Loading [MathJax]/jax/element/mml/optable/GeneralPunctuation.js

User interface language: English | Español

Date November 2018 Marks available 4 Reference code 18N.1.AHL.TZ0.H_7
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number H_7 Adapted from N/A

Question

Consider the curves C1 and C2 defined as follows

C1:xy=4x>0

C2:y2x2=2x>0

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

[4]
a.

Let P(a, b) be the unique point where the curves C1 and C2 intersect.

Show that the tangent to C1 at P is perpendicular to the tangent to C2 at P.

[2]
b.

Markscheme

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

C1:y+xdydx=0      (M1)

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

 

dydx=yx      A1

Note: Accept 4x2

 

C2:2ydydx2x=0      (M1)

dydx=xy      A1

Note: Accept ±x2+x2

 

[4 marks]

a.

substituting a and b for x and y      M1

product of gradients at P is (ba)(ab)=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
Show 40 related questions
Topic 5 —Calculus

View options