IB Mathematics SL 5.1 Derivative interpreted as gradient function AI SL Paper 2 - Exam Style Questions - New Syllabus

(i) Differentiate \(f(x)= -\dfrac{x^2}{50} + 2x + 30\): \[ f'(x)= -\frac{2x}{50} + 2 = -\frac{x}{25} + 2 = 2 – 0.04x. \]

(ii) Furthest north \(\Rightarrow\) maximum of \(f(x)\) on \([0,70]\). Set derivative to zero: \[ f'(x)=0 \;\Rightarrow\; 2 – \frac{x}{25}=0 \;\Rightarrow\; x=50. \] Then \[ f(50)= -\frac{50^2}{50}+2(50)+30 = -50 + 100 + 30 = 80. \] So the point is \(\boxed{(50,\,80)}\).

(i) Area of the region under the curve and above \(y=0\) from \(x=0\) to \(x=70\): \[ \text{Area}=\int_{0}^{70}\Big(-\frac{x^2}{50}+2x+30\Big)\,dx. \]

(ii) Evaluate: \[ \int\!\Big(-\frac{x^2}{50}+2x+30\Big)dx = -\frac{x^3}{150}+x^2+30x. \] Hence \[ \text{Area}=\left[-\frac{x^3}{150}+x^2+30x\right]_{0}^{70} = -\frac{70^3}{150}+70^2+30\cdot 70. \] Compute each term: \[ 70^3=343{,}000,\quad \frac{343{,}000}{150}=2286.\overline{6}, \quad 70^2=4900,\quad 30\cdot 70=2100. \] Therefore \[ \text{Area}= -2286.\overline{6}+4900+2100 = 4713.\overline{3}\text{ m}^2=\frac{14140}{3}\text{ m}^2 \approx \boxed{4710\text{ m}^2}\ \text{(to 3 s.f.)}. \]

(i) Percentage error (underestimate by \(11.4\text{ m}^2\) compared to true area \(A\)): \[ \%\text{ error} = \frac{11.4}{4713.33\ldots}\times 100\% \approx \boxed{0.242\%}. \]

(ii) To reduce error using the trapezoidal rule: either decrease the subinterval width (i.e. use a finer partition) or equivalently increase the number of intervals.

Let the left side of the square be at \(x=E_x\). With \([GH]\) on \(x=70\), the square’s side length is the horizontal width \(70 – E_x\). For \(F\) (the top-left corner) to lie on the boundary \(y=f(x)\), we must have the vertical side length equal to the function value at \(x=E_x\):

Equate side length = height: \[ 70 – x = f(x) = -\frac{x^2}{50} + 2x + 30. \] Rearranging: \[ -\frac{x^2}{50} + 3x – 40 = 0. \] Multiply by \(50\): \[ -x^2 + 150x – 2000 = 0 \;\Longrightarrow\; x^2 – 150x + 2000 = 0. \] Solve: \[ x=\frac{150 \pm \sqrt{150^2-4\cdot 2000}}{2} =\frac{150 \pm \sqrt{14500}}{2}. \] The feasible solution in \([0,70]\) is \[ E_x \approx \boxed{14.8}\ \text{m} \quad (14.7920\ldots). \]

Largest area: The square’s side length is \(s=70 – E_x \approx 55.208\) m, hence \[ \text{Area}_{\max}=s^2\approx (55.208)^2 \approx \boxed{3050\text{ m}^2} \quad (3047.92\ldots\ \text{exact with }E_x). \]