Home / IB Mathematics SL 5.1 Derivative interpreted as gradient function AI HL Paper 1- Exam Style Questions

IB Mathematics SL 5.1 Derivative interpreted as gradient function AI HL Paper 1- Exam Style Questions

IB Mathematics SL 5.1 Derivative interpreted as gradient function AI HL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AI : HL Paper -1 : All Topics
Question

A company is designing a new carpet. The intended design of the carpet is in the shape of a rectangle with a semi-circle at each end.

The width of the rectangle is \( y \) metres, and the diameter of each semi-circle is \( x \) metres, with \( x > 0 \) and \( y \geq 0 \).

Carpet diagram

The company has decided that the perimeter of the carpet will be 20 metres and would like to maximize its area.

(a) Find an expression for the perimeter in terms of \( x \) and \( y \) [2]

(b) Show that the area, \( A \) (m²), of the carpet can be expressed as: \( A = 10x – \frac{\pi x^2}{4} \) [2]

(c) Find \( \frac{dA}{dx} \) [2]

(d) Hence, find the exact value of \( x \) for which the area is a maximum [2]

▶️ Answer/Explanation
Markscheme

(a)
\(\pi x + 2y = 20\)

Two semicircles: Arc length = \(\pi x\)
Two straight sides: \(2y\)
Total perimeter: \(\pi x + 2y = 20\)

Result: \(\pi x + 2y = 20\) [2]

(b)
\(A = 10x – \frac{\pi x^2}{4}\)

Rectangle area: \(A_{\text{rect}} = x \times y\)
Two semicircles (full circle): \(A_{\text{circle}} = \pi \left( \frac{x}{2} \right)^2 = \frac{\pi x^2}{4}\)
Total area: \(A = x y + \frac{\pi x^2}{4}\)
Perimeter constraint: \(\pi x + 2y = 20 \implies y = \frac{20 – \pi x}{2}\)
Substitute: \(A = x \times \frac{20 – \pi x}{2} + \frac{\pi x^2}{4}\)
Simplify: \(A = \frac{20x – \pi x^2}{2} + \frac{\pi x^2}{4} = \frac{40x – 2\pi x^2 + \pi x^2}{4} = 10x – \frac{\pi x^2}{4}\)

Result: \(A = 10x – \frac{\pi x^2}{4}\) [2]

(c)
\(\frac{dA}{dx} = 10 – \frac{\pi x}{2}\)

Given: \(A = 10x – \frac{\pi x^2}{4}\)
Differentiate: \(\frac{dA}{dx} = \frac{d}{dx} \left( 10x \right) – \frac{d}{dx} \left( \frac{\pi x^2}{4} \right)\)
Compute: \(\frac{d}{dx} (10x) = 10\), \(\frac{d}{dx} \left( -\frac{\pi x^2}{4} \right) = -\frac{\pi}{4} \times 2x = -\frac{\pi x}{2}\)
Thus: \(\frac{dA}{dx} = 10 – \frac{\pi x}{2}\)

Result: \(\frac{dA}{dx} = 10 – \frac{\pi x}{2}\) [2]

(d)
\(x = \frac{20}{\pi}\)

Set derivative to zero: \(10 – \frac{\pi x}{2} = 0\)
Solve: \(\frac{\pi x}{2} = 10 \implies \pi x = 20 \implies x = \frac{20}{\pi}\)
Verify maximum: \(\frac{d^2A}{dx^2} = -\frac{\pi}{2} < 0\)
Check: At \(x = \frac{20}{\pi}\), \(y = \frac{20 – \pi \cdot \frac{20}{\pi}}{2} = 0\), which is valid (\(y \geq 0\))

Result: \(x = \frac{20}{\pi}\) [2]

Question

The gradient of the curve \( f(x) = \ln x \) at point \( x = a \) is defined by the limit

\( m_a = \lim_{h \to 0} \frac{\ln(a+h) – \ln a}{h} \)

For example, the gradient at x = 1 is \( m_1 \) since \( \lim_{h \to 0} \frac{\ln(1+h) – \ln 1}{h} = 1 \).

(a) By using the graph mode on your GDC, find

(i) \( \lim_{h \to 0} \frac{\ln(2+h) – \ln 2}{h} \) and hence the gradient \( m_2 \) [2]

(ii) \( \lim_{h \to 0} \frac{\ln(5+h) – \ln 5}{h} \) and hence the gradient \( m_5 \) [2]

(b) Find the gradient \( m_{10} \) of the curve at \( x = 10 \) [2]

(c) Deduce the value of the gradient \( m_a \) in terms of \( a \) in general [2]

▶️ Answer/Explanation
Markscheme

(a)(i)
\(\lim_{h \to 0} \frac{\ln(2+h) – \ln 2}{h} = 0.5\), \( m_2 = 0.5 = \frac{1}{2} \)

Rewrite: \( \ln(2+h) – \ln 2 = \ln\left(\frac{2+h}{2}\right) = \ln\left(1 + \frac{h}{2}\right) \)
Limit: \( \frac{\ln\left(1 + \frac{h}{2}\right)}{h} = \frac{1}{2} \cdot \frac{\ln\left(1 + \frac{h}{2}\right)}{\frac{h}{2}} \)
As \( h \to 0 \), let \( u = \frac{h}{2} \), so \( \lim_{u \to 0} \frac{\ln(1+u)}{u} = 1 \)
Thus: \( \frac{1}{2} \times 1 = \frac{1}{2} = 0.5 \)
Gradient: \( m_2 = 0.5 = \frac{1}{2} \)
GDC: Graph \( y = \frac{\ln(2+h) – \ln 2}{h} \), observe \( y \to 0.5 \) as \( h \to 0 \)
Alternative: \( f(x) = \ln x \), \( f'(x) = \frac{1}{x} \), at \( x = 2 \): \( f'(2) = \frac{1}{2} = 0.5 \)

Result: \(\lim_{h \to 0} \frac{\ln(2+h) – \ln 2}{h} = 0.5\), \( m_2 = 0.5 = \frac{1}{2} \) [2]

(a)(ii)
\(\lim_{h \to 0} \frac{\ln(5+h) – \ln 5}{h} = 0.2\), \( m_5 = 0.2 = \frac{1}{5} \)

Rewrite: \( \ln(5+h) – \ln 5 = \ln\left(\frac{5+h}{5}\right) = \ln\left(1 + \frac{h}{5}\right) \)
Limit: \( \frac{\ln\left(1 + \frac{h}{5}\right)}{h} = \frac{1}{5} \cdot \frac{\ln\left(1 + \frac{h}{5}\right)}{\frac{h}{5}} \)
As \( h \to 0 \), let \( u = \frac{h}{5} \), so \( \lim_{u \to 0} \frac{\ln(1+u)}{u} = 1 \)
Thus: \( \frac{1}{5} \times 1 = \frac{1}{5} = 0.2 \)
Gradient: \( m_5 = 0.2 = \frac{1}{5} \)
GDC: Graph \( y = \frac{\ln(5+h) – \ln 5}{h} \), observe \( y \to 0.2 \) as \( h \to 0 \)
Alternative: \( f'(x) = \frac{1}{x} \), at \( x = 5 \): \( f'(5) = \frac{1}{5} = 0.2 \)

Result: \(\lim_{h \to 0} \frac{\ln(5+h) – \ln 5}{h} = 0.2\), \( m_5 = 0.2 = \frac{1}{5} \) [2]

(b)
\( m_{10} = 0.1 = \frac{1}{10} \)

Rewrite: \( \ln(10+h) – \ln 10 = \ln\left(\frac{10+h}{10}\right) = \ln\left(1 + \frac{h}{10}\right) \)
Limit: \( \frac{\ln\left(1 + \frac{h}{10}\right)}{h} = \frac{1}{10} \cdot \frac{\ln\left(1 + \frac{h}{10}\right)}{\frac{h}{10}} \)
As \( h \to 0 \), let \( u = \frac{h}{10} \), so \( \lim_{u \to 0} \frac{\ln(1+u)}{u} = 1 \)
Thus: \( \frac{1}{10} \times 1 = \frac{1}{10} = 0.1 \)
Gradient: \( m_{10} = 0.1 = \frac{1}{10} \)
Alternative: \( f'(x) = \frac{1}{x} \), at \( x = 10 \): \( f'(10) = \frac{1}{10} = 0.1 \)

Result: \( m_{10} = 0.1 = \frac{1}{10} \) [2]

(c)
\( m_a = \frac{1}{a} \)

Rewrite: \( \ln(a+h) – \ln a = \ln\left(\frac{a+h}{a}\right) = \ln\left(1 + \frac{h}{a}\right) \)
Limit: \( \frac{\ln\left(1 + \frac{h}{a}\right)}{h} = \frac{1}{a} \cdot \frac{\ln\left(1 + \frac{h}{a}\right)}{\frac{h}{a}} \)
As \( h \to 0 \), let \( u = \frac{h}{a} \), so \( \lim_{u \to 0} \frac{\ln(1+u)}{u} = 1 \)
Thus: \( \frac{1}{a} \times 1 = \frac{1}{a} \)
Gradient: \( m_a = \frac{1}{a} \)
Alternative: \( f'(x) = \frac{1}{x} \), at \( x = a \): \( f'(a) = \frac{1}{a} \)

Result: \( m_a = \frac{1}{a} \) [2]

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