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 2x2=sin3y 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 dydx.
Find the equation of the tangent to the curve at the point (14, 5π6).
Find the area of R.
The region R is now rotated about the y-axis, through 2π radians, to form a solid.
By writing sin3y as (1−cos2y)siny, show that the volume of the solid formed is 2π3.
Markscheme
valid attempt to differentiate implicitly (M1)
4x=3sin2ycosydydx A1A1
dydx=4x3sin2ycosy A1
[4 marks]
at (14, 5π6), dydx=4x3sin2ycosy=13(12)2(−√32) (M1)
⇒dydx=−83√3(=−1.54) A1
hence equation of tangent is
y−5π6=−1.54(x−14) OR y=−1.54x+3.00 (M1)A1
Note: Accept y=−1.54x+3.
[4 marks]
x=√12sin3y (M1)
∫π0√12sin3ydy (A1)
=1.24 A1
[3 marks]
use of volume =∫πx2dy (M1)
=∫π012πsin3ydy A1
=12π∫π0(siny−sinycos2y)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(1−13)−12π(−1+13) A1
=2π3 AG
Note: Do not accept decimal answer equivalent to 2π3.
[6 marks]