20N.2.AHL.TZ0.H_8

pestleMathematicsAAHLPaper 220N· sl-5-6-differentiating-polynomials-n-e-q-chain-product-and-quotient-rulessource ↗

A small bead is free to move along a smooth wire in the shape of the curve $y = \frac{10}{3 - 2 e^{- 0.5 x}} (x \geq 0)$ .

Find an expression for $\frac{d y}{d x}$ .

[3]

At the point on the curve where $x = 4$ , it is given that $\frac{d y}{d t} = - 0.1 {m s}^{- 1}$

Find the value of $\frac{d x}{d t}$ at this exact same instant.

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid attempt to use chain rule or quotient rule (M1)

$\frac{d y}{d x} = \frac{- 10 e^{- 0.5 x}}{{(3 - 2 e^{- 0.5 x})}^{2}}$ OR $\frac{d y}{d x} = - 10 e^{- 0.5 x} {(3 - 2 e^{- 0.5 x})}^{- 2}$ A1A1

[3 marks]

Note: Award A1 for numerator and A1 for denominator, or A1 for each part if the second alternative given.

valid attempt to use chain rule $(eg \frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t})$ (M1)

$\frac{d x}{d t} = - 0.1 \div \frac{- 10 e^{- 2}}{{(3 - 2 e^{- 2})}^{2}} (= - 0.1 \div - 0.181676 \dots)$ or equivalent (A1)

$= 0.550428 \dots$

$\frac{d x}{d t} = 0.550 ({ms}^{- 1})$ A1

[3 marks]

Examiners’ report

[N/A]