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Date May 2021 Marks available 2 Reference code 21M.2.AHL.TZ1.4
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number 4 Adapted from N/A

Question

Charlotte decides to model the shape of a cupcake to calculate its volume.

From rotating a photograph of her cupcake she estimates that its cross-section passes through the points (0, 3.5), (4, 6), (6.5, 4), (7, 3) and (7.5, 0), where all units are in centimetres. The cross-section is symmetrical in the x-axis, as shown below:

She models the section from (0, 3.5) to (4, 6) as a straight line.

Charlotte models the section of the cupcake that passes through the points (4, 6), (6.5, 4), (7, 3) and (7.5, 0) with a quadratic curve.

Charlotte thinks that a quadratic with a maximum point at (4, 6) and that passes through the point (7.5, 0) would be a better fit.

Believing this to be a better model for her cupcake, Charlotte finds the volume of revolution about the x-axis to estimate the volume of the cupcake.

Find the equation of the line passing through these two points.

[2]
a.

Find the equation of the least squares regression quadratic curve for these four points.

[2]
b.i.

By considering the gradient of this curve when x=4, explain why it may not be a good model.

[1]
b.ii.

Find the equation of the new model.

[4]
c.

Write down an expression for her estimate of the volume as a sum of two integrals.

[4]
d.i.

Find the value of Charlotte’s estimate.

[1]
d.ii.

Markscheme

y=58x+72   y=0.625x+3.5                  A1A1


Note:
Award A1 for 0.625x, A1 for 3.5.
Award a maximum of A0A1 if not part of an equation.


[2 marks]

a.

y=-0.975x2+9.56x-16.7                  (M1)A1

y=-0.974630x2+9.55919x-16.6569


[2 marks]

b.i.

gradient of curve is positive at x=4                 R1


Note: Accept a sensible rationale that refers to the gradient.


[1 mark]

b.ii.

METHOD 1

let y=ax2+bx+c

differentiating or using x=-b2a                       (M1)

8a+b=0

substituting in the coordinates
7.52a+7.5b+c=0                       (A1)
42a+4b+c=6                       (A1)

solve to get
y=-2449x2+19249x-9049  OR  y=-0.490x2+3.92x-1.84                       A1


Note: Use of quadratic regression with points using the symmetry of the graph is a valid method.

 

METHOD 2

y=ax-42+6                       (M1)

0=a7.5-42+6                       (M1)

a=-2449                       (A1)

y=-2449x-42+6  OR  y=-0.490x-42+6                       A1

 

[4 marks]

c.

π0458x+3.52dx+π47.5-2449x-42+62dx                       (M1)(M1) (M1)A1


Note: Award (M1)(M1)(M1)A0 if π is omitted but response is otherwise correct. Award (M1) for an integral that indicates volume, (M1) for their part (a) within their volume integral, (M1) for their part (b)(i) within their volume integral, A1 for their correct two integrals with all correct limits.

 

[4 marks]

d.i.

501 cm3  501.189                      A1

 

[1 mark]

d.ii.

Examiners report

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

Syllabus sections

Topic 2—Functions » SL 2.5—Modelling functions
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Topic 4—Statistics and probability » AHL 4.13—Non-linear regression
Topic 2—Functions
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