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Date November 2019 Marks available 4 Reference code 19N.2.AHL.TZ0.H_11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number H_11 Adapted from N/A

Question

The following diagram shows part of the graph of 2 x 2 = si n 3 y for 0 y π .

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 d y d x .

[4]
a.i.

Find the equation of the tangent to the curve at the point  ( 1 4 5 π 6 ) .

[4]
a.ii.

Find the area of R .

[3]
b.

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

By writing  si n 3 y as  ( 1 co s 2 y ) sin y , show that the volume of the solid formed is 2 π 3 .

[6]
c.

Markscheme

valid attempt to differentiate implicitly       (M1)

4 x = 3 si n 2 y cos y d y d x        A1A1

d y d x = 4 x 3 si n 2 y cos y        A1

[4 marks]

a.i.

at  ( 1 4 5 π 6 ) d y d x = 4 x 3 si n 2 y cos y = 1 3 ( 1 2 ) 2 ( 3 2 )        (M1)

d y d x = 8 3 3 ( = 1.54 )        A1

hence equation of tangent is

y 5 π 6 = 1.54 ( x 1 4 )   OR   y = 1.54 x + 3.00        (M1)A1

Note: Accept  y = 1.54 x + 3

[4 marks]

a.ii.

x = 1 2 si n 3 y        (M1)

0 π 1 2 si n 3 y d y        (A1)

= 1.24        A1

[3 marks]

b.

use of volume  = π x 2 d y        (M1)

= 0 π 1 2 π si n 3 y d y        A1

= 1 2 π 0 π ( sin y sin y co s 2 y ) d y

Note: Condone absence of limits up to this point.

reasonable attempt to integrate       (M1)

= 1 2 π [ cos y + 1 3 co s 3 y ] 0 π        A1A1

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

= 1 2 π ( 1 1 3 ) 1 2 π ( 1 + 1 3 )  A1

= 2 π 3        AG

Note: Do not accept decimal answer equivalent to  2 π 3 .

[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 5.4—Tangents and normal
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Topic 5 —Calculus » AHL 5.14—Implicit functions, related rates, optimisation
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