Applications: Polar Curves

The pick-up pattern is a polar graph: the distance r from the pole in each direction says how strongly the microphone responds to sound from that direction. Everything asked can be read from the equation. The microphone hears nothing where r is zero. A polar graph is symmetric about the initial line when replacing θ by −θ leaves r unchanged. A condition on the output is an inequality in r, and here it becomes an inequality in cos θ, which has to be solved over one full turn rather than undone like an equation.

1. The output is zero where r = 0: 12(1 + cos θ) = 0 gives cos θ = −1, and over one full turn that happens only at θ = π. Everywhere else cos θ > −1, so r > 0.
2. (a) The microphone picks up nothing from θ = π, directly behind it. For the symmetry, replace θ by −θ: since cos(−θ) = cos θ, r(−θ) = r(θ), so the point of the pattern at angle −θ is the mirror image in the initial line of the point at angle θ.
3. For the monitor the condition is 12(1 + cos θ) ≤ 14. Multiplying both sides by 2 gives 1 + cos θ ≤ 12, and subtracting 1 gives cos θ ≤ −12.
4. Over one turn, cos θ = −12 at θ = 2π3 and at θ = 4π3. Between those two angles the cosine is smaller still, reaching −1 at θ = π, and outside them it is larger. So the condition holds for 2π3 ≤ θ ≤ 4π3.
5. (b) The monitor can stand in any direction from θ = 2π3 round to θ = 4π3, a range spanning 4π3 − 2π3 = 2π3 radians, which is 120^°. Check: at θ = 2π3, r = 12(1 − 12) = 14 exactly, and by the symmetry of part (a) the range is centered on θ = π.

answer(a) θ = π (about 3.142 radians), directly behind the microphone; the pattern is symmetric about the initial line because cos(−θ) = cos θ, so r(−θ) = r(θ); (b) 2π/3 ≤ θ ≤ 4π/3 (from about 2.094 to 4.189 radians), a range spanning 2π/3 radians, or 120°

techniqueSketching Polar Curves · Polar Coordinates

A rose curve is read from where r is greatest, which gives the petal tips, and from where r is zero, which is where each petal closes up at the pole. Between two neighboring zeros r keeps one sign and the curve draws exactly one petal, so that interval is the one to integrate over. The area of a polar region is half the integral of r squared with respect to θ. Integrating round a whole turn is the usual slip: with three petals, r is negative for half of every turn, and those stretches retrace petals already drawn.

1. The tips are where r is greatest, r = 3, which needs cos 3θ = 1. Over one turn, 3θ = 0, 2π or 4π, so θ = 0, 2π3 or 4π3: three tips, each 3 meters from the center and a third of a turn apart.
2. (a) The tips point along θ = 0, 2π3 and 4π3. The petal on the initial line closes up at the pole on each side where r = 0: cos 3θ = 0 gives 3θ = ±π2, so θ = ±π6. That petal is traced exactly once as θ runs from −π6 to π6.
3. The area of that petal is 12∫_−π/6^π/6 9cos² 3θ   dθ. Writing cos² 3θ = 12(1 + cos 6θ) turns it into 94[θ + sin 6θ6]_−π/6^π/6.
4. Since sin π = sin(−π) = 0, one petal has area 94 × π3 = 3π4 ≈ 2.356 square meters. The three petals are the same shape, so the bed covers 3 × 3π4 = 9π4 ≈ 7.069 square meters.
5. (b) Spread 0.2 meters deep, the compost needed is 9π4 × 0.2 = 9π20 ≈ 1.41 cubic meters. Check: the circle through the three tips has area 9π, and the bed is a quarter of it, which fits three narrow petals with wide gaps between them.

answer(a) the tips point along θ = 0, 2π/3 and 4π/3, each 3 meters from the center, and the petal on the initial line is traced once for −π/6 ≤ θ ≤ π/6; (b) each petal is 3π/4 ≈ 2.356 square meters and the bed 9π/4 ≈ 7.069 square meters, so 9π/20 ≈ 1.41 cubic meters of compost

techniqueThe Area of a Polar Region · Sketching Polar Curves · Integrating sin² and cos²

examsH2

On a spiral, one direction θ is met again every full turn, at θ + 2π, θ + 4π and so on, so the gap along a ray is the change in r over one extra turn. For the area, the integral of half r squared gives the area swept out by the radius as θ runs over an interval. Sweeping the second turn covers everything inside it, which is the region inside the first turn together with the strip, so the strip is the difference of two sweeps.

1. A ray from the pole in the direction θ meets the path at r = 0.5θ, and again one full turn later at r = 0.5(θ + 2π) = 0.5θ + π.
2. (a) The two crossings differ by π meters whatever the direction, so neighboring turns of the path are always π ≈ 3.14 meters apart along a ray.
3. As θ runs from 0 to 2π, the radius sweeps out everything inside the first turn. That area is 12∫₀^2π 0.25θ²   dθ = [θ³24]₀^2π = 8π³24 = π³3 square meters.
4. As θ runs from 2π to 4π, the radius sweeps out everything inside the second turn, which is the first turn's region together with the strip. That area is [θ³24]_2π^4π = 64π³ − 8π³24 = 7π³3 square meters.
5. (b) The strip is the difference, 7π³3 − π³3 = 2π³ ≈ 62.0 square meters. Check: the strip is π meters wide all the way round, and the line along its middle, r = 0.5θ + π2, is on average π meters from the pole, so it is roughly 2π × π = 2π² meters long, and π × 2π² = 2π³.

answer(a) π ≈ 3.14 meters, the same along every ray; (b) 2π³ ≈ 62.0 square meters

techniqueThe Area of a Polar Region · Sketching Polar Curves

A straight edge running east and west touches the outline where the outline's tangent runs east and west, which is where the distance north, y = r sin θ, stops increasing. So y is written as a function of θ and its derivative is set to zero. The drain is not the middle of the egg, so the point due north of the drain is not the answer. The cover is the whole region inside the outline, whose area is half the integral of r squared round a full turn.

1. A point of the outline is y = rsin θ = (3.5 + 1.5cos θ)sin θ meters north of the drain. The paving edge touches the pool where y is greatest, which is where the tangent runs east and west and dydθ = 0.
2. By the product rule, dydθ = −1.5sin² θ + (3.5 + 1.5cos θ)cos θ. Writing sin² θ = 1 − cos² θ and setting it equal to zero gives 3cos² θ + 3.5cos θ − 1.5 = 0, or 6cos² θ + 7cos θ − 3 = 0.
3. This factorizes as (3cos θ − 1)(2cos θ + 3) = 0. A cosine cannot be −1.5, so cos θ = 13, and on the north side sin θ = √1 − 19 = 2√23.
4. (a) There r = 3.5 + 1.5 × 13 = 4, so y = 4 × 2√23 = 8√23 ≈ 3.77 meters: the paving edge is laid 3.77 meters north of the drain. Check: due north of the drain, at θ = π2, the pool reaches only 3.5 meters, so the northernmost point lies a little east of north, where the egg is wider.
5. The cover's area is 12∫₀^2π(3.5 + 1.5cos θ)²   dθ = 12∫₀^2π(12.25 + 10.5cos θ + 2.25cos² θ)dθ. Round a full turn, cos θ integrates to 0 and cos² θ integrates to π.
6. (b) The area is 12(12.25 × 2π + 2.25π) = 13.375π ≈ 42.0 square meters. Check: a circle of radius 3.5 meters about the drain has area 12.25π ≈ 38.5 square meters, and the egg reaches as far beyond that circle toward the east as it falls short of it toward the west. A band farther from the drain covers more area than a band of the same width nearer to it, so the gain in the east outweighs the loss in the west.

answer(a) 8√2/3 ≈ 3.77 meters north of the drain, touching the pool where cos θ = 1/3; (b) 13.375π ≈ 42.0 square meters

techniqueThe Rate of Change of a Polar Function · Parametric Differentiation · The Area of a Polar Region

examsH2

A polar curve is a pair of parametric equations in θ: x = r cos θ and y = r sin θ, with r itself a function of θ. The gradient of the curve is dy/dθ divided by dx/dθ, each found by the product rule. The rate of change dr/dθ is part of both, but on its own it only says how fast the distance from the pole is changing. The tangent line then gives where the kart goes, and the signs of dx/dθ and dy/dθ say which way along the line it moves.

1. At θ = π4, r = 50 + 20cos π2 = 50 meters, so the kart lets go at x = y = 50 × √22 = 25√2 ≈ 35.36 meters. The rate of change of r there is drdθ = −40sin 2θ = −40.
2. By the product rule, dxdθ = drdθcos θ − rsin θ = −40 − 50√2 = −90√2 and dydθ = drdθsin θ + rcos θ = −40 + 50√2 = 10√2.
3. (a) The gradient is dydx = dy/dθdx/dθ = 10−90 = −19. The karts go counterclockwise, so θ is increasing, x is falling and y is rising: the kart slides toward the upper left, almost parallel to the initial line.
4. The tangent line is y − 25√2 = −19(x − 25√2). The row of cones lies along x = 0 with y > 0, and there y = 25√2 + 25√29 = 250√29 ≈ 39.28 meters.
5. (b) The kart crosses the row of cones 39.28 meters from the post. The circuit crosses the same ray at r = 50 + 20cos π = 30 meters, so the kart is 39.28 − 30 = 9.28 meters beyond the circuit. Check: across its slide the kart moves 35.36 meters to the left, and a gradient of −19 lifts it 35.369 = 3.93 meters, from 35.36 to 39.28.

answer(a) dy/dx = −1/9 ≈ −0.111; (b) 250√2/9 ≈ 39.28 meters from the post, which is 9.28 meters beyond the circuit, since the circuit crosses that ray 30 meters from the post

techniqueThe Rate of Change of a Polar Function · Parametric Differentiation · Tangents and Normals to Curves

examsH2

Every reach is measured from the sprinkler, so the edge is wanted as a polar equation with the sprinkler at the pole. Substituting x = r cos θ and y = r sin θ into the Cartesian equation and using cos²θ + sin²θ = 1 gives it. The watered region is then swept out by the radius between the two angles of the sweep, and its area is half the integral of r squared between them.

1. Substituting x = rcos θ and y = rsin θ into x² + y² = 6x gives r²cos² θ + r²sin² θ = 6rcos θ. Since cos² θ + sin² θ = 1, this is r² = 6rcos θ.
2. (a) Dividing by r gives r = 6cos θ. Nothing is lost by the division: the curve r = 6cos θ still passes through O, at θ = π2. In the direction θ = π3 the reach needed is 6 × 12 = 3 meters.
3. The watered region is swept out by the radius from θ = −π4 to θ = π4, so its area is 12∫_−π/4^π/4 36cos² θ   dθ = 9∫_−π/4^π/4(1 + cos 2θ) dθ.
4. That is 9[θ + 12sin 2θ]_−π/4^π/4 = 9(π2 + 1), since the two sine terms are 12 and −12.
5. (b) The sprinkler waters 9π2 + 9 ≈ 23.1 square meters. Check: the two end rays meet the edge at (3, 3) and (3, −3), so the region is a triangle with a base of 6 and a height of 3, area 9, together with the half of the lawn beyond x = 3, area 9π2.

answer(a) r = 6 cos θ, and the reach needed at θ = π/3 is 3 meters; (b) 9π/2 + 9 ≈ 23.1 square meters

techniquePolar Coordinates · The Area of a Polar Region · Integrating sin² and cos²

examsH2

Putting x = r cos θ into the equation of the track gives r as a function of the angle the camera points in. Its derivative dr/dθ is the rate at which the distance changes for each radian the camera turns. The focus has to keep up in time, and the chain rule turns the rate per radian into a rate per second by multiplying by the radians turned each second.

1. On the track, x = rcos θ = 30, so the track is r = 30cos θ = 30sec θ, for −π2 < θ < π2.
2. (a) At θ = π3, cos θ = 12, so the focus distance is r = 30 × 2 = 60 meters.
3. The rate of change of r with the angle is drdθ = 30sec θtan θ. At θ = π3 this is 30 × 2 × √3 = 60√3 ≈ 103.9 meters per radian.
4. (b) By the chain rule, drdt = drdθ × dθdt = 60√3 × 0.08 = 4.8√3 ≈ 8.31 meters per second.
5. Check: along the track the sprinter is y = 30tan θ meters from its nearest point, so the sprinter's speed is 30sec² θ × 0.08 = 30 × 4 × 0.08 = 9.6 meters per second, a sprinter's pace. Only the part of that speed along the line from the camera changes r, and that part is 9.6sin π3 = 4.8√3, the same.

answer(a) r = 30 sec θ, and the focus distance at θ = π/3 is 60 meters; (b) 4.8√3 ≈ 8.31 meters per second

techniqueThe Rate of Change of a Polar Function · Polar Coordinates · Related Rates

examsO-Level · H2

The old edge is the circle r = 3 and the new edge is the limaçon r = 3 + 2cos θ, both about the same pole. Where their values of r are equal they cross, and between two crossings one edge lies outside the other. The area between them there is half the integral of the outer r squared less the inner r squared, which is not the same as the square of the gap between them.

1. The edges meet where 3 + 2cos θ = 3, that is where cos θ = 0: at θ = π2 and θ = 3π2, straight out to either side of the line to the town.
2. For −π2 < θ < π2 the cosine is positive and the new edge lies outside the old one. The area between them is 12∫_−π/2^π/2[(3 + 2cos θ)² − 3²]dθ = 12∫_−π/2^π/2(12cos θ + 4cos² θ)dθ.
3. (a) Over that half turn cos θ integrates to 2 and cos² θ integrates to π2, so the area newly reached is 12(24 + 2π) = 12 + π ≈ 15.14 square kilometers.
4. For π2 < θ < 3π2 the cosine is negative and the old circle lies outside. The area between them is 12∫_π/2^3π/2(−12cos θ − 4cos² θ)dθ, and over this half turn cos θ integrates to −2 and cos² θ to π2.
5. (b) The area no longer reached is 12(24 − 2π) = 12 − π ≈ 8.86 square kilometers. Check: the whole new coverage is 12∫₀^2π(3 + 2cos θ)²   dθ = 11π and the old circle is 9π, and the area gained less the area lost, (12 + π) − (12 − π) = 2π, is exactly their difference.

answer(a) the edges meet at θ = π/2 and θ = 3π/2, and the new antenna newly reaches 12 + π ≈ 15.14 square kilometers; (b) 12 − π ≈ 8.86 square kilometers

techniqueThe Area of a Polar Region · Area Between Two Curves · Sketching Polar Curves

examsH2

Converting to Cartesian form uses x = r cos θ and r² = x² + y². The equation is first arranged so that r stands alone on one side, because then a single squaring removes the root that r stands for. For part (b), the rim is given by its polar coordinates, and x = r cos θ, y = r sin θ turn it into a point whose coordinates give the width and the depth directly.

1. Multiplying out, r − rcos θ = 1.2. Since rcos θ = x, this is r = x + 1.2, and since r² = x² + y², squaring both sides gives x² + y² = x² + 2.4x + 1.44.
2. (a) The x² terms cancel, leaving y² = 2.4x + 1.44 = 2.4(x + 0.6), a parabola whose axis is the initial line. Its vertex, the center of the dish, is where y = 0: x = −0.6, so the center of the dish is 0.6 meters behind the receiver. The polar equation agrees: at θ = π, r = 1.22 = 0.6.
3. At the rim, θ = 2π3 and cos θ = −12, so r = 1.21.5 = 0.8 meters. In Cartesian coordinates that point is x = 0.8cos 2π3 = −0.4 and y = 0.8sin 2π3 = 0.4√3, and by symmetry the rim point at θ = 4π3 is (−0.4, −0.4√3).
4. (b) The dish is 2 × 0.4√3 = 0.8√3 ≈ 1.39 meters wide across its rim. The rim lies in the plane x = −0.4 and the center at x = −0.6, so the dish is 0.6 − 0.4 = 0.2 meters deep. Check: the rim point satisfies the Cartesian equation, since (0.4√3)² = 0.48 and 2.4(−0.4 + 0.6) = 0.48.

answer(a) y² = 2.4(x + 0.6), a parabola, and the center of the dish is 0.6 meters behind the receiver; (b) the dish is 0.8√3 ≈ 1.39 meters wide and 0.2 meters deep

techniquePolar Coordinates · Sketching Polar Curves

The second measurement fixes the constant k. After that, the arm is a polar curve like any other, and its direction at a point comes from treating x = r cos θ and y = r sin θ as parametric equations in θ. The circle about the center through the same point runs at right angles to the radius, so the pitch angle is measured from that direction, not from the radius.

1. At θ = π, 4e^kπ = 9, so e^kπ = 94 and k = 1πln 94 ≈ 0.2581.
2. (a) One full turn after the initial line, r = 4e^2kπ = 4(e^kπ)² = 4 × 8116 = 20.25 kiloparsecs. Each half turn multiplies the distance from the center by 94.
3. Where the arm crosses the initial line, θ = 0, r = 4 and drdθ = 4ke^kθ = 4k. By the product rule, dxdθ = drdθcos θ − rsin θ = 4k and dydθ = drdθsin θ + rcos θ = 4.
4. The tangent's gradient is dydx = 44k = 1k ≈ 3.874, so the tangent makes an angle of tan⁻¹1k with the initial line. The circle about the center through (4, 0) has a vertical tangent there, at π2 to the initial line.
5. (b) The pitch angle is the angle between those two directions, π2 − tan⁻¹1k = tan⁻¹ k = tan⁻¹ 0.2581 ≈ 0.253 radians, about 14.5^°. Check: at θ = π, where r = 9, the same working gives dxdθ = −9k and dydθ = −9, so the gradient is again 1k against a vertical circle tangent, and the pitch angle is the same there.

answer(a) k = (1/π) ln(9/4) ≈ 0.2581, and one full turn on the arm is 81/4 = 20.25 kiloparsecs from the center; (b) tan⁻¹ k ≈ 0.253 radians, about 14.5°

techniqueThe Rate of Change of a Polar Function · Parametric Differentiation · The Natural Logarithm

examsH2