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[lania]

 

Nedenfor er en eksamen laget i https://claude.ai/, som gir mulighet til å laste ned gjennom pdf fil (som en forsovidt og så kan gjøre i chatGPT), og tilhørende løsnings forslag.

 

Spørsmålet er egentlig litt filosofisk: Burde en gidde å selv stå for arbeidet med å lage repetitive oppgave på en eksamen, og skal en selv bare stå for de oppgaver som er mer kreativ (som forsovidt også kan lages av claude.ai)?

AI er ganske skremmende. Min mening er at en burde ha mulighet til å reservere seg mot AI, og da heller få tildelt en vanlig eksamen; videre burde en ha mulighet til å kreve AI-generert eksamen, hvis mange nok mener det er god grunn til dette.

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Calculus Examination

Total Points: 100
Time Allowed: 2 hours
Instructions: Show all work for full credit. No calculators permitted unless specified.

Part I: Limits and Continuity (20 points)

1. (5 points) Evaluate the following limit: lim(x→3) (x² - 9)/(x - 3)

2. (5 points) Find the value of c that makes the following function continuous on ℝ: f(x) = { x² + 2x if x < 1 cx + 4 if x ≥ 1 }

3. (5 points) Determine if the following limit exists. If it does, find its value: lim(x→0) sin(3x)/x

4. (5 points) Evaluate: lim(x→∞) (1 + 2/x)ˣ

Part II: Derivatives (25 points)

5. (5 points) Find the derivative of f(x) = x³ln(x) - 2x²sin(x)

6. (5 points) If f(x) = tan²(3x), find f'(x).

7. (5 points) Find the equation of the tangent line to the curve y = xeˣ at the point (0,0).

8. (5 points) Use implicit differentiation to find dy/dx for the curve: x²y + xy² = 6

9. (5 points) If f(x) = ∫₀^(x²) cos(t²) dt, find f'(x) using the Fundamental Theorem of Calculus.

Part III: Applications of Derivatives (20 points)

10. (5 points) Find the absolute maximum and minimum values of f(x) = x³ - 3x² - 9x + 7 on the interval [-2, 4].

11. (5 points) A particle moves along a line with position function s(t) = t³ - 6t² + 9t where s is measured in meters and t in seconds. Find: a) The velocity at time t = 2 b) The time(s) when the particle is at rest c) When the particle is moving in the positive direction

12. (5 points) Use Newton's method with the initial approximation x₁ = 2 to find x₂ and x₃ as approximations to the solution of the equation x³ - 2x - 5 = 0.

13. (5 points) Use differentials to approximate the value of √17.

Part IV: Integration (20 points)

14. (5 points) Evaluate: ∫ (x² + 1)/x³ dx

15. (5 points) Evaluate: ∫ sin³(x)cos²(x) dx

16. (5 points) Evaluate the definite integral: ∫₀¹ xe^(x²) dx

17. (5 points) Find the area of the region bounded by the curves y = x² and y = 4 - x².

Part V: Applications of Integration (15 points)

18. (5 points) Find the volume of the solid obtained by rotating the region bounded by y = x², y = 0, and x = 2 about the x-axis.

19. (5 points) A particle moves along a line with velocity function v(t) = t² - 4t + 3 meters per second. If the particle is at position s = 5 when t = 2, find its position function s(t).

20. (5 points) Find the average value of the function f(x) = sin(x) on the interval [0, π].

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Bonus Question: (5 points) Find the general solution to the differential equation: dy/dx = (y² + xy)/(x² + xy)

End of Examination

 

 

Calculus Examination Solution Manual

Part I: Limits and Continuity (20 points)

1. Evaluate: lim(x→3) (x² - 9)/(x - 3)

Solution: The expression is in indeterminate form 0/0 since direct substitution gives (9-9)/(3-3) = 0/0. We can factor the numerator: (x² - 9)/(x - 3) = ((x - 3)(x + 3))/(x - 3) = x + 3 for x ≠ 3

Therefore: lim(x→3) (x² - 9)/(x - 3) = lim(x→3) (x + 3) = 3 + 3 = 6

2. Find value of c for continuous function:

f(x) = { x² + 2x if x < 1 cx + 4 if x ≥ 1 }

Solution: For f(x) to be continuous at x = 1, the left and right limits must be equal: lim(x→1⁻) f(x) = lim(x→1⁺) f(x)

Left limit: lim(x→1⁻) (x² + 2x) = 1² + 2(1) = 1 + 2 = 3 Right limit: lim(x→1⁺) (cx + 4) = c(1) + 4 = c + 4

For continuity: 3 = c + 4 Therefore: c = -1

3. Determine if limit exists: lim(x→0) sin(3x)/x

Solution: Rewrite as: lim(x→0) sin(3x)/x = lim(x→0) [3·sin(3x)/(3x)] We know that lim(u→0) sin(u)/u = 1

Let u = 3x, then as x→0, u→0 lim(x→0) sin(3x)/x = 3·lim(u→0) sin(u)/u = 3·1 = 3

Therefore, the limit exists and equals 3.

4. Evaluate: lim(x→∞) (1 + 2/x)ˣ

Solution: This is in the form of the definition of e: lim(x→∞) (1 + 1/x)ˣ = e

We can rewrite our limit: lim(x→∞) (1 + 2/x)ˣ = lim(x→∞) [(1 + 2/x)^(x/2)]² = [lim(x→∞) (1 + 2/x)^(x/2)]²

Let y = x/2, then as x→∞, y→∞ [lim(y→∞) (1 + 1/(y/2))^y]² = [lim(y→∞) (1 + 2/y)^y]² = e²

Therefore: lim(x→∞) (1 + 2/x)ˣ = e² ≈ 7.389

Part II: Derivatives (25 points)

5. Find the derivative of f(x) = x³ln(x) - 2x²sin(x)

Solution: Using the product rule and chain rule:

For x³ln(x): d/dx[x³ln(x)] = x³·(1/x) + ln(x)·(3x²) = x² + 3x²ln(x) = x²(1 + 3ln(x))

For 2x²sin(x): d/dx[2x²sin(x)] = 2x²·cos(x) + sin(x)·(4x) = 2x²cos(x) + 4xsin(x)

Therefore: f'(x) = x²(1 + 3ln(x)) - 2x²cos(x) - 4xsin(x) = x²(1 + 3ln(x) - 2cos(x)) - 4xsin(x)

6. If f(x) = tan²(3x), find f'(x)

Solution: Using the chain rule: f'(x) = d/dx[tan²(3x)] = 2tan(3x)·d/dx[tan(3x)] = 2tan(3x)·sec²(3x)·d/dx[3x] = 2tan(3x)·sec²(3x)·3 = 6tan(3x)sec²(3x)

Since sec²(θ) = 1 + tan²(θ), we can also write: f'(x) = 6tan(3x)(1 + tan²(3x)) = 6tan(3x) + 6tan³(3x)

7. Find equation of tangent line to y = xeˣ at (0,0)

Solution: First, find the derivative: y' = d/dx[xeˣ] = eˣ + x·d/dx[eˣ] = eˣ + xeˣ = eˣ(1 + x)

At point (0,0): y'(0) = e⁰(1 + 0) = 1·1 = 1

Using point-slope form of a line: y - y₀ = m(x - x₀) Where (x₀,y₀) = (0,0) and m = 1

The equation of the tangent line is: y - 0 = 1(x - 0) y = x

8. Use implicit differentiation to find dy/dx for x²y + xy² = 6

Solution: Differentiating both sides with respect to x: d/dx[x²y + xy² = 6]

For x²y: d/dx[x²y] = 2xy + x²(dy/dx) For xy²: d/dx[xy²] = y² + x·2y(dy/dx) = y² + 2xy(dy/dx)

Therefore: 2xy + x²(dy/dx) + y² + 2xy(dy/dx) = 0 x²(dy/dx) + 2xy(dy/dx) = -2xy - y² (dy/dx)(x² + 2xy) = -2xy - y² dy/dx = (-2xy - y²)/(x² + 2xy) dy/dx = -y(2x + y)/(x² + 2xy)

9. If f(x) = ∫₀^(x²) cos(t²) dt, find f'(x) using the Fundamental Theorem of Calculus

Solution: By the Fundamental Theorem of Calculus: If F(x) = ∫ₐ^x g(t)dt, then F'(x) = g(x)

Here f(x) = ∫₀^(x²) cos(t²) dt, which means: f'(x) = cos((x²)²)·d/dx[x²] = cos(x⁴)·2x

Therefore: f'(x) = 2x·cos(x⁴)

Part III: Applications of Derivatives (20 points)

10. Find absolute max/min values of f(x) = x³ - 3x² - 9x + 7 on [-2, 4]

Solution: Step 1: Find critical points by setting f'(x) = 0 f'(x) = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x - 3)(x + 1) f'(x) = 0 when x = 3 or x = -1

Step 2: Evaluate f at critical points and endpoints f(-2) = (-2)³ - 3(-2)² - 9(-2) + 7 = -8 - 12 + 18 + 7 = 5 f(-1) = (-1)³ - 3(-1)² - 9(-1) + 7 = -1 - 3 + 9 + 7 = 12 f(3) = (3)³ - 3(3)² - 9(3) + 7 = 27 - 27 - 27 + 7 = -20 f(4) = (4)³ - 3(4)² - 9(4) + 7 = 64 - 48 - 36 + 7 = -13

Therefore: Absolute maximum: f(-1) = 12 Absolute minimum: f(3) = -20

11. Particle with position s(t) = t³ - 6t² + 9t

Solution: a) Velocity at time t = 2: v(t) = s'(t) = 3t² - 12t + 9 v(2) = 3(2)² - 12(2) + 9 = 12 - 24 + 9 = -3 m/s

b) Times when particle is at rest: v(t) = 0 3t² - 12t + 9 = 0 3(t² - 4t + 3) = 0 t² - 4t + 3 = 0

Using quadratic formula: t = (4 ± √(16-12))/2 = (4 ± √4)/2 = (4 ± 2)/2 t = 3 or t = 1

c) Particle moves in positive direction when v(t) > 0 3t² - 12t + 9 > 0 For t < 1 or t > 3, the velocity is positive Therefore, the particle moves in the positive direction when t ∈ (-∞, 1) ∪ (3, ∞)

12. Newton's method for x³ - 2x - 5 = 0 with x₁ = 2

Solution: For Newton's method, we use the formula: xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)

f(x) = x³ - 2x - 5 f'(x) = 3x² - 2

For x₁ = 2: f(2) = 2³ - 2(2) - 5 = 8 - 4 - 5 = -1 f'(2) = 3(2)² - 2 = 12 - 2 = 10 x₂ = 2 - (-1/10) = 2 + 0.1 = 2.1

For x₂ = 2.1: f(2.1) = (2.1)³ - 2(2.1) - 5 = 9.261 - 4.2 - 5 = 0.061 f'(2.1) = 3(2.1)² - 2 = 3(4.41) - 2 = 13.23 - 2 = 11.23 x₃ = 2.1 - (0.061/11.23) = 2.1 - 0.005433 ≈ 2.095

Therefore: x₂ = 2.1 and x₃ ≈ 2.095

13. Use differentials to approximate √17

Solution: We can use the formula: f(a + h) ≈ f(a) + f'(a)·h

Let f(x) = √x, a = 16, and h = 1 (since 17 = 16 + 1) f'(x) = 1/(2√x)

f(16) = √16 = 4 f'(16) = 1/(2√16) = 1/(2·4) = 1/8

Therefore: √17 ≈ √16 + (1/8)·1 = 4 + 0.125 = 4.125

Part IV: Integration (20 points)

14. Evaluate: ∫ (x² + 1)/x³ dx

Solution: ∫ (x² + 1)/x³ dx = ∫ (x²/x³ + 1/x³) dx = ∫ (1/x + 1/x³) dx = ∫ x⁻¹ dx + ∫ x⁻³ dx = ln|x| + (-1/2)x⁻² + C = ln|x| - 1/(2x²) + C

15. Evaluate: ∫ sin³(x)cos²(x) dx

Solution: Let u = sin(x), then du = cos(x) dx This gives: cos(x) dx = du

∫ sin³(x)cos²(x) dx = ∫ u³cos(x)cos(x) dx = ∫ u³ cos²(x) dx

We know sin²(x) + cos²(x) = 1, so cos²(x) = 1 - sin²(x) = 1 - u² Therefore: ∫ u³(1 - u²) du = ∫ (u³ - u⁵) du = u⁴/4 - u⁶/6 + C = sin⁴(x)/4 - sin⁶(x)/6 + C

16. Evaluate: ∫₀¹ xe^(x²) dx

Solution: Let u = x², then du = 2x dx, or x dx = du/2

∫₀¹ xe^(x²) dx = ∫₀¹ (1/2)e^(x²)(2x dx) = (1/2)∫₀¹ e^u du

When x = 0, u = 0; when x = 1, u = 1

(1/2)∫₀¹ e^u du = (1/2)[e^u]₀¹ = (1/2)(e¹ - e⁰) = (1/2)(e - 1) = (e - 1)/2

17. Find area between y = x² and y = 4 - x²

Solution: First, find intersection points by setting the equations equal: x² = 4 - x² 2x² = 4 x² = 2 x = ±√2

The area is: ∫₍₋√2₎^√2 [(4 - x²) - x²] dx = ∫₍₋√2₎^√2 [4 - 2x²] dx = [4x - (2x³/3)]₍₋√2₎^√2 = [4√2 - (2(√2)³/3)] - [4(-√2) - (2(-√2)³/3)] = [4√2 - (2·2√2/3)] - [-4√2 - (-2·2√2/3)] = [4√2 - (4√2/3)] - [-4√2 + (4√2/3)] = [4√2(1 - 1/3)] - [-4√2(1 - 1/3)] = [(8√2/3)] - [-(8√2/3)] = 16√2/3 ≈ 7.54 square units

Part V: Applications of Integration (15 points)

18. Volume from rotating y = x², y = 0, x = 2 around x-axis

Solution: Using the disk method, the volume is: V = π∫ₐ^b [f(x)]² dx

For rotation around the x-axis from x = 0 to x = 2: V = π∫₀^2 (x²)² dx = π∫₀^2 x⁴ dx = π[x⁵/5]₀^2 = π(2⁵/5 - 0) = π(32/5) = 32π/5 ≈ 20.11 cubic units

19. Particle with v(t) = t² - 4t + 3, s(2) = 5

Solution: To find position function, integrate the velocity: s(t) = ∫ v(t) dt = ∫(t² - 4t + 3) dt = t³/3 - 2t² + 3t + C

Given s(2) = 5: 5 = (2³/3) - 2(2)² + 3(2) + C 5 = 8/3 - 8 + 6 + C 5 = 8/3 - 2 + C C = 5 - 8/3 + 2 = 7 - 8/3 = (21-8)/3 = 13/3

Therefore: s(t) = t³/3 - 2t² + 3t + 13/3

20. Average value of f(x) = sin(x) on [0, π]

Solution: The average value is given by: f_avg = (1/(b-a))∫ₐ^b f(x) dx

For f(x) = sin(x) on [0, π]: f_avg = (1/π)∫₀^π sin(x) dx = (1/π)[-cos(x)]₀^π = (1/π)[-cos(π) - (-cos(0))] = (1/π)[-(−1) - (-1)] = (1/π)[1 + 1] = 2/π ≈ 0.637

Bonus Question (5 points)

Find the general solution to: dy/dx = (y² + xy)/(x² + xy)

Solution: This is a homogeneous differential equation. Let y = vx where v is a function of x. Then dy/dx = v + x(dv/dx)

Substituting: v + x(dv/dx) = (v²x² + vx²)/(x² + vx²) v + x(dv/dx) = (v²x + vx)/(x + vx) v + x(dv/dx) = v(vx + x)/(x + vx) = v

Therefore: x(dv/dx) = 0, which means dv/dx = 0 This implies v is constant: v = C where C is any constant

Since y = vx, the general solution is: y = Cx

Where C is an arbitrary constant of integration.

Endret 28. april av lania

[The Avatar]

lania skrev (1 time siden):

 

Spørsmålet er egentlig litt filosofisk: Burde en gidde å selv stå for arbeidet med å lage repetitive oppgave på en eksamen, og skal en selv bare stå for de oppgaver som er mer kreativ (som forsovidt også kan lages av claude.ai)?

AI er ganske skremmende. Min mening er at en burde ha mulighet til å reservere seg mot AI, og da heller få tildelt en vanlig eksamen; videre burde en ha mulighet til å kreve AI-generert eksamen, hvis mange nok mener det er god grunn til dette.

Hva tenker dere?

Tenker at dette er eit godt eksempel på områder der KI gjerne kan brukast.
Å generere slike matteoppgåver med fasit sparar mykje arbeidstid som kan brukast til meir nyttige ting som å gi personleg oppfølging til studenten.
Teknologien opner for at ein kan generere mange ulike versjonar slik at det blir mindre sannsynleg at oppgåva du får på eksamen er lik ei oppgåve som har blitt gjort på ein tidlegare eksamen.
Og kanskje viktigast av alt, sidan dette er matematikk så er det veldig enkelt for læraren å kvalitetsjekke at oppgåvene gir meining og at fasiten stemmer. Så lenge den kvalitetsjekken faktisk blir utført så skal det ikkje være fare for at studenten får eksamensoppgåver med feil.