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Date November 2019 Marks available 6 Reference code 19N.2.AHL.TZ0.H_11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Show that Question number H_11 Adapted from N/A

Question

The following diagram shows part of the graph of 2x2=sin3y for 0yπ.

The shaded region R is the area bounded by the curve, the y-axis and the lines y=0 and y=π.

Using implicit differentiation, find an expression for dydx.

[4]
a.i.

Find the equation of the tangent to the curve at the point (145π6).

[4]
a.ii.

Find the area of R.

[3]
b.

The region R is now rotated about the y-axis, through 2π radians, to form a solid.

By writing sin3y as (1cos2y)siny, show that the volume of the solid formed is 2π3.

[6]
c.

Markscheme

valid attempt to differentiate implicitly       (M1)

4x=3sin2ycosydydx       A1A1

dydx=4x3sin2ycosy       A1

[4 marks]

a.i.

at (145π6)dydx=4x3sin2ycosy=13(12)2(32)       (M1)

dydx=833(=1.54)       A1

hence equation of tangent is

y5π6=1.54(x14)  OR  y=1.54x+3.00       (M1)A1

Note: Accept y=1.54x+3

[4 marks]

a.ii.

x=12sin3y       (M1)

π012sin3ydy       (A1)

=1.24       A1

[3 marks]

b.

use of volume =πx2dy       (M1)

=π012πsin3ydy       A1

=12ππ0(sinysinycos2y)dy

Note: Condone absence of limits up to this point.

reasonable attempt to integrate       (M1)

=12π[cosy+13cos3y]π0       A1A1

Note: Award A1 for correct limits (not to be awarded if previous M1 has not been awarded) and A1 for correct integrand.

=12π(113)12π(1+13)  A1

=2π3       AG

Note: Do not accept decimal answer equivalent to 2π3.

[6 marks]

c.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 5 —Calculus » SL 5.5—Integration introduction, areas between curve and x axis
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Topic 1—Number and algebra
Topic 5 —Calculus

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