Home / IB Mathematics SL 5.2 Increasing and decreasing function AI HL Paper 1- Exam Style Questions

IB Mathematics SL 5.2 Increasing and decreasing function AI HL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AI HL Paper 1 - All Topics

Question

An environmental researcher is using a rectangular model to track the dimensions of an ice shelf. The width (\(x\) km) is estimated to be expanding at a constant rate of \(10\) km per year, while the length (\(y\) km) is shrinking at a steady rate of \(5\) km per year. Time, \(t\), is recorded in years, and the surface area, \(A\), is given in \(\text{km}^2\).
At the start of the study (\(t=0\)), the width is \(75\) km and the length is \(40\) km.
(a) Determine the value of \(\frac{dA}{dt}\) at \(t=0\).
(b) Decide if the surface area of the ice shelf is expanding or contracting at \(t=0\). Justify your conclusion.
(c) Calculate the net change in the surface area of the shelf from \(t=0\) to \(t=1\).

Most-appropriate topic codes:

• AHL 5.9: Related rates of change — part (a)
• SL 5.2: Increasing and decreasing functions — part (b)
• SL 2.5: Modelling with linear functions — part (c)
▶️ Answer/Explanation
Detailed solution

(a)
Area \(A = xy\). We use the product rule for differentiation with respect to \(t\):
\(\frac{dA}{dt} = x \frac{dy}{dt} + y \frac{dx}{dt}\)
Given: \(\frac{dx}{dt} = 10\), \(\frac{dy}{dt} = -5\).
At \(t=0\), \(x=75\) and \(y=40\).
\(\frac{dA}{dt} = 75(-5) + 40(10)\)
\(\frac{dA}{dt} = -375 + 400\)
\(\frac{dA}{dt} = 25 \, \text{km}^2/\text{year}\)

(b)
The area is increasing because \(\frac{dA}{dt} = 25\), which is a positive value.

(c)
We can find the dimensions at \(t=1\).
\(x(1) = 75 + 10(1) = 85\)
\(y(1) = 40 – 5(1) = 35\)
Area at \(t=1\): \(A(1) = 85 \times 35 = 2975 \, \text{km}^2\).
Area at \(t=0\): \(A(0) = 75 \times 40 = 3000 \, \text{km}^2\).
Change in area \(= A(1) – A(0) = 2975 – 3000 = -25\).
The area decreases by \(25 \, \text{km}^2\).

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