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### Transcript of leaving certificate HigHer level Active Maths 1...Active Maths I SBN 978 - 1 - 78090 - 531 ......

• 2r

y 2 -

y 1x 2

- x

1

m =

r2

Michael Keating, Derek Mulvany and James OLoughlin

Special Advisors: Oliver Murphy, Colin Townsend and Jim McElroy

Active Maths 4

I SBN 9 7 8 - 1 - 7 8 0 9 0 - 5 3 1 - 0

9 7 8 1 7 8 0 9 0 5 3 1 0folensonline.ie

leaving certificate HigHer level

Book 1

2714

Supplementary Material for 2015 exam and

onwards

• Chapter 13 Differential Calculus I

13.10 Implicit Differentiation 1

Editor: Priscilla OConnorDesigner: Liz WhiteLayout: CompuscriptIllustrations: Compuscript

ISBN: 978-1-78090-531-0

Michael Keating, Derek Mulvany, James OLoughlin and Colin Townsend, 2014

Folens Publishers, Hibernian Industrial Estate, Greenhills Road, Tallaght, Dublin 24, Ireland

All rights reserved No part of this publication may be reproduced or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission from the publisher The publisher reserves the right to change, without notice, at any time, the specification of this product, whether by change of materials, colours, bindings, format, text revision or any other characteristic

Contents

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Learning Outcomes Use differentiation to find the slope of a tangent to a circle

13.10 impliCit DifferentiationThe functions that we have met so far can be described by expressing one variable explicitly in terms of another variable, for example:

y = ______

x2 + 5 or y = 2x2 cos x

Sometimes, we deal with equations in x and y in which the relation between x and y is implicit.

Eq. (1) x 2 + y 2 = 36

or

Eq. (2) x 2 + y 2 + 2x + 2y 34 = 0

The graph of Eq. 1 above is a circle with centre (0,0) and radius length 6, while the graph of Eq. 2 is a circle with centre (1,1) and radius length 6 (See Active Maths 4, Book 2, Chapter 11).

On our course we need to be able to use differentiation to find the slope of a tangent to a circle. Therefore, we will need to know how to differentiate equations such as Eq. (1) and Eq. (2) above.

Worked example 13.28

Use differentiation to find the slope of the tangent to the circle x 2 + y 2 = 25 at the point (3,4).

SolutionDifferentiate both sides of the equation x 2 + y 2 = 25 with respect to x.

d(x2 + y2)

_________ dx

= d(25)

_____ dx

d(x2)

____ dx

+ d(y2)

____ dx

= d(25)

_____ dx

d(x2)

____ dx

= 2x

d(25)

_____ dx

= 0 (as 25 is a constant)

____ dx

?

Use the Chain Rule here:

Write d(y2)

____ dx

= d(y2)

____ d()

d()

___ dx

Replace () with y

d(y2)

____ dx

= d(y2)

____ dy

dy

___ dx

= 2y dy

___ dx

Therefore:

2x + 2y d ___ dx

= 0

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Worked example 13.29

(i) Use differentiation to find the slope of the tangent t to the circle x 2 + y 2 + 6x 2y 15 = 0 at the point (6,3).

(ii) Hence, find the equation of t.

Solve this equation for dy

___ dx

:

2y dy

___ dx

= 2x

dy

___ dx

= 2x ___ 2y

dy

___ dx

= x __ y

Now find the value of dy

___ dx

at (3,4). This will be the required slope.

dy

___ dx

|(3,4)

= 3 __ 4

Alternative Method:

Write y explicitly in terms of x:

x 2 + y 2 = 25

y 2 = 25 x 2

y = _______

25 x 2

y = _______

25 x 2 is the semicircle of radius 5 units and centre (0,0) on or above the x-axis.

y = _______

25 x 2 is the semicircle of radius 5 units and centre (0,0) on or below the x-axis.

As we are looking for the slope of the tangent at the point (3,4), we will use y =

_______ 25 x 2

y = (25 x 2 ) 1 __ 2

Use the Chain Rule:

y = u 1 __ 2 u = 25 x 2

dy

___ du

= 1 __ 2 u 1 __ 2

du ___

dx = 2x

dy

___ dx

= dy

___ du

du ___ dx

= 1 __ 2 u 1 __ 2 (2x)

= x ___ u

1 __ 2

= x ________

_______ 25 x 2

= x __ y

The required slope is dy

___ dx

|(3,4)

= 3 __ 4

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iSolution (i) x 2 + y 2 + 6x 2y 15 = 0

2x + 2y dy

___ dx

+ 6 2 dy

___ dx

+ 0 = 0

(2y 2) dy

___ dx

= 2x 6

(y 1) dy

___ dx

= x 3

dy

___ dx

= x 3 _______ y 1

dy

___ dx

|(6,3) = (6) 3 _________ 3 1 = 6 3 _____ 4

= 3 ___ 4

Slope = 3 __ 4

(ii) Point (6,3) Slope = 3 __ 4

y (3) = 3 __ 4 (x (6))

4(y + 3) = 3(x + 6)

4y + 12 = 3x 18

t: 3x + 4y + 30 = 0

exercise 13.9

1. For each of the following circles, express dy

___ dx

in terms of x and y.

(i) x 2 + y 2 = 16 (iv) 3 x 2 + 3 y 2 = 64

(ii) x 2 + y 2 = 49 (v) a x 2 + a y 2 = b (where a, b are of equal sign)

(iii) 2 x 2 + 2 y 2 = 81

2. For each of the following circles, express

dy

___ dx

in terms of x and y.

(i) x 2 + y 2 + 2x 2y + 1 = 0

(ii) x 2 + y 2 + 4x 6y + 11 = 0

(iii) 2 x 2 + 2 y 2 2x 8y 1 = 0

(iv) 5 x 2 + 5 y 2 + 2x 2y 1 = 0

(v) 7 x 2 + 7 y 2 + 2x 2y 3 = 0

3. For each of the following circles, express

dy

___ dx

in terms of x and y.

(i) (x 3 ) 2 + (y + 2 ) 2 = 14

(ii) (x 8 ) 2 + (y 2 ) 2 = 36

(iii) (x + 3 ) 2 + (y 9 ) 2 = 24

(iv) (x 5 ) 2 + (y + 3 ) 2 = 25

(v) (x 7 ) 2 + (y + 6 ) 2 = 49

4. Use differentiation to find the slope of the tangent to the circle x 2 + y 2 = 100 at the point (6,8).

5. Use differentiation to find the slope of the tangent to the circle x 2 + y 2 6x 2y 3 = 0 at the point (5,4).

6. Use differentiation to find the slope of the tangent to the circle (x 1 ) 2 + (y + 2 ) 2 = 5 at the point (3,3). Hence, find the equation of the tangent to the circle containing the point (3,3).

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i7. The graphs of the function f(x) = x 2 + 2x 3 and the circle (x 7 ) 2 + (y 1 ) 2 = 20 are shown.

AC is a tangent to the circle at A(3,3) and a tangent to f(x) at C. B is the centre of the circle.

(i) Use differentiation to find the slope of the tangent AC.

(ii) Find the equation of AC.

(iii) Use two different methods to find the co-ordinates of the point C.

(iv) Show that |AC| : |AB| = 3 : 2.

(v) Show that the area of the circle is more than four times greater than the area of the triangle ABC.

8. c is the circle x 2 + y 2 + 8x + 14 = 0, and d is the circle x 2 + y 2 4x 8y 30 = 0. a is a tangent to c at the point F(3,1).

(i) Use differentiation to find the slope of a.

(ii) Find the equation of a.

(iii) Show that a contains the centre of d.

(iv) Find the co-ordinates of E, if E is an endpoint of the diameter [EF].

(v) A tangent f is drawn from E to the circle c. Show that the equation of f is 79x 119y + 518 = 0.

(vi) Find to the nearest degree the measure of the acute angle between the tangents a and f.

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irevision exercises extra Questions

13A.

The circle (x 3) 2 + (y 3) 2 = 5 has centre A and contains the points B and C.

(i) Write the equation of this circle in the form x 2 + y 2 + ex + fy + g = 0, where e, f, g R.

(ii) Express the slope of a tangent to this circle at a point on the circle in terms of x and y.

(iii) Hence, calculate the slope of the tangent BD.

(iv) The point C is vertically above the centre point A. Find the slope of the tangent CD.

(v) Find the co-ordinates of the point D.

(vi) Calculate the radius of the circle. Hence, calculate the area of the kite ABDC.

13B. C is the circle x 2 + y 2 4x 2y + 1 = 0.

(i) Verify that the point A(3, 1 + __

3 ) lies on C.

(ii) Use differentiation to find the slope of the tangent to C at the point A.

(iii) Show that the equation of the tangent at A

can be written as y = 1 ___

__ 3 x + 2

__ 3 + 1.

(iv) Sketch the circle and tangent.

(v) Find the area of the triangle enclosed between the tangent at A and the x and y axes.