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Date November 2018 Marks available 2 Reference code 18N.1.AHL.TZ0.H_7
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Show that Question number H_7 Adapted from N/A

Question

Consider the curves  C 1 and  C 2  defined as follows

C 1 : x y = 4 x > 0

C 2 : y 2 x 2 = 2 x > 0

Using implicit differentiation, or otherwise, find  d y d x  for each curve in terms of  x and  y .

[4]
a.

Let P( a , b ) be the unique point where the curves C 1 and C 2 intersect.

Show that the tangent to C 1 at P is perpendicular to the tangent to C 2 at P.

[2]
b.

Markscheme

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

C 1 : y + x d y d x = 0       (M1)

Note: M1 is for use of both product rule and implicit differentiation.

 

d y d x = y x       A1

Note: Accept  4 x 2

 

C 2 : 2 y d y d x 2 x = 0       (M1)

d y d x = x y       A1

Note: Accept  ± x 2 + x 2

 

[4 marks]

a.

substituting  a and  b for  x and  y       M1

product of gradients at P is  ( b a ) ( a b ) = 1 or equivalent reasoning       R1

Note: The R1 is dependent on the previous M1

 

so tangents are perpendicular       AG

 

[2 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 —Calculus » AHL 5.14—Implicit functions, related rates, optimisation
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Topic 5 —Calculus

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