangle m/s2^2. From t=1t=1 to t=2t=2, displacement is r(2)−r(1)=⟨5,2⟩−⟨2,−2⟩=⟨3,4⟩\mathbf r(2)-\mathbf r(1)=\langle5,2\rangle-\langle2,-2\rangle=\langle3,4\rangle m.

Displacement is the net position change, so opposite motions can cancel; total distance cannot cancel because speed is nonnegative. In the example, ∥⟨3,4⟩∥=5\|\langle3,4\rangle\|=5 m is the magnitude of displacement, not automatically the distance traveled along the curved path.

9.7 Defining Polar Coordinates and Differentiating in Polar Form

Syllabus
2020
Topic
9.7
Level
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Treat a Polar Curve as a Parametric Curve

For a polar curve r=f(θ)r=f(\theta), write x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta. Both coordinates depend on θ\theta, so differentiate each with the product rule and then use the parametric derivative rule.

\frac{dx}{d\theta}=\frac{dr}{d\theta}\cos\theta-r\sin\theta,\qquad \frac{dy}{d\theta}=\frac{dr}{d\theta}\sin\theta+r\cos\theta

\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta},\qquad \frac{d^2y}{dx^2}=\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\bigg/\frac{dx}{d\theta}

  1. Find dr/dθdr/d\theta.
  2. Substitute rr and dr/dθdr/d\theta into the formulas for dx/dθdx/d\theta and dy/dθdy/d\theta.
  3. Divide to find dy/dxdy/dx and simplify before evaluating a requested angle.
  4. For d2y/dx2d^2y/dx^2, differentiate the entire slope with respect to θ\theta, then divide by dx/dθdx/d\theta again.

Example: let r=2cos⁡θr=2\cos\theta. Then dr/dθ=−2sin⁡θdr/d\theta=-2\sin\theta, dx/dθ=−2sin⁡(2θ)dx/d\theta=-2\sin(2\theta), and dy/dθ=2cos⁡(2θ)dy/d\theta=2\cos(2\theta). Therefore dy/dx=−cot⁡(2θ)dy/dx=-\cot(2\theta). At θ=π/4\theta=\pi/4, the slope is 00. Also,
d2ydx2=2csc⁡2(2θ)−2sin⁡(2θ)=−csc⁡3(2θ),\frac{d^2y}{dx^2}=\frac{2\csc^2(2\theta)}{-2\sin(2\theta)}=-\csc^3(2\theta),
so at θ=π/4\theta=\pi/4 the second derivative is −1-1, indicating local concave-down behavior with respect to xx.

Do not use dr/dθdr/d\theta as the Cartesian slope: rr measures radial distance, not vertical position. The quotient for dy/dxdy/dx requires dx/dθ≠0dx/d\theta\ne0; when dx/dθ=0dx/d\theta=0, analyze the component derivatives separately because the tangent may be vertical or the point may require further investigation.

9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Syllabus
2020
Topic
9.8
Level
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Accumulate Polar Area as Thin Sectors

A small change dθd\theta sweeps out a thin sector of radius r=f(θ)r=f(\theta). Because a sector with angle dθd\theta has area approximately 12r2dθ\tfrac12r^2d\theta, adding all sectors gives the area bounded by one polar curve.

A=\frac12\int_{\alpha}^{\beta}[r(\theta)]^2,d\theta

  1. Identify the curve r=f(θ)r=f(\theta) and the region inside it.
  2. Choose [α,β][\alpha,\beta] so the desired region is traced exactly once; use symmetry only when the chosen interval covers a known fraction.
  3. Square the entire expression for rr, multiply the integral by 1/21/2, and evaluate.
  4. Check that the result is nonnegative and that no lobe or sector was counted twice.

Example: the cardioid r=1+cos⁡θr=1+\cos\theta is traced once for 0≤θ≤2π0\le\theta\le2\pi. Its enclosed area is
A=12∫02π(1+cos⁡θ)2 dθ=12∫02π(1+2cos⁡θ+cos⁡2θ) dθ.A=\frac12\int_0^{2\pi}(1+\cos\theta)^2\,d\theta=\frac12\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta.
Over a full period, ∫02π1 dθ=2π\int_0^{2\pi}1\,d\theta=2\pi, ∫02π2cos⁡θ dθ=0\int_0^{2\pi}2\cos\theta\,d\theta=0, and ∫02πcos⁡2θ dθ=π\int_0^{2\pi}\cos^2\theta\,d\theta=\pi. Therefore A=12(3π)=3π2A=\tfrac12(3\pi)=\tfrac{3\pi}{2} square units.

Do not compute ∫r dθ\int r\,d\theta: polar area depends on r2r^2 and includes the factor 1/21/2. Also, bounds describe a traversal, not merely the visible width of a sketch; if the interval traces the same region twice, the integral double-counts its area.

9.9 Finding the Area of the Region Bounded by Two Polar Curves

Syllabus
2020
Topic
9.9
Level
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Subtract Inner Sectors from Outer Sectors

For a fixed angle θ\theta, the region between two polar curves runs radially from an inner radius to an outer radius. Its thin-slice area is therefore the outer sector minus the inner sector, which produces a difference of squared radii.

A=\frac12\int_{\alpha}^{\beta}\left([r_{\text{outer}}(\theta)]^2-[r_{\text{inner}}(\theta)]^2\right),d\theta

  1. Solve r1(θ)=r2(θ)r_1(\theta)=r_2(\theta) to find candidate intersection angles.
  2. Use the desired region and a test angle to decide which curve is farther from the pole.
  3. Integrate outer squared minus inner squared between consecutive relevant boundaries.
  4. Split the integral wherever the curves exchange outer and inner roles, then add the positive region areas.

Example: find the area inside r=2cos⁡θr=2\cos\theta but outside r=1r=1. Intersections satisfy 2cos⁡θ=12\cos\theta=1, giving θ=±π/3\theta=\pm\pi/3. On [−π/3,π/3][-\pi/3,\pi/3], 2cos⁡θ2\cos\theta is the outer radius. Thus
A=12∫−π/3π/3(4cos⁡2θ−1) dθ=π3+32A=\frac12\int_{-\pi/3}^{\pi/3}(4\cos^2\theta-1)\,d\theta=\frac{\pi}{3}+\frac{\sqrt3}{2}
square units. The integrand is nonnegative on this interval, consistent with the chosen outer and inner curves.

Do not subtract the radii first and then square: (router−rinner)2(r_{\text{outer}}-r_{\text{inner}})^2 is not a sector-area difference. Also, one curve may not remain outer across the whole region; an unsplit integral can create negative contributions or cancel genuine area.