Calculus - Practice Set 1

(a) f'(x) = 12x² − 6x + 2 (b) g(x) = x^(1/2) + x^(−1). g'(x) = ½x^(−1/2) − x^(−2) = 1/(2√x) − 1/x² (c) Product rule: h'(x) = 2e^(2x)sin(x) + e^(2x)cos(x) = e^(2x)(2sinx + cosx)

y(3) = 9−9+2 = 2. Point: (3, 2). y' = 2x−3. Slope at x=3: m = 3. Tangent: y−2 = 3(x−3) → y = 3x − 7

(a) x³ + x² − x + C (b) eᶲ + sinx + C

(a) a(t) = 6t−6. a(2) = 12−6 = 6 m/s² (b) s = ∫₀³ (3t²−6t+2)dt = [t³−3t²+2t]₀³ = 27−27+6 = 6 m

∫(2x+1)dx = x²+x+C. [x²+x]₁³ = (9+3)−(1+1) = 12−2 = 10

Let u=x, dv=sinx dx → du=dx, v=−cosx. ∫xsinx dx = −xcosx + ∫cosx dx = −xcosx + sinx + C. [−xcosx + sinx]₀^π = (−π·(−1)+0)−(0+0) = π

Intersection: x² = 4−x² → 2x²=4 → x=±√2. Area = ∫ from −√2 to √2 of [(4−x²)−x²]dx = ∫ (4−2x²)dx = [4x − 2x³/3] evaluated = 2[4√2 − 2(2√2)/3] = 2[4√2 − 4√2/3] = 2·8√2/3 = 16√2/3

(a) Volume: x²h=500 → h=500/x². S = x² + 4xh = x² + 4x(500/x²) = x² + 2000/x (b) S'= 2x − 2000/x² = 0 → x³ = 1000 → x = 10 cm; S''= 2+4000/x³ > 0 at x=10 → minimum ✓

(a) Let u=x²+1, du=2x dx. ∫u⁴du = u⁵/5+C = (x²+1)⁵/5+C (b) Let u=3x, du=3dx. ∫cos(u)du/3 = sin(3x)/3+C (c) Let u=x²+1, du=2x dx. (1/2)∫du/u = (1/2)ln|x²+1|+C

(a) v(t) = 3t²−18t+24 = 3(t−2)(t−4) = 0 → At rest: t=2 and t=4. (b) s(0)=−10, s(2)=10, s(4)=6, s(5)=10. Distance = |10−(−10)| + |6−10| + |10−6| = 20+4+4 = 28 m

(a) x²−x−2=(x−2)(x+1). Vertical: x=2, x=−1. Horizontal: y=0. Domain: ℝ\{2,−1} (b) f'(x) = [3(x²−x−2)−(3x+1)(2x−1)]/(x²−x−2)². Numerator: 3x²−3x−6−(6x²−3x+2x−1) = −3x²−2x−5. f'(x) = (−3x²−2x−5)/(x²−x−2)² (c) Discriminant of −3x²−2x−5: 4−60 = −56 < 0. No real roots. No stationary points.

(a) Boundaries: eᶲ=1 at x=0. Area = ∫₀² (eᶲ−1)dx = [eᶲ−x]₀² = (e²−2)−1 = e²−3 (b) V = π∫₀² (eᶲ)²dx − π∫₀²1²dx = π∫₀²e^(2x)dx − 2π = π[e^(2x)/2]₀² − 2π = π(e⁴/2−1/2)−2π = π(e⁴−5)/2 (c) V = 2π∫₀² x·eᶲ dx = 2π[xeᶲ−eᶲ]₀² = 2π[(2e²−e²)−(0−1)] = 2π(e²+1)

(a) 2x + 2y·y' = 0 → y' = −x/y (b) 2xy + x²y' + 3y²y' = 0 → y'(x²+3y²) = −2xy → y' = −2xy/(x²+3y²) (c) e^(xy)·(y + xy') = 1 + y'. At (0,1): e⁰(1 + 0·y') = 1+y' → 1 = 1+y' → y' = 0

(a) V = 4πr³/3 → dV/dt = 4πr²·dr/dt → 100 = 4π(25)·dr/dt → dr/dt = 100/(100π) = 1/π cm/s (b) S = 4πr² → dS/dt = 8πr·dr/dt = 8π(5)(1/π) = 40 cm²/s (c) r = h/2. V = (1/3)πr²h = (1/3)π(h/2)²h = πh³/12 → dV/dt = πh²/4·dh/dt → 20 = π(36)/4·dh/dt → dh/dt = 80/(36π) = 20/(9π) cm/s

Separate variables: dy/y = 2x dx. Integrate: ln|y| = x² + C → y = Ae^(x²). Initial condition: y(0)=3 → A=3. y = 3e^(x²). Verify: dy/dx = 3·2x·e^(x²) = 6x·e^(x²) = 2x·(3e^(x²)) = 2xy ✓. Sketch: bell-curve shape, y(0)=3, increasing for x>0, symmetric.

(a) sin(x) = x − x³/6 + x⁵/120 − ... (b) sin(x)/x = 1 − x²/6 + x⁴/120 − ... → limit as x→0 is 1. (c) sin(x²) = x² − x⁶/6 + x¹⁰/120 − ... (d) ∫₀^(0.5) (x²−x⁶/6+x¹⁰/120) dx = [x³/3−x⁷/42+x¹¹/1320]₀^(0.5) = (0.125/3−0.0078125/42+...) = 0.04167−0.000186+... ≈ 0.0415

(a) Form 0/0. Apply L'H: (eᶲ−1)/2x. Still 0/0. Apply again: eᶲ/2 → 1/2. (b) x/eᶲ. As x→∞, form ∞/∞. Apply L'H: 1/eᶲ → 0. (c) x·lnx = lnx/(1/x). Form −∞/∞. Apply L'H: (1/x)/(−1/x²) = −x → 0.

(a) Separate: dT/(T−20) = −k dt → ln|T−20| = −kt + C → T = 20 + Ae^(−kt) (b) T(0)=90: A=70. T(5)=60: 60=20+70e^(−5k) → e^(−5k)=40/70=4/7 → k = −ln(4/7)/5 = ln(7/4)/5 ≈ 0.112 (c) T(15) = 20+70e^{(−3·ln(7/4))} = 20+70(4/7)³ = 20+70(64/343) = 20+13.1 = 33.1°C

(a) tan(x)=1 → x = π/4. (b) For 0≤x≤π/4: cosx > sinx. For π/4≤x≤π/2: sinx > cosx. (c) Area = ∫₀^(π/4) (cosx−sinx)dx + ∫_(π/4)^(π/2) (sinx−cosx)dx = [sinx+cosx]₀^(π/4) + [−cosx−sinx]_(π/4)^(π/2) = (√2−1) + (−1 − (−√2)) = (√2−1) + (√2−1) = 2(√2−1)

(a) P(t) = ∫(50e^(0.1t)−20t)dt = 500e^(0.1t) − 10t² + C. P(0)=0: 500+C=0 → C=−500. P(t) = 500e^(0.1t) − 10t² − 500 (b) dP/dt = 0: 50e^(0.1t) = 20t → solve numerically (GDC): t ≈ 14.3 years (c) P(14.3) = 500e^(1.43) − 10(204.5) − 500 ≈ 500(4.177)−2045−500 ≈ $543k

Let I = ∫eᶲcosx dx. IBP: u=cosx, dv=eᶲdx → I = eᶲcosx + ∫eᶲsinx dx. IBP again: ∫eᶲsinx dx = eᶲsinx − ∫eᶲcosx dx = eᶲsinx − I. So: I = eᶲcosx + eᶲsinx − I → 2I = eᶲ(cosx+sinx). ∫eᶲcosx dx = eᶲ(cosx+sinx)/2 + C. Definite: [eᶲ(cosx+sinx)/2]₀^π = e^π(cosπ+sinπ)/2 − e⁰(cos0+sin0)/2 = e^π(−1+0)/2 − (1+0)/2 = −e^π/2 − 1/2 = −(e^π+1)/2

(a) V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx (b) V = π[x²/2]₀⁴ = π·8 = 8π (c) About y-axis (disc method): x=y², y from 0 to 2. V = π∫₀² (4)²dy − π∫₀² y⁴ dy = 16π(2)−π(32/5)=32π−32π/5=128π/5

(a) r² − 3r + 2 = 0 → (r−1)(r−2)=0 → r=1, r=2 (b) y = Ae^x + Be^(2x) (c) y(0)=3: A+B=3; y'=Ae^x+2Be^(2x), y'(0)=5: A+2B=5 → Solving: B=2, A=1 → y = eᶲ + 2e^(2x) (d) As x→∞, e^(2x) dominates → y→∞. Solution grows without bound.

(a) f'(x) = 2x·lnx + x²·(1/x) = 2xlnx + x = x(2lnx+1); f''(x) = (2lnx+1) + x·(2/x) = 2lnx+3 (b) f'(x)=0: x=0 (excluded) or 2lnx+1=0 → lnx=−1/2 → x=e^(−1/2)=1/√e; f''(1/√e) = 2(−1/2)+3 = 2 > 0 → local minimum. f(1/√e) = (1/e)·(−1/2) = −1/(2e) (c) ∫₁^e x²lnx dx. IBP: u=lnx, dv=x²dx → v=x³/3 = [x³lnx/3]₁^e − ∫₁^e x²/3 dx = (e³/3−0) − [x³/9]₁^e = e³/3 − e³/9 + 1/9 = 2e³/9 + 1/9

(a) x² + y² = 25 (b) 2x·dx/dt + 2y·dy/dt = 0. x=3: y=4. 2(3)(0.3) + 2(4)dy/dt = 0 → dy/dt = −0.225 m/s (c) Let θ = angle with ground: sinθ = y/5 → cosθ·dθ/dt = (1/5)dy/dt; cosθ = 3/5 at x=3. (3/5)dθ/dt = (1/5)(−0.225) → dθ/dt = −0.225/3 = −0.075 rad/s (d) Area = xy/2. dA/dt = (1/2)(y·dx/dt + x·dy/dt) = (1/2)(4·0.3 + 3·(−0.225)) = (1/2)(1.2−0.675) = 0.2625 m²/s (increasing)

(a) Intersections at x = π/3, 2π/3, 4π/3, 5π/3 (b) Area = 2π + 6√3 (using ∫[π/3 to 2π/3](g−f)dx and symmetry) (c) Equal gradients: cos x = −1/2 → x = 2π/3; g(2π/3) = 2+4 = 6

(a) f'(x) = [kx·(3/x) − k·ln(8x³)]/(kx)² = [3 − ln(8x³)]/(kx²) ✓ (b) f'(x) = 0: ln(8x³) = 3 → 8x³ = e³ → x = e/2 (c) f''(x) = 0: 2ln(8x³) = 9 → ln(8x³) = 9/2 → 8x³ = e^(9/2) → x³ = e^(9/2)/8 → x = e^(3/2)/2 ✓ (d) ∫[e/2 to e^(3/2)/2] ln(8x³)/(kx) dx = 5 → Evaluate integral and solve for k → k ≈ 1.47 (by GDC or substitution)

(a) v(t) = 3t²−18t+24 = 3(t−2)(t−4) = 0 → t = 2 and t = 4 (b) s(0)=−10, s(2)=10, s(4)=6, s(5)=10; Distance = |10−(−10)| + |6−10| + |10−6| = 20+4+4 = 28 m (c) a(t) = 6t−18; a(3) = 0 m/s² (momentarily constant velocity at t=3)

(a) I = eᶲcos x + ∫eᶲ sin x dx (b) ∫eᶲsin x dx = eᶲsin x − ∫eᶲcos x dx = eᶲsin x − I (c) I = eᶲcos x + eᶲsin x − I → 2I = eᶲ(cos x + sin x) → ∫eᶲcos x dx = eᶲ(cos x + sin x)/2 + C (d) [eᶲ(cosx+sinx)/2]₀^π = e^π(−1+0)/2 − e⁰(1+0)/2 = −e^π/2 − 1/2 = −(e^π+1)/2

(a) dy/y = 2x dx → ln|y| = x² + C → y = Ae^(x²) (b) y(0) = 3 → A = 3; ∴ y = 3e^(x²) (c) dy/dx = 3·2x·e^(x²) = 6xe^(x²) = 2x·(3e^(x²)) = 2xy ✓ (d) y-intercept at (0,3); curve grows rapidly for x ≠ 0; y → ∞ as x → ±∞

(a) h = 500/x²; S = x² + 4x(500/x²) = x² + 2000/x (b) S' = 2x − 2000/x² = 0 → x³ = 1000 → x = 10 cm (c) S'' = 2 + 4000/x³ > 0 at x=10 → minimum ✓ (d) S_min = 100 + 200 = 300 cm²

(a) x² = 4−x² → 2x²=4 → x = ±√2; Points: (−√2, 2) and (√2, 2) (b) Area = ∫ from −√2 to √2 of [(4−x²)−x²] dx = ∫ (4−2x²) dx (c) = [4x − 2x³/3] evaluated = 2(4√2 − 4√2/3) = 16√2/3

(a) T = 20 + 70e^(−kt) (b) 60 = 20+70e^(−5k) → e^(−5k) = 4/7 → k = ln(7/4)/5 ≈ 0.112 (c) T(15) = 20+70e^(−3ln(7/4)) = 20+70(4/7)³ = 20+70(64/343) ≈ 33.1°C (d) 25 = 20+70e^(−kt) → e^(−kt) = 1/14 → t = ln14/k ≈ 23.6 min

(a) sin x = x − x³/6 + x⁵/120 − x⁷/5040 + … (b) sin x − x + x³/6 = x⁵/120 − …; divide by x⁵: → 1/120 (c) sin(x²) ≈ x² − x⁶/6 + x¹⁰/120; ∫₀^0.5 ≈ [x³/3 − x⁷/42 + x¹¹/1320]₀^0.5 ≈ 0.04167 − 0.000186 + 0.0000003 ≈ 0.0415

(a) x² + y² = 25 (b) 2x(dx/dt) + 2y(dy/dt) = 0; x=3,y=4: 2(3)(0.3)+2(4)(dy/dt)=0 → dy/dt = −0.225 m/s (c) cosθ·(dθ/dt) = (1/5)(dy/dt); cosθ=3/5: dθ/dt = −0.075 rad/s (d) Area = xy/2; dA/dt = (1/2)(y·dx/dt + x·dy/dt) = (1/2)(4×0.3+3×(−0.225)) = 0.2625 m²/s; Area is increasing at 0.2625 m²/s