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Flashcards · AP® Calculus BC · Unit 9

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

62 cards on topics 9.1 to 9.9 (BC only). Pick a topic or study the whole unit.

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9.1 Defining and Differentiating Parametric Equations

How do you find dy/dx for a parametric curve x = f(t), y = g(t)?
dy/dx = (dy/dt) ÷ (dx/dt), valid where dx/dt ≠ 0. It comes from the chain rule dy/dt = (dy/dx)(dx/dt), and is usually a function of t.
For x = 2t − 1, y = t³, find dy/dx at t = 2.
dy/dt = 3t² and dx/dt = 2, so dy/dx = 3t²/2 = 6 at t = 2.
Where does a parametric curve have horizontal or vertical tangents?
Horizontal: dy/dt = 0 and dx/dt ≠ 0. Vertical: dx/dt = 0 and dy/dt ≠ 0. If both are 0, there is no conclusion from the formula.
For x = t² − 6t, y = 2t + 1, where is the tangent vertical?
dx/dt = 2t − 6 = 0 at t = 3, and dy/dt = 2 ≠ 0. The point is (9 − 18, 7) = (−9, 7).
For x = cos t, y = sin t, find the slope at t = π/6.
dy/dx = cos t ÷ (−sin t) = (√3/2) ÷ (−1/2) = −√3.
How do you find the slope at a point (x₀, y₀) on a parametric curve?
Solve x(t) = x₀ and y(t) = y₀ for t first, then substitute t. A curve that crosses itself can give two values of t, each with its own slope.
How do you write the tangent line to a parametric curve at t = a?
y − y(a) = m(x − x(a)), with m = dy/dx at t = a. Take the point and the slope from the same t.

9.2 Second Derivatives of Parametric Equations

How do you find d²y/dx² for a parametric curve?
d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt). Find dy/dx, simplify it, differentiate with respect to t, then divide by dx/dt again.
Why is (d²y/dt²) ÷ (d²x/dt²) not the second derivative?
It is a false shortcut. For x = 2t, y = t² it gives 2 ÷ 0, but dy/dx = t, d/dt (t) = 1, and 1 ÷ 2 = ½ is correct.
For x = t², y = 4t³, find d²y/dx² at t = 1.
dy/dx = 12t²/(2t) = 6t. d/dt (6t) = 6, and 6 ÷ (2t) = 3/t, which is 3 at t = 1.
For x = cos t, y = sin t, is the curve concave up or down at t = π/2?
dy/dx = −cot t, d/dt = csc²t, and csc²t ÷ (−sin t) = −1/sin³t = −1. Concave down: the top of the circle.
For x = 2t + 1, y = t² + t, find d²y/dx².
dy/dx = (2t + 1)/2, d/dt = 1, and 1 ÷ 2 = ½.
What does the sign of d²y/dx² mean for a tangent-line estimate on a parametric curve?
Positive: concave up, tangent below the curve, so the estimate is too small. Negative: concave down, so the estimate is too big.

9.3 Finding Arc Lengths of Curves Given by Parametric Equations

What is the arc length of a parametric curve for a ≤ t ≤ b?
L = ∫ (a to b) √((dx/dt)² + (dy/dt)²) dt, for a smooth curve traced once. The limits are values of t.
Where does the parametric arc length formula come from?
A tiny piece has legs Δx ≈ x′(t)Δt and Δy ≈ y′(t)Δt, so by Pythagoras Δs ≈ √(x′² + y′²)Δt. Adding the pieces gives the integral.
Find the length of x = 3 cos t, y = 3 sin t for 0 ≤ t ≤ π/2.
(dx/dt)² + (dy/dt)² = 9 sin²t + 9 cos²t = 9, so the integrand is 3 and L = 3π/2 ≈ 4.712.
Write the length integral for x = t³, y = e^(2t), 0 ≤ t ≤ 1.
∫ (0 to 1) √(9t⁴ + 4e^(4t)) dt, evaluated with a calculator.
For x = ½t², y = ⅓(2t + 1)^(3/2), what is the arc length integrand?
dx/dt = t and dy/dt = (2t + 1)^(1/2), so the sum of squares is t² + 2t + 1 = (t + 1)² and the integrand is t + 1.
What goes wrong if a curve is traced twice on the interval?
The integral counts every pass, so it gives twice the length. Make sure the point does not go back over the curve.
What mistakes are common with parametric arc length?
No square root or no squaring; x- or y-limits instead of t; lost chain factors (d/dt sin 3t = 3 cos 3t); splitting the root; degree mode.

9.4 Defining and Differentiating Vector-Valued Functions

How do you differentiate a vector-valued function r(t) = ⟨x(t), y(t)⟩?
Component by component: r′(t) = ⟨x′(t), y′(t)⟩ and r″(t) = ⟨x″(t), y″(t)⟩. If r is position, r′ is velocity and r″ is acceleration.
Find r′(t) for r(t) = ⟨teᵗ, cos(3t)⟩.
⟨eᵗ + teᵗ, −3 sin(3t)⟩, using the product rule and the chain rule inside the components.
How do you find the tangent slope of a vector-valued function?
dy/dx = y′(t)/x′(t) with x′(t) ≠ 0. Horizontal tangent: r′ = ⟨nonzero, 0⟩. Vertical tangent: r′ = ⟨0, nonzero⟩.
r(t) = ⟨t² − 4t, t³⟩. Which way is the point moving at t = 1?
r′(1) = ⟨2 − 4, 3⟩ = ⟨−2, 3⟩: to the left and up.
For r(t) = ⟨2t, t³⟩, find r″(1) and d²y/dx² at t = 1.
r″(1) = ⟨0, 6⟩. dy/dx = 3t²/2, d/dt = 3t, divided by 2 gives d²y/dx² = 3t/2 = 1.5. They are different things: a vector and a number.
What is the domain of a vector-valued function?
The values of t where both components are defined, not just one.
What does r′(t) = ⟨0, 0⟩ tell you about the tangent?
Nothing on its own. It is not a vertical tangent; you must investigate further.

9.5 Integrating Vector-Valued Functions

How do you integrate a vector-valued function?
Component by component: ∫ ⟨f(t), g(t)⟩ dt = ⟨F(t) + C₁, G(t) + C₂⟩. The two constants are usually different.
Evaluate ∫ (0 to π/2) ⟨cos t, 2t⟩ dt.
⟨[sin t] (0 to π/2), [t²] (0 to π/2)⟩ = ⟨1, π²/4⟩. A definite integral of a vector is a vector.
r′(t) = ⟨4t, 3⟩ and r(0) = ⟨1, −2⟩. Find r(2).
r(2) = r(0) + ∫ (0 to 2) ⟨4t, 3⟩ dt = ⟨1, −2⟩ + ⟨8, 6⟩ = ⟨9, 4⟩.
How do you find position from velocity for a vector function?
r(b) = r(a) + ∫ (a to b) v(t) dt. The integral alone is only the net change r(b) − r(a).
r′(t) = ⟨1/t, 2t⟩ for t > 0 and r(1) = ⟨0, 3⟩. Find r(e).
r(t) = ⟨ln t + C₁, t² + C₂⟩; r(1) = ⟨0, 3⟩ gives C₁ = 0, C₂ = 2. So r(e) = ⟨1, e² + 2⟩.
How do you work up from acceleration to position?
v(b) = v(a) + ∫ (a to b) a(t) dt, then integrate v for position. You need both the initial velocity and the initial position.
What mistakes are common with vector integrals?
One shared constant for both components; using t = 0 when the condition is at another time; calling the net change the position; adding the components into one number; missing chain factors (∫ sin 3t dt = −⅓ cos 3t).

9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

What are velocity, acceleration and speed for a particle with position ⟨x(t), y(t)⟩?
v(t) = ⟨x′(t), y′(t)⟩; a(t) = ⟨x″(t), y″(t)⟩; speed = √((x′)² + (y′)²), a non-negative number.
r(t) = ⟨t², 3t − t³⟩. Find the speed at t = 2.
v(t) = ⟨2t, 3 − 3t²⟩, so v(2) = ⟨4, −9⟩ and the speed is √(16 + 81) = √97 ≈ 9.849.
How do you find total distance travelled by a particle in the plane?
∫ (a to b) √((x′)² + (y′)²) dt, the integral of speed. Displacement is ∫ v(t) dt = r(b) − r(a), a vector.
v(t) = ⟨cos t, −sin t⟩ for 0 ≤ t ≤ π. Find the displacement and the distance.
Displacement ∫ (0 to π) v dt = ⟨0, −2⟩. Speed is 1, so the distance is π.
When is a particle moving in the plane at rest?
Only when x′(t) = 0 and y′(t) = 0 at the same time. One zero component is not enough.
How do you tell whether the speed of a particle in the plane is increasing?
Check the sign of x′x″ + y′y″ (velocity dotted with acceleration): positive means speed is increasing. The size of a(t) does not decide it.
A particle has constant velocity ⟨4, 3⟩ for 0 ≤ t ≤ 2. How far does it travel?
Speed √(16 + 9) = 5, so the distance is 5 × 2 = 10.

9.7 Defining Polar Coordinates and Differentiating in Polar Form

How do you convert between polar and rectangular coordinates?
x = r cos θ, y = r sin θ, r² = x² + y². Check the quadrant when finding θ. (6, 7π/6) is (−3√3, −3).
How do you find dy/dx for a polar curve r = f(θ)?
Treat θ as the parameter: x = f(θ) cos θ, y = f(θ) sin θ. Then dx/dθ = f′ cos θ − f sin θ, dy/dθ = f′ sin θ + f cos θ, and dy/dx = (dy/dθ) ÷ (dx/dθ).
For r = 1 + cos θ, find dy/dx at θ = π/2.
r = 1, r′ = −1. dx/dθ = (−1)(0) − (1)(1) = −1 and dy/dθ = (−1)(1) + (1)(0) = −1, so dy/dx = 1.
For r = 5 sin θ, find dy/dx at θ = π/6.
r = 5/2, r′ = 5√3/2. dx/dθ = 15/4 − 5/4 = 5/2 and dy/dθ = 5√3/4 + 5√3/4 = 5√3/2, so dy/dx = √3.
Is dr/dθ the slope of a polar curve? Is tan θ?
Neither. dr/dθ is how fast r changes; tan θ is the slope of the ray to the point. The tangent slope is (dy/dθ) ÷ (dx/dθ).
When is a polar point moving away from the pole?
When r and dr/dθ have the same sign (|r| increasing). If r = −1 and dr/dθ = 2, the signs are opposite, so it is moving towards the pole.
Which points name the same location in polar coordinates?
(r, θ), (r, θ + 2π) and (−r, θ + π) are all the same point. Negative r means go backwards along the ray.

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

What is the area of a polar region bounded by r = f(θ) and the rays θ = α and θ = β?
A = ½ ∫ (α to β) r² dθ, from thin sectors of area ½r²Δθ, provided the region is swept once from α to β.
Find the area bounded by r = 5 and the rays θ = 0 and θ = π/2.
½ ∫ (0 to π/2) 25 dθ = 25π/4, a quarter disc.
Find the area of one petal of r = sin 4θ.
The petal lies between consecutive zeros θ = 0 and θ = π/4: ½ ∫ (0 to π/4) sin²4θ dθ = ½ × ½ × π/4 = π/16.
Find the area swept out by r = eᶿ for 0 ≤ θ ≤ 1.
½ ∫ (0 to 1) e^(2θ) dθ = (e² − 1)/4 ≈ 1.597.
Why does ½ ∫ (0 to 2π) (6 cos θ)² dθ not give the area of the circle r = 6 cos θ?
The circle is traced once on [0, π], giving 9π. On [0, 2π] it is traced twice, so the integral gives 18π, double the area. Sketch to choose limits.
Which identities do you need for polar areas by hand?
sin²kθ = ½(1 − cos 2kθ) and cos²kθ = ½(1 + cos 2kθ).
What mistakes are common with polar area?
Forgetting the ½ or the square; losing the middle term 2ab cos θ in (a + b cos θ)²; using 0 to 2π by habit; wrong zeros for a petal; degree mode.

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

What is the area between two polar curves?
½ ∫ (α to β) (R² − r²) dθ, with R the outer curve and r the inner curve, R ≥ r ≥ 0 on [α, β]. Each slice is a big sector minus a small sector.
Why is ½ ∫ (R − r)² dθ wrong?
Square each radius first, then subtract: ½ ∫ (R² − r²) dθ.
Find the area between r = 4 and r = 2.
½ ∫ (0 to 2π) (16 − 4) dθ = 12π.
Where do r = 2 + 2 sin θ and r = 3 meet?
2 + 2 sin θ = 3 gives sin θ = ½, so θ = π/6 and θ = 5π/6.
How do you find the area of the region inside both of two polar curves?
No subtraction: split at each intersection angle and use ½ ∫ (nearer curve)² dθ on each piece. Using the farther curve is a common mistake.
Find the area inside r = 2 + 2 cos θ and outside r = 3.
They meet where cos θ = ½. Area = ½ ∫ (−π/3 to π/3) [(2 + 2 cos θ)² − 9] dθ = 9√3/2 − π.
Find the area between r = 2θ and r = θ for 0 ≤ θ ≤ π.
½ ∫ (0 to π) (4θ² − θ²) dθ = ½ ∫ (0 to π) 3θ² dθ = π³/2 ≈ 15.503.

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