IB Mathematics SL 3.4 length of an arc area of a sector AI HL Paper 2- Exam Style Questions- New Syllabus

(a)(i)
    The diagram shows a right triangle with adjacent side = 4m and hypotenuse = 4.5m
    Using cosine:
    \( \cos(\theta) = \frac{4}{4.5} = \frac{8}{9} \)
    \( \theta = \arccos\left(\frac{8}{9}\right) \approx 27.266…° \)
    This is half of angle AÔB, so:
    \( \angle AÔB = 2 \times 27.266…° \approx 54.532…° \)
    \( \approx 54.5° \) or 0.952 radians (M1)(A1)(A1)
Result:
54.5° (or 0.952 radians)

(a)(ii)
    Sector area:
    \( \text{Area} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 4.5^2 \times 0.951764… \)
    \( = \frac{1}{2} \times 20.25 \times 0.951764… \approx 9.63661 \, \text{m}^2 \) (M1)(A1)
    Triangle area:
    \( \text{Area} = \frac{1}{2} \times 4.5^2 \times \sin(54.532…°) \)
    \( = \frac{1}{2} \times 20.25 \times 0.814379… \approx 8.24621 \, \text{m}^2 \) (M1)(A1)
    Segment area:
    \( 9.63661 – 8.24621 = 1.39040 \, \text{m}^2 \)
    \( \approx 1.39 \, \text{m}^2 \) (A1)
Result:
1.39 \( \text{m}^2 \)

(b)
    Method 1:

Circle area \( \pi \times 4.5^2 = 63.617… \, \text{m}^2 \) (A1)
    Four segment areas: \( 4 \times 1.39040 = 5.56160 \, \text{m}^2 \) (A1)
    Reachable area: \( 63.617… – 5.56160 \approx 58.055… \, \text{m}^2 \) (M1)(A1)
    \( \approx 58.1 \, \text{m}^2 \)
    Method 2: 4 triangles + 4 sectors = \( 58.055… \, \text{m}^2 \) (A1)(A1)(M1)(A1)
Result:
58.1 \( \text{m}^2 \)

(c)
    Rate of grass consumption: \( \frac{dV}{dt} = 0.3 t e^{-t} \)
    Differentiate:
    \( \frac{d}{dt} (0.3 t e^{-t}) = 0.3 \left[ e^{-t} \cdot 1 + t \cdot (-e^{-t}) \right] \)
    \( = 0.3 e^{-t} (1 – t) \)
    Set to zero: \( 0.3 e^{-t} (1 – t) = 0 \)
    \( 1 – t = 0 \)
    \( t = 1 \)
    Second derivative: \( \frac{d}{dt} \left[ 0.3 e^{-t} (1 – t) \right] = 0.3 e^{-t} (t – 2) \)
    At \( t = 1 \): \( 0.3 e^{-1} (1 – 2) < 0 \) (M1)
    \( t = 1 \) hour (A1)
Result:
\( t = 1 \, \text{hour} \)

(d)
    Total volume \( V = \int_0^8 0.3 t e^{-t} \, dt \) (M1)
    Integration by parts: \( u = t \), \( dv = e^{-t} \, dt \)
    \( du = dt \), \( v = -e^{-t} \)
    \( \int t e^{-t} \, dt = -t e^{-t} – e^{-t} + C \)
    Evaluate from 0 to 8:
    \( \left[ -t e^{-t} – e^{-t} \right]_0^8 = (-8e^{-8} – e^{-8}) – (-0 – 1) \)
    \( = -9e^{-8} + 1 \) (A1)
    Multiply by 0.3: \( V = 0.3 (1 – 9e^{-8}) \approx 0.3 (1 – 0.003019…) \)
    \( \approx 0.3 \times 0.99698… \approx 0.299094… \, \text{m}^3 \) (A1)
    \( \approx 0.299 \, \text{m}^3 \)
Result:
0.299 \( \text{m}^3 \)