Implicit Differentiation

An introduction to implicit differentiation – An investigation 1. a. Find the centre and radius of the circle 𝑥 2 + 𝑦 2 − 6𝑥 + 4𝑦 + 9 = 0 b. Find the equation of the tangent at (1, -2) c. Find the equation of the tangent at (3, 0) d. Find the equation of the tangent when 𝑥 =

23 5

and 𝑦 < −1

2. How many turning points has the curve 𝑥 2 + 𝑦 3 = 4?

3. How many turning points has the curve 𝑡𝑎𝑛𝑦 = 𝑥 2 ?

© www.teachitmaths.co.uk 2017

28112

Page 1 of 5

An introduction to implicit differentiation – An investigation Teacher notes

An investigative worksheet designed to make students think about the nature of curves. 1. The gradient function is not always needed to find the gradient of tangents to a curve, as in parts b and c. When the gradient of the function is required, as in part d, the problem arises as to how to get around the difficulty of making 𝑦 the subject of the equation. a. 𝑥 2 + 𝑦 2 − 6𝑥 + 4𝑦 + 9 = 0 (𝑥 − 3)2 − 9 + (𝑦 + 2)2 − 4 + 9 = 0 (𝑥 − 3)2 + (𝑦 + 2)2 = 4 Centre (3, -2), radius 2 3 The circle 𝑥 2 + 𝑦 2 = 4, has been translated by the vector ( ) −2

b. Tangent, 𝑥 = 1

c. Tangent, 𝑦 = 0

b. 𝑥 2 + 𝑦 2 − 6𝑥 + 4𝑦 + 9 = 0 2𝑥 + 2𝑦 (

𝑑𝑦 𝑑𝑦 ) − 6 + 4( ) = 0 𝑑𝑥 𝑑𝑥

𝑑𝑦 6 − 2𝑥 3 − 𝑥 = ≡ 𝑑𝑥 2𝑦 + 4 𝑦 + 2 © www.teachitmaths.co.uk 2017

28112

Page 2 of 5

An introduction to implicit differentiation – An investigation When 𝑥 =

23 5

2

23

, ( 5 − 3) + (𝑦 + 2)2 = 4 ⇒ 𝑦 =

−16 5

, in this case and 𝑦 =

−4 5

23 3− 𝑑𝑦 5 ⇒ the gradient of the tangent is 4 = 𝑑𝑥 −16 + 2 3 5

4

The equation of the tangent is 𝑦 = 3 𝑥 + 𝑐 ⇒ 3𝑦 = 4𝑥 + 3𝑐 Substitute 𝑥 =

23 5

,𝑦 =

−16 5

⇒ 3𝑐 = −28

⇒ 3𝑦 − 4𝑥 + 28 = 0

© www.teachitmaths.co.uk 2017

28112

Page 3 of 5

An introduction to implicit differentiation – An investigation 2. When I ask my students for the number of turning points [of the ‘Darth Vader’s hat’ curve], the modal answer is 5

𝑥2 + 𝑦3 = 4 ⇒ 2𝑥 + 3𝑦 2 ( ⇒

𝑑𝑦 )=0 𝑑𝑥

𝑑𝑦 −2𝑥 = 𝑑𝑥 3𝑦 2

Turning points occur when the gradient is zero 3 The gradient is zero when the numerator is zero, ⇒ 𝑥 = 0, ⇒ 𝑦 = √4 There is only one turning point 3. 𝑡𝑎𝑛𝑦 = 𝑥 2 ⇒ 𝑠𝑒𝑐 2 𝑦 ( ⇒

𝑑𝑦 ) = 2𝑥 𝑑𝑥

𝑑𝑦 2𝑥 = 𝑑𝑥 𝑠𝑒𝑐 2 𝑦

Again, it appears there is only one turning point, when 𝑥 = 0

© www.teachitmaths.co.uk 2017

28112

Page 4 of 5

An introduction to implicit differentiation – An investigation However,

𝑑𝑦 2𝑥 = ≡ 2𝑥𝑐𝑜𝑠 2 𝑦 2 𝑑𝑥 𝑠𝑒𝑐 𝑦

Hence, there are an infinite number of turning points, when 𝑥 = 0 and 𝑐𝑜𝑠𝑦 = 0

© www.teachitmaths.co.uk 2017

28112

Page 5 of 5