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Date November 2016 Marks available 6 Reference code 16N.2.AHL.TZ0.H_6
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number H_6 Adapted from N/A

Question

An earth satellite moves in a path that can be described by the curve 72.5x2+71.5y2=1 where x=x(t) and y=y(t) are in thousands of kilometres and t is time in seconds.

Given that dxdt=7.75×105 when x=3.2×103, find the possible values of dydt.

Give your answers in standard form.

Markscheme

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

METHOD 1

substituting for x and attempting to solve for y (or vice versa)     (M1)

y=(±)0.11821    (A1)

EITHER

145x+143ydydx=0 (dydx=145x143y)    M1A1

OR

145xdxdt+143ydydt=0    M1A1

THEN

attempting to find dxdt (dydt=145(3.2×103)143((±)0.11821)×(7.75×105))     (M1)

dydt=±2.13×106    A1

 

Note: Award all marks except the final A1 to candidates who do not consider ±.

 

METHOD 2

y=(±)172.5x271.5    M1A1

dydx=(±)0.0274    (M1)(A1)

dydt=(±)0.0274×7.75×105    (M1)

dydt=±2.13×106    A1

 

Note: Award all marks except the final A1 to candidates who do not consider ±.

 

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » AHL 5.14—Setting up a DE, solve by separating variables
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Topic 5—Calculus

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