User interface language: English | Español

Date May 2014 Marks available 1 Reference code 14M.2.sl.TZ1.5
Level SL only Paper 2 Time zone TZ1
Command term Interpret Question number 5 Adapted from N/A

Question

The population of deer in an enclosed game reserve is modelled by the function \(P(t) = 210\sin (0.5t - 2.6) + 990\), where \(t\) is in months, and \(t = 1\) corresponds to 1 January 2014.

Find the number of deer in the reserve on 1 May 2014.

[3]
a.

Find the rate of change of the deer population on 1 May 2014.

[2]
b(i).

Interpret the answer to part (i) with reference to the deer population size on 1 May 2014.

[1]
b(ii).

Markscheme

\(t = 5\)     (A1)

correct substitution into formula     (A1)

eg     \(210\sin (0.5 \times 5 - 2.6) + 990,{\text{ }}P(5)\)

\(969.034982 \ldots \)

969 (deer) (must be an integer)     A1     N3

[3 marks]

a.

evidence of considering derivative     (M1)

eg     \(P'\)

\(104.475\)

\(104\) (deer per month)     A1     N2

[2 marks]

b(i).

(the deer population size is) increasing     A1     N1

[1 mark]

b(ii).

Examiners report

[N/A]
a.
[N/A]
b(i).
[N/A]
b(ii).

Syllabus sections

Topic 6 - Calculus » 6.1 » Derivative interpreted as gradient function and as rate of change.
Show 56 related questions

View options