Tutorial sheets 2–7: hints, then solutions

The three tests (Thomas §11.4). A curve is symmetric about the x-axis if replacing (r, θ) by (r, −θ) or by (−r, π−θ) gives an equivalent equation; about the y-axis if (r, π−θ) or (−r, −θ) does; about the origin if (−r, θ) or (r, θ+π) does.

Replace θ by −θ: 2 sin(−θ/2) = −2 sin(θ/2) = −r. So whenever (r, θ) is on the curve, so is (−r, −θ). That is the second form of the y-axis test ⟹ symmetric about the y-axis.

Replace θ by θ + 2π (same direction, so the same point-name family): 2 sin(θ/2 + π) = −2 sin(θ/2) = −r. So (−r, θ+2π) = (−r, θ) is on the curve — the origin test passes ⟹ symmetric about the origin. (The direct substitutions (−r, θ) and (r, θ+π) both fail — 2 sin(θ/2 + π/2) = 2 cos(θ/2) ≠ ±r — which is exactly why the tests are "sufficient, not necessary": a curve can have a symmetry the direct test misses.)

Two symmetries give the third: reflect in the y-axis then through the origin and you have reflected in the x-axis ⟹ symmetric about the x-axis too. Final: all three symmetries.

Origin: replace r by −r: (−r)² = r² = sin θ, the same equation ⟹ symmetric about the origin.

y-axis: replace θ by π−θ: r² = sin(π−θ) = sin θ, same equation ⟹ symmetric about the y-axis.

x-axis: the direct test (r, −θ) gives r² = −sin θ — fails. But the second form (−r, π−θ) gives r² = sin(π−θ) = sin θ ✓ ⟹ symmetric about the x-axis (also forced by the first two). All three symmetries. Note sin θ ≥ 0 is required, so the curve only exists for 0 ≤ θ ≤ π, with r = ±√(sin θ) supplying the lower half — a lemniscate-like figure-eight through the origin.

Shape first. r = a + b sin θ with a = ½ < b = 1: a limaçon with an inner loop, symmetric about the y-axis (sin is unchanged by θ → π−θ).

Zeros. ½ + sin θ = 0 ⟹ sin θ = −½ ⟹ θ = 7π/6, 11π/6. Between them r < 0: that stretch is the inner loop.

Plotting the negative values: at θ = 3π/2, r = −½ means go ½ unit in the opposite direction — straight up — landing at (0, ½). So the inner loop sits inside the big loop, above the origin, with its tip at (0, ½); the outer loop's top is at (0, 3/2); the curve crosses the x-axis at (±½, 0) and passes through the origin twice (at 7π/6 and 11π/6), tangent to those rays. Sketch:

A = ½ ∫₀^2π (2 − cos θ)² dθ = ½ ∫₀^2π (4 − 4 cos θ + cos²θ) dθ.

Power-reduce: cos²θ = ½ + ½ cos 2θ, so the integrand is 9/2 − 4 cos θ + ½ cos 2θ. Over a full period the cos θ and cos 2θ terms integrate to 0, leaving ½ · (9/2)(2π) = 9π/2.

dr/dθ = 2 cos θ − 2 sin θ. Then r² + r′² = 4(sin θ + cos θ)² + 4(cos θ − sin θ)² = 4(1 + 2 sin θ cos θ) + 4(1 − 2 sin θ cos θ) = 8.

L = ∫₀^π/2 √8 dθ = 2√2 · π/2 = π√2. (Sanity: the curve is the circle of radius √2 centred at (1,1); this quarter-sweep of θ traces the half of it, so ½ · 2π√2 = π√2 ✓.)

r = √(2 cos²θ) = √2 |cos θ| = √2 cos θ for −π/2 ≤ θ ≤ π/2. Then r′ = −√2 sin θ and r² + r′² = 2 cos²θ + 2 sin²θ = 2.

L = ∫_−π/2^π/2 √2 dθ = √2 · π = π√2. (The curve is the circle r = √2 cos θ of diameter √2, traced once — circumference π · √2 ✓.)

2 cos θ = 2 sin θ ⟹ θ = π/4. For 0 ≤ θ ≤ π/4 the inner boundary is r = 2 sin θ; by symmetry in the line y = x the other half matches.

A = 2 · ½ ∫₀^π/4 (2 sin θ)² dθ = ∫₀^π/4 4 sin²θ dθ = ∫₀^π/4 2(1 − cos 2θ) dθ = [2θ − sin 2θ]₀^π/4 = π/2 − 1.

The lens: below the dashed ray its edge is the sin-circle, above it the cos-circle — hence the split at θ = π/4.

Crossing: 4 sin θ = 3/sin θ ⟹ sin²θ = ¾ ⟹ θ = π/3, 2π/3.

Split: for 0 ≤ θ ≤ π/3 and 2π/3 ≤ θ ≤ π the region's outer edge is the circle; for π/3 ≤ θ ≤ 2π/3 it is the line (the circle pokes above y = 3 there). By symmetry the two circle pieces are equal.

Circle pieces: 2 · ½ ∫₀^π/3 16 sin²θ dθ = ∫₀^π/3 8(1 − cos 2θ) dθ = [8θ − 4 sin 2θ]₀^π/3 = 8π/3 − 4 · (√3/2) = 8π/3 − 2√3.

Line piece: ½ ∫_π/3^2π/3 9 csc²θ dθ = (9/2)[−cot θ]_π/3^2π/3 = (9/2)(1/√3 + 1/√3) = 9/√3 = 3√3.

Total: 8π/3 − 2√3 + 3√3 = 8π/3 + √3 ≈ 10.11. (Check: whole circle is 4π ≈ 12.57; the cap above y = 3 of a radius-2 circle with height 1 has area 4π/3 − √3 ≈ 2.46; difference ≈ 10.11 ✓.)

Outer edge: the circle on the side pieces (0→π/3 and 2π/3→π), the line between them — exactly the three integrals in the solution.

Zeros: 1 + 2 sin θ = 0 ⟹ θ = −π/6, 7π/6 (equivalently 11π/6). Outer loop: −π/6 ≤ θ ≤ 7π/6. Inner loop: 7π/6 ≤ θ ≤ 11π/6.

One antiderivative serves both: ½(1 + 2 sin θ)² = ½(1 + 4 sin θ + 4 sin²θ) = ½(3 + 4 sin θ − 2 cos 2θ), using 4 sin²θ = 2 − 2 cos 2θ. So F(θ) = ½(3θ − 4 cos θ − sin 2θ).

Outer: F(7π/6) − F(−π/6). At 7π/6: ½(7π/2 + 2√3 − √3/2) = ½(7π/2 + 3√3/2). At −π/6: ½(−π/2 − 2√3 + √3/2) = ½(−π/2 − 3√3/2). Difference: ½(4π + 3√3) = 2π + 3√3/2.

Inner: F(11π/6) − F(7π/6). At 11π/6: ½(11π/2 − 2√3 + √3/2) = ½(11π/2 − 3√3/2). Difference from the value at 7π/6: ½(2π − 3√3) = π − 3√3/2.

Shaded: (2π + 3√3/2) − (π − 3√3/2) = π + 3√3 ≈ 8.34.

The zero-rays at 7π/6 and 11π/6 bound both loops — the same limits that appear in both integrals.

v = ⟨49·32(1+t)^1/2, −49·32(1−t)^1/2, 13⟩ = ⟨23√(1+t), −23√(1−t), 13⟩ (the minus on the j term is the chain rule on 1 − t).

a = ⟨13(1+t)^−1/2, 13(1−t)^−1/2, 0⟩ (the two minus signs on the j term cancel).

At t = 0: v(0) = ⟨23, −23, 13⟩, a(0) = ⟨13, 13, 0⟩. Dot product: 2/9 − 2/9 + 0 = 0. Perpendicular ⟹ angle π/2.

Point: r(π) = ⟨cos 2π, sin 2π, cos 2π⟩ = ⟨1, 0, 1⟩.

Direction: v(t) = ⟨−2 sin 2t, 2 cos 2t, −2 sin 2t⟩, so v(π) = ⟨0, 2, 0⟩.

Line: x = 1, y = 2s, z = 1 (any nonzero multiple of the direction is fine: y = s also full marks). The tangent is parallel to the y-axis.

v(t) = ⟨−1, 2t, 1/t⟩. Tangent line at t: ⟨−t − s, t² + 2ts, ln t + s/t⟩.

x: −t − s = 2 ⟹ s = −t − 2.

y: t² + 2t(−t − 2) = −5 ⟹ −t² − 4t + 5 = 0 ⟹ t² + 4t − 5 = 0 ⟹ (t + 5)(t − 1) = 0. Since ln t needs t > 0, t = 1 (and s = −3).

z check: ln 1 + (−3)/1 = −3 ✓. So t = 1. (Always verify the third equation — two equations fixed the unknowns; the third is the test that the point really lies on the line.)

i: tan(−t) = −tan t, odd ⟹ ∫ = 0.

j: ∫_−π/4^π/4 sec²t dt = [tan t]_−π/4^π/4 = 1 − (−1) = 2.

k: t cos t is odd × even = odd ⟹ 0. (Without the shortcut: by parts, [t sin t + cos t] over the symmetric interval also gives 0.)

Result: ⟨0, 2, 0⟩ = 2j.

v(t) = −t(i + j + k) + C; v(0) = C = 0.

r(t) = −t²2(i + j + k) + D; r(0) = D = 10(i + j + k).

r(t) = (10 − t²/2)(i + j + k). Check: r″ = −(i+j+k) ✓.

Direction: (3, 0, 3) − (1, −1, 2) = ⟨2, 1, 1⟩, length √(4+1+1) = √6. So v(0) = 2⟨2, 1, 1⟩/√6 = (2/√6)⟨2, 1, 1⟩. (Note the acceleration ⟨2,1,1⟩ is along the same line — that's what keeps the motion straight.)

v(t) = ⟨2, 1, 1⟩t + v(0) = (t + 2/√6)⟨2, 1, 1⟩.

r(t) = (t²/2 + 2t/√6)⟨2, 1, 1⟩ + ⟨1, −1, 2⟩, using r(0) = ⟨1, −1, 2⟩. Componentwise, with 2/√6 = √6/3:

r(t) = ⟨1 + (2√6/3)t + t², −1 + (√6/3)t + t²/2, 2 + (√6/3)t + t²/2⟩. Check: r″ = ⟨2, 1, 1⟩ ✓, |v(0)| = (2/√6)·√6 = 2 ✓.

v = ⟨12 cos t, 12 sin t, 5⟩, |v| = √(144 + 25) = 13, constant.

The start point: 12 sin t = 0, −12 cos t = −12, 5t = 0 ⟹ t = 0.

Distance 13π at speed 13 takes |Δt| = π; decreasing t ⟹ t = −π. r(−π) = ⟨12 sin(−π), −12 cos(−π), −5π⟩ = (0, 12, −5π).

v = ⟨−3, −2, 6⟩, |v| = √(9 + 4 + 36) = √49 = 7. s(t) = ∫₀^t 7 dτ = 7t. Length on [0, 1]: s(1) = 7.

v = ⟨−½ sin(t/2), ½ cos(t/2), ½⟩, |v| = √(¼ + ¼) = √2/2.

L = ∫₀^4π (√2/2) dt = 4π · √2/2 = 2π√2. (It's the unit helix ⟨cos u, sin u, u⟩ with u = t/2 — same wire, half the speed, same length as the notes' helix ✓.)

v = ⟨e^t sin t + e^t cos t, e^t cos t − e^t sin t, 0⟩ = e^t⟨sin t + cos t, cos t − sin t, 0⟩.

|v| = e^t √((sin t + cos t)² + (cos t − sin t)²) = e^t √(1 + 2 sin t cos t + 1 − 2 sin t cos t) = e^t√2.

T = ⟨sin t + cos t, cos t − sin t, 0⟩ / √2 — the e^t cancels.

dT/dt = ⟨cos t − sin t, −sin t − cos t, 0⟩ / √2, and by the same identity |dT/dt| = √2/√2 = 1.

κ = |dT/dt| / |v| = 1/(e^t√2) = e^−t/√2 — the spiral straightens as it grows.

N = ⟨cos t − sin t, −(sin t + cos t), 0⟩ / √2. Check T·N: [(sin+cos)(cos−sin) − (cos−sin)(sin+cos)]/2 = 0 ✓.

v = ⟨2/t, −1 + 1/t²⟩ = (1/t²)⟨2t, 1 − t²⟩. |v| = (1/t²)√(4t² + 1 − 2t² + t⁴) = (1/t²)√((1 + t²)²) = (1 + t²)/t².

T = ⟨2t, 1 − t²⟩ / (1 + t²). Quotient rule on each component: dT/dt = ⟨2 − 2t², −4t⟩ / (1 + t²)².

At t = 1: |v| = 2, T = ⟨1, 0⟩, dT/dt = ⟨0, −4⟩/4 = ⟨0, −1⟩, so |dT/dt| = 1, κ = 1/2, ρ = 2, N = ⟨0, −1⟩.

Centre: (0, −2) + 2⟨0, −1⟩ = (0, −4). x² + (y + 4)² = 4. (Sanity: at t = 1 the curve is at its highest point moving horizontally and bending downward — a circle below it, tangent at (0, −2), is right.)

v = ⟨1, 2t⟩, |v| = √(1 + 4t²), T = ⟨1, 2t⟩(1 + 4t²)^−1/2.

Differentiate (product rule with d/dt[(1 + 4t²)^−1/2] = −4t(1 + 4t²)^−3/2): dT/dt = [⟨0, 2⟩(1 + 4t²) − 4t⟨1, 2t⟩] (1 + 4t²)^−3/2 = ⟨−4t, 2⟩ (1 + 4t²)^−3/2.

At t = 1: |v| = √5; dT/dt = ⟨−4, 2⟩/5^3/2, magnitude √20 / (5√5) = 2√5/(5√5) = 2/5. So κ = (2/5)/√5 = 2/(5√5), ρ = 5√5/2, N = ⟨−4, 2⟩/√20 = ⟨−2, 1⟩/√5 (up and to the left — into the concave side ✓).

Centre: (1, 1) + (5√5/2)·⟨−2, 1⟩/√5 = (1, 1) + (5/2)⟨−2, 1⟩ = (−4, 7/2).

x = −4 + (5√5/2) cos s, y = 7/2 + (5√5/2) sin s, 0 ≤ s ≤ 2π. Check it passes through (1, 1): distance from centre = √(25 + 25/4) = √(125/4) = 5√5/2 ✓.

For a plane curve ⟨x(t), y(t)⟩ the definition κ = |dT/dt|/|v| simplifies (Thomas §13.4 exercises, or §13.5's |v × a|/|v|³) to

Here x′ = −a sin t, x″ = −a cos t, y′ = b cos t, y″ = −b sin t. Numerator: |(−a sin t)(−b sin t) − (b cos t)(−a cos t)| = ab(sin²t + cos²t) = ab.

(using cos² = 1 − sin²). Since a² − b² > 0, the denominator is smallest when sin²t = 0 (t = 0, π: the points (±a, 0), the ends of the major axis) and largest when sin²t = 1 (t = π/2, 3π/2: the points (0, ±b), ends of the minor axis). Hence κ_max = ab/b³ = a/b² on the major axis, κ_min = ab/a³ = b/a² on the minor axis. (The lesson's widget shows exactly this for a = 2, b = 1: κ from 2 down to ¼.)

(a) f = y/x². Domain: x ≠ 0 — the plane minus the y-axis. Range: all reals (fix x = 1, let y vary). Level curves y/x² = c ⟹ y = cx²: parabolas through the origin with the vertex removed (for c = 0, the x-axis minus the origin). Boundary: the y-axis. Open (none of the boundary belongs to it).

(b) f = xy. Domain: the whole plane. Range: all reals. Level curves xy = c: hyperbolas with the axes as asymptotes (for c = 0, the two axes themselves). Boundary: empty. The whole plane is both open and closed.

(c) f = ln(x² + y² − 1). Domain: x² + y² > 1 — the outside of the unit circle, circle excluded. Range: all reals (the argument runs over (0, ∞)). Level curves ln(x² + y² − 1) = c ⟹ x² + y² = 1 + e^c: circles centred at the origin of radius √(1 + e^c) > 1. Boundary: the unit circle. Open.

Let u = x² + y². As (x,y) → (0,0), u → 0⁺, and the function is exactly sin u / u — the one limit you memorised in the limits lesson (tool 3).

Limit = 1. (Numerically: at r = 0.01 the value is 0.99999999833….)

|y| and |z| are continuous, so their sum is, so the quotient is continuous wherever |y| + |z| ≠ 0. A sum of two non-negative numbers is zero only when both are: y = 0 and z = 0 — the entire x-axis.

Continuous at every point of space except the x-axis {(x, 0, 0)} — where h isn't even defined. Note x is unrestricted: the "hole" is a line, not a point.

Path 1, y = 0: the function becomes (x − 1)/(x − 1) = 1 for every x ≠ 1 → limit 1.

Path 2, x = 1: the function becomes (e^y − 1)/(e^y − 1 + y). Divide top and bottom by y: [(e^y−1)/y] / [(e^y−1)/y + 1]. Since (e^y − 1)/y → 1 as y → 0 (that ratio is the slope of e^y at 0, i.e. the derivative e⁰ = 1 — same idea as sin u/u), this heads to 1/(1+1) = 1/2.

Two paths, two different answers (1 vs ½) → the limit does not exist. (Numerically along x = 1: 0.5126, 0.5013, 0.5001 at y = 0.1, 0.01, 0.001.)

Line y = mx (x ≠ 0): f = 2x²(mx) / (x⁴ + m²x²) = 2mx³ / [x²(x² + m²)] = 2mx / (x² + m²).

For m ≠ 0: numerator → 0 while the denominator → m² ≠ 0, so the ratio → 0. For m = 0 (the x-axis): f = 0 identically. Vertical line x = 0: f = 0/y² = 0.

Along every straight line the limit is 0 — which is what the sheet asks; done.

Beyond the ask, but worth knowing: along the parabola y = x², f = 2x⁴/(2x⁴) = 1 at every point — so the two-dimensional limit still does not exist. "Zero along every line" is not "the limit is zero"; curves count too. This is the twin of the Module 3 lesson's x²y/(x⁴ + y²) example, doubled.

Continuous: continuity at (x₀,y₀) means, by definition, that the limit exists and equals the value there. So the limit is 3 — no computation needed.

Not continuous: nothing at all. Both failure modes happen: the limit can exist but differ from 3 — e.g. g = 0 everywhere except g(x₀,y₀) = 3, where the limit is 0 — or the limit can fail to exist entirely, as in Q10 below. The value at one point never constrains the approach.

0 ≤ |x cos(1/y)| ≤ |x| · 1 = |x|, and |x| → 0 as (x,y) → (0,0). Both fences collapse to 0, so the limit exists and is 0 — the oscillation of cos(1/y) (which has no limit of its own) is squeezed out.

One care point for full marks: the function is undefined where y = 0, so the approach is understood within the domain y ≠ 0 — say so in one line.

The inside, in polar (x = r cos θ, y = r sin θ): (r³cos³θ − r³sin³θ)/r² = r(cos³θ − sin³θ). Since |cos³θ − sin³θ| ≤ 2, the inside is trapped in [−2r, 2r] → 0 whatever θ does.

The outside: cos is continuous, so the limit passes through it: cos(inside) → cos 0.

Limit = 1. (Numerically along a fixed direction: 0.99956, 0.999996, 0.99999996 at r = 0.1, 0.01, 0.001.)

(3x² − x²y² + 3y²)/(x²+y²) = 3 − x²y²/(x²+y²).

The crumb, in polar: x²y² = r⁴ cos²θ sin²θ, and cos²θ sin²θ = (sin 2θ)²/4 ≤ 1/4. So 0 ≤ x²y²/(x²+y²) ≤ r⁴/(4r²) = r²/4 → 0.

Hence the argument of ln → 3, and since ln is continuous at 3, f → ln 3. Define f(0,0) = ln 3 ≈ 1.0986, and the extension is continuous. (Numerically: 1.09784, 1.09860, 1.098612 at r = 0.1, 0.01, 0.001.)

f(0,0,0) = 0/1 = 0. Let ρ = √(x²+y²+z²). Each coordinate is at most the distance: |x|, |y|, |z| ≤ ρ, so |x+y+z| ≤ 3ρ. The denominator is ≥ 1. Therefore |f − 0| ≤ 3ρ.

Demand 3ρ < 0.015, i.e. ρ < 0.005: δ = 0.005 (= ε/3) works — the ε–δ game from the limits lesson, with the slope-3 role played by the bound 3. (Spot check: the largest |f| sampled on the sphere ρ = 0.005 is ≈ 0.0087 < 0.015 ✓.)

(a) On y = x with x ≠ 0: xy = x² ≠ 0, so f ≡ 0 along the path → limit 0.

(b) On y = 0: xy = 0 at every point, so f ≡ 1 → limit 1.

(c) Two paths into (0,0) give different limits (0 vs 1), so the limit does not exist. Continuity at the origin requires that limit to exist and equal f(0,0) = 1; the first requirement already fails → not continuous at (0,0). (Geometry: the axes carry the value 1 into every neighbourhood of the origin while the rest of the plane carries 0.)

Both paths end at the origin, carrying different constant values — that disagreement is the whole proof.

Where the formula comes from (once). Along the curve, F(x, y(x)) = 0 for every x. Differentiate both sides with respect to x by the chain rule: F_x · 1 + F_y · dy/dx = 0, so dy/dx = −F_x/F_y wherever F_y ≠ 0.

1. F_x = e^y + y cos(xy) (the xe^y term gives e^y; sin(xy) gives cos(xy) times the inner derivative y).

2. F_y = xe^y + x cos(xy) + 1.

3. At (0, ln 2): e^{ln 2} = 2 and cos 0 = 1, so F_x = 2 + ln 2 and F_y = 0 + 0 + 1 = 1. (Check the point is on the curve: 0 + sin 0 + ln 2 − ln 2 = 0 ✓.)

4. dy/dx = −(2 + ln 2)/1 = −(2 + ln 2) ≈ −2.693.

Why the formula. Treat z = z(x, y) in F(x, y, z) = 0 and differentiate with respect to x, holding y fixed: F_x + F_z · ∂z/∂x = 0. Same for y.

1. F_x = cos(x+y) + cos(x+z), F_y = cos(x+y) + cos(y+z), F_z = cos(y+z) + cos(x+z).

2. At (π, π, π) every sum is 2π and cos 2π = 1, so F_x = F_y = F_z = 2 (and F_z ≠ 0, so the formula applies). The point is on the surface: sin 2π × 3 = 0 ✓.

3. ∂z/∂x = −2/2 = −1 and ∂z/∂y = −2/2 = −1.

1. w_x = w_y = w_z = 2(x + y + z) (power rule; the inside has slope 1 in each letter).

2. The bottom edges. Each is a one-variable derivative with respect to r, with s frozen as a constant (like a number such as 5):

 • x = r − s: the derivative of r is 1 and the derivative of the constant s is 0, so x_r = 1.

 • y = cos(r + s): this is cos of a bracket, so use the school chain rule, derivative of the outside × derivative of the inside. The outside gives cos(□) → −sin(□); the inside r + s has derivative 1 (since s is frozen). So y_r = −sin(r + s) · 1 = −sin(r + s).

 • z = sin(r + s): the same steps with sin(□) → cos(□), so z_r = cos(r + s).

 (The greyed edges down to s in the walkthrough follow the same way with r frozen: x_s = −1, y_s = −sin(r + s), z_s = cos(r + s).)

3. So ∂w/∂r = 2(x + y + z)(1 − sin(r+s) + cos(r+s)).

4. At r = 1, s = −1: r + s = 0, so x = 2, y = cos 0 = 1, z = sin 0 = 0, and ∂w/∂r = 2(3)(1 − 0 + 1) = 12.

1. Volume. V_a = bc = 6, V_b = ac = 3, V_c = ab = 2. So dV/dt = 6(1) + 3(1) + 2(−3) = 3 m³/sec (increasing).

2. Surface area. S_a = 2(b + c) = 10, S_b = 2(a + c) = 8, S_c = 2(a + b) = 6. So dS/dt = 10(1) + 8(1) + 6(−3) = 0 m²/sec (momentarily unchanging).

3. Diagonal. D_a = a/D (chain rule on the square root), similarly for b, c; D = √14. So dD/dt = (1·1 + 2·1 + 3·(−3))/√14 = −6/√14 ≈ −1.60 m/sec: the diagonals are decreasing.

(a) ∂w/∂r = f_x x_r + f_y y_r = f_x cos θ + f_y sin θ. And ∂w/∂θ = f_x(−r sin θ) + f_y(r cos θ); divide by r. ✓

(b) Write A = ∂w/∂r, B = (1/r) ∂w/∂θ, c = cos θ, s = sin θ. The equations are A = f_xc + f_ys and B = −f_xs + f_yc. Multiply the first by c, the second by s, subtract: the f_y terms cancel and Ac − Bs = f_x(c² + s²) = f_x. Multiply the first by s, the second by c, add: As + Bc = f_y. So

(c) (Ac − Bs)² + (As + Bc)² = A²(c² + s²) + B²(s² + c²) + (−2ABcs + 2ABsc) = A² + B²: the cross terms cancel and cos² + sin² = 1. With B² = (1/r²)(∂w/∂θ)² that is the claim. ✓

1. x′ = −sin t, y′ = cos t, z′ = 1.

2. Where the formula comes from: f depends on x, y, z, and each of those depends on t. That's a tree with f on top, x, y, z in the middle and t at the bottom, with three paths, so df/dt = f_x x′ + f_y y′ + f_z z′. The partials f_x = cos t, f_y = sin t, f_z = t² + t − 2 are given in the question, so copy them (f itself is never given). The derivatives x′, y′, z′ come from step 1.

 df/dt = cos t(−sin t) + sin t(cos t) + (t² + t − 2)(1) = t² + t − 2. The first two terms are −sin t cos t and +sin t cos t, the same product with opposite signs, so they cancel.

3. t² + t − 2 = (t + 2)(t − 1) = 0 gives t = 1 or t = −2. These are the only places f can have an extreme value on the curve: (cos 1, sin 1, 1) and (cos 2, −sin 2, −2) (using cos(−2) = cos 2, sin(−2) = −sin 2).

4. Bonus classification: d²f/dt² = 2t + 1 is 3 > 0 at t = 1 (a local minimum along the helix) and −3 < 0 at t = −2 (a local maximum along the helix).

1. x′ = −2√2 sin t, y′ = √2 cos t.

2. Same tree as Q6, with two paths: dT/dt = T_x x′ + T_y y′ = y · x′ + x · y′, using the given T_x = y and T_y = x. On the ellipse, y = √2 sin t and x = 2√2 cos t, so substitute them (and use √2 · 2√2 = 4):

 dT/dt = (√2 sin t)(−2√2 sin t) + (2√2 cos t)(√2 cos t) = −4 sin²t + 4 cos²t = 4(cos²t − sin²t) = 4 cos 2t (take out the common 4, then the double-angle identity cos²t − sin²t = cos 2t; worth tabbing for the open-book mid-sem).

3. cos 2t = 0. Cosine (the x-coordinate on the unit circle) is 0 at the top and bottom of the circle, π/2 and 3π/2, and again every full turn after that. Because 0 ≤ t ≤ 2π, the angle 2t runs from 0 to 4π, which is two full turns, so take every zero up to 4π: 2t = π/2, 3π/2, 5π/2, 7π/2. Divide by 2: t = π/4, 3π/4, 5π/4, 7π/4. (Solving only once around would give π/4 and 3π/4 and miss half the points.)

4. Differentiate 4 cos 2t again (chain rule: cos(□) → −sin(□), times the inside's derivative 2): d²T/dt² = 4 · (−sin 2t) · 2 = −8 sin 2t. At t = π/4 and 5π/4, 2t = π/2 and 5π/2, where sin = 1, so it is −8: negative means the graph bends down like a cap, a maximum. At 3π/4 and 7π/4, 2t = 3π/2 and 7π/2, where sin = −1, so it is +8: bends up like a cup, a minimum.

5. The points: substitute each t into x = 2√2 cos t, y = √2 sin t. At π/4, cos = sin = √2/2, so x = 2√2 · √2/2 = 2 and y = √2 · √2/2 = 1: (2, 1). At the other three angles cos and sin are still ±√2/2; only the signs change with the quadrant: 3π/4 (cos −, sin +) → (−2, 1); 5π/4 (both −) → (−2, −1); 7π/4 (cos +, sin −) → (2, −1). (a) Max temperature at (2, 1) and (−2, −1); min at (−2, 1) and (2, −1).

(b) T = xy − 2 = (2√2 cos t)(√2 sin t) − 2 = 4 sin t cos t − 2 = 2 sin 2t − 2 (using sin 2t = 2 sin t cos t, so 4 sin t cos t = 2 sin 2t; and its partials really are y and x). Since sin 2t runs from −1 to 1, T runs from −4 to 0. At the max points 2 sin 2t = 2 → max T = 0; at the min points → min T = −4.

On the dashed curves xy = 2 and xy = −2, T is constant (0 and −4). They just touch the ellipse at the four answer points — no hotter or colder level curve reaches the ellipse.

1. Quotient rule, (top′ · bottom − top · bottom′)/bottom². For g_x: top′ = 1, bottom′ = y, so g_x = [(xy + 2) − (x − y)y]/(xy + 2)² = (2 + y²)/(xy + 2)².

2. For g_y: top′ = −1, bottom′ = x, so g_y = [−(xy + 2) − (x − y)x]/(xy + 2)² = −(x² + 2)/(xy + 2)².

3. At (1, −1): xy + 2 = 1, so ∇g = (3, −3).

4. D_ug = (3, −3) · (12, 5)/13 = (36 − 15)/13 = 21/13 ≈ 1.615.

1. ∇h = (2x/(x²+y²−1), 2y/(x²+y²−1) + 1, 6). At (1, 1, 0) the log's argument is 1, so ∇h = (2, 3, 6).

2. |∇h| = √(4 + 9 + 36) = 7.

3. Increases fastest along u = (2, 3, 6)/7, at rate 7; decreases fastest along −u = (−2, −3, −6)/7, at rate −7.

1. Quotient rule: f_x = [(x + y) − (x − y)]/(x + y)² = 2y/(x + y)² and f_y = [−(x + y) − (x − y)]/(x + y)² = −2x/(x + y)². At (−½, 3/2), x + y = 1, so ∇f = (3, 1) and |∇f| = √10.

(a) u = (3, 1)/√10, D = √10. (b) u = −(3, 1)/√10, D = −√10. (c) Perpendicular to (3, 1): u = (1, −3)/√10 or (−1, 3)/√10, D = 0.

(d) 3u₁ + u₂ = −2 gives u₂ = −2 − 3u₁. Substitute into u₁² + u₂² = 1: u₁² + 4 + 12u₁ + 9u₁² = 1, i.e. 10u₁² + 12u₁ + 3 = 0, so u₁ = (−12 ± √24)/20 = (−6 ± √6)/10 and u₂ = −2 − 3u₁ = (−2 ∓ 3√6)/10: u = ((−6 − √6)/10, (−2 + 3√6)/10) or ((−6 + √6)/10, (−2 − 3√6)/10). (Two answers exist because −2 lies between −√10 and √10.)

(e) u₂ = 1 − 3u₁; substituting: 10u₁² − 6u₁ = 0, so u₁ = 0 or 3/5: u = (0, 1) or (3/5, −4/5).

Every unit direction lives on the circle. The dashed lines 3u₁ + u₂ = 1, 0, −2 are perpendicular to ∇f; where each meets the circle are the answers to (e), (c), (d). The value can never leave [−√10, √10].

1. Reason. Assuming f is differentiable at P₀, D_uf = ∇f · u for every unit u. The two given directions are not parallel, so they pin down both components of ∇f.

2. From −2j (unit (0, −1)): −b = −3, so b = 3. From i + j (unit (1, 1)/√2): (a + b)/√2 = 2√2, so a + b = 4 and a = 1. So ∇f(P₀) = (1, 3).

3. Unit direction of −i − 2j: (−1, −2)/√5. D_uf = (1·(−1) + 3·(−2))/√5 = −7/√5 ≈ −3.13.

(a) Along the x-axis f(h, 0) = 0, so [f(h, 0) − 0]/h = 0 → 0: f_x(0, 0) = 0. Likewise f(0, h) = 0 gives f_y(0, 0) = 0.

(b) f(hu₁, hu₂) = h²u₁u₂/(h²(u₁² + u₂²)) = u₁u₂ (a constant along each line). So the quotient is u₁u₂/h, which has a limit as h → 0 only if u₁u₂ = 0. D_uf(0, 0) exists only for u = (±1, 0) and (0, ±1), with value 0.

(c) The formula needs f differentiable at the point. This f is not even continuous at (0, 0): along y = x it equals x²/(2x²) = ½ ≠ 0 = f(0, 0). Using ∇f · u = (0, 0) · u anyway would wrongly predict 0 in every direction, when in fact the directional derivative doesn't exist off the axes.

1. F_x = −π sin(πx) − 2xy + ze^xz, F_y = −x² + z, F_z = xe^xz + y.

2. At (0, 1, 2): F_x = 0 − 0 + 2·1 = 2, F_y = 0 + 2 = 2, F_z = 0 + 1 = 1. So ∇F = (2, 2, 1). (The point is on the surface: 1 − 0 + 1 + 2 = 4 ✓.)

3. Plane: 2(x − 0) + 2(y − 1) + 1(z − 2) = 0, i.e. 2x + 2y + z = 4.

4. Normal line: x = 2t, y = 1 + 2t, z = 2 + t.

1. ∇F = (2x, 2y, 0) = (2√2, 2√2, 0) and ∇G = (2x, 2y, −1) = (2√2, 2√2, −1) at the point.

2. Cross product (a × b = (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁)), one component at a time:

 i: 2√2·(−1) − 0·2√2 = −2√2;  j: 0·2√2 − 2√2·(−1) = 2√2;  k: 2√2·2√2 − 2√2·2√2 = 0.

3. Direction (−2√2, 2√2, 0), or more simply (−1, 1, 0). x = √2 − t, y = √2 + t, z = 4.

4. Sanity check: the two surfaces meet where z = x² + y² = 4, the circle of radius 2 at height 4. Its tangent at (√2, √2) is horizontal and at right angles to the radius (1, 1): exactly (−1, 1, 0) ✓.

1. ∇F = (1, 2y, 1) = (1, 2, 1), ∇G = (0, 1, 0).

2. i: 2·0 − 1·1 = −1;  j: 1·0 − 1·0 = 0;  k: 1·1 − 2·0 = 1. Direction (−1, 0, 1).

3. x = ½ − t, y = 1, z = ½ + t.

4. Check: putting y = 1 into the first surface gives x + z = 1, a straight line in the plane y = 1. So the tangent line is the curve itself, and every point (½ − t, 1, ½ + t) satisfies x + z = 1 ✓.

1. Critical points. f_x = 4y − 4x³ = 0 and f_y = 4x − 4y³ = 0 give y = x³ and x = y³. Substituting: x = x⁹, i.e. x(x⁸ − 1) = 0, so x = 0, ±1: the points (0, 0), (1, 1), (−1, −1). All three lie on the edge y = x, not strictly inside, so they're picked up in step 2.

2. Edge y = x (−2 ≤ x ≤ 2): f = 4x² − 2x⁴ + 16; derivative 8x − 8x³ = 8x(1 − x)(1 + x) → x = 0, ±1, values 16, 18, 18; endpoints x = ±2: 16 − 32 + 16 = 0.

3. Edge y = −2 (−2 ≤ x ≤ 2): f = −8x − x⁴ − 16 + 16 = −8x − x⁴; derivative −8 − 4x³ = 0 → x³ = −2, x = −∛2 ≈ −1.26 (inside [−2, 2]). Value: 8∛2 − (∛2)⁴ = 8∛2 − 2∛2 = 6∛2 ≈ 7.56 (because (∛2)⁴ = (∛2)³·∛2 = 2∛2). Endpoints: x = −2 → 16 − 16 = 0; x = 2 → −16 − 16 = −32.

4. Edge x = 2 (−2 ≤ y ≤ 2): f = 8y − 16 − y⁴ + 16 = 8y − y⁴; derivative 8 − 4y³ = 0 → y = ∛2, value 8∛2 − 2∛2 = 6∛2. Endpoints: y = −2 → −32; y = 2 → 0.

5. Compare: absolute max 18 at (1, 1) and (−1, −1); absolute min −32 at (2, −2). (A brute-force grid over R agrees.)

Every candidate on the triangle with its value. The winners sit on the slanted edge (max) and at the lower-right corner (min): nothing strictly inside R competes.

1. f_x = 6x² − 18x = 6x(x − 3) → x = 0 or 3. f_y = 6y² + 6y − 12 = 6(y + 2)(y − 1) → y = 1 or −2. Critical points: (0, 1), (0, −2), (3, 1), (3, −2).

2. f_xx = 12x − 18, f_yy = 12y + 6, f_xy = 0, so D = (12x − 18)(12y + 6).

3. Local max f(0, −2) = 20; local min f(3, 1) = −34; saddle points at (0, 1) and (3, −2) (value −7 at both).

Critical points are where a line f_x = 0 (dashed vertical: x = 0, 3) crosses a line f_y = 0 (dashed horizontal: y = 1, −2): four crossings, four points.

1. f_y = 0 gives x = −2; then 2(−2) + y + 3 = 0 gives y = 1. One critical point: (−2, 1).

2. f_xx = 2, f_yy = 0, f_xy = 1, so D = 2·0 − 1² = −1 < 0.

3. Saddle point at (−2, 1), where f = 3; no local maximum or minimum. (Even though f_xx > 0: once D < 0, the sign of f_xx doesn't matter.)