IB Mathematics AHL 5.9 The derivatives of sin x-AI HL Paper 2- Exam Style Questions- New Syllabus

(a)
    \(k = A e^{-\frac{c}{T}}\)
    Differentiate using chain rule: \(\frac{dk}{dT} = A \cdot e^{-\frac{c}{T}} \cdot \frac{d}{dT} \left(-\frac{c}{T}\right)\)
    \(\frac{d}{dT} \left(-\frac{c}{T}\right) = -c \cdot (-T^{-2}) = \frac{c}{T^2}\)
    \(\frac{dk}{dT} = A e^{-\frac{c}{T}} \cdot \frac{c}{T^2}\)
    Since \(A > 0\), \(c > 0\), \(T > 0\), and \(e^{-\frac{c}{T}} > 0\), \(\frac{dk}{dT} > 0\)
Result:
\(\frac{dk}{dT}\) is always positive

(b)
    As \(T \to 0\), \(\frac{c}{T} \to \infty\), \(e^{-\frac{c}{T}} \to 0\), so \(k \to 0\)
    As \(T \to \infty\), \(\frac{c}{T} \to 0\), \(e^0 = 1\), so \(k \to A\)
    From (a), \(\frac{dk}{dT} > 0\), so \(k\) increases with \(T\)

Result:
Graph: Increasing curve, starts at (0, 0), approaches \(k = A\) asymptotically

(c)(i)
    \(\ln k = \ln (A e^{-\frac{c}{T}}) = \ln A – \frac{c}{T}\)
    Let \(x = \frac{1}{T}\), so \(\ln k = -c x + \ln A\)
    Gradient is \(-c\)
Result:
\(-c\)

(c)(ii)
    From \(\ln k = -c \cdot \frac{1}{T} + \ln A\)
    Y-intercept when \(\frac{1}{T} = 0\) is \(\ln A\)
Result:
\(\ln A\)

(d)
    Convert data: e.g., \(T = 590\), \(k = 5 \times 10^{-4}\), \(\frac{1}{T} = \frac{1}{590} \approx 0.001695\), \(\ln k = \ln (5 \times 10^{-4}) \approx -7.601\)
    Regression line: \(\ln k = m \cdot \frac{1}{T} + b\)
    From data, slope \(m \approx -13383.1 \approx -13400\), intercept \(b \approx 15.0107 \approx 15.0\)

Result:
\(\ln k = -13400 \cdot \frac{1}{T} + 15.0\)

(e)(i)
    From (c)(i), gradient = \(-c\)
    From (d), gradient \(\approx -13400\), so \(c \approx 13400\) (or 13383.1 exactly)
Result:
13400

(e)(ii)
    From (c)(ii), y-intercept = \(\ln A\)
    From (d), y-intercept \(\approx 15.0\) (or 15.0107), so \(\ln A \approx 15.0\)
    \(A = e^{15.0} \approx 3,269,017 \approx 3,300,000\) (or \(e^{15.0107} \approx 3,304,258\))
Result:
3300000