finding the maximum distance between two curves2 + v. the distance pq was then defined by the...

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Page 16 Learning and Teaching Mathematics, No. 20, 2016, pp. 16-17 Finding the Maximum Distance between Two Curves Deepak Mavani 1 & Beena Mavani 2 Khanyisa High School, Mthatha 1 [email protected] 2 [email protected] Nothing ever becomes real until it is experienced – John Keats Consider the graphs of the functions () = 2 + 3 2 −1 and () = − 2 +4. P is a point on and Q is a point on such that QP is parallel to the -axis. What is the maximum length of QP between the points of intersection of the two graphs? This is a fairly standard question that can be answered using a number of different approaches. The length of QP is given by () − () which simplifies to: () = −2 2 3 2 +5 The points of intersection of and can be found by setting () = 0 since the distance QP will be zero at these points. Solving () = 0 gives = −2 and = 5 4 . In order to determine the maximum length of QP all we need to do is determine the turning point of the parabola represented by (). We can do this by (i) completing the square and writing () in the form ( − ) 2 + from which the turning point can be read, (ii) determining the axis of symmetry of and using it to find the corresponding -value, or (iii) setting ′() = 0 to determine the point at which the gradient is zero and then finding the corresponding -value. Using any of these methods gives the turning point of to be (− 3 8 ; 169 32 ). The maximum length of QP is thus 5,28 to two decimal places. Irrespective of the approach taken, the most critical aspect is for pupils to appreciate that the expression () = () − () represents the length of the line segment QP. The use of dynamic geometry software such as GeoGebra is a useful way of adding a more visual element to reinforce this. We began by drawing the two original parabolas, () = 2 + 3 2 −1 and () = − 2 +4. The distance PQ was then defined by the function () = −2 2 3 2 +5. We then set E to be a point on between = −2 and = 5 4 , i.e. between the points of intersection of and . A slider was then created so that the -value of point E could be adjusted for ∈ [−2; 5 4 ]. The file was set up so that as the slider is moved it not only traces the curve of () − () but also the line segment QP.

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Page 1: Finding the Maximum Distance between Two Curves2 + v. The distance PQ was then defined by the function ( ) =− t 2 − 3 2 + w. We then set E to be a point on between and 5 4, i.e

Page 16

Learning and Teaching Mathematics, No. 20, 2016, pp. 16-17

Finding the Maximum Distance between Two Curves

Deepak Mavani1 & Beena Mavani2

Khanyisa High School, Mthatha 1 [email protected] 2 [email protected]

Nothing ever becomes real until it is experienced – John Keats

Consider the graphs of the functions 𝑝(𝑥) = 𝑥2 +3

2𝑥 − 1 and 𝑔(𝑥) = −𝑥2 + 4. P is a point on 𝑝 and Q is a

point on 𝑔 such that QP is parallel to the 𝑦-axis. What is the maximum length of QP between the points of

intersection of the two graphs?

This is a fairly standard question that can be answered

using a number of different approaches. The length of QP

is given by 𝑔(𝑥) − 𝑝(𝑥) which simplifies to:

𝐿(𝑥) = −2𝑥2 −3

2𝑥 + 5

The points of intersection of 𝑝 and 𝑔 can be found by

setting 𝐿(𝑥) = 0 since the distance QP will be zero at these

points. Solving 𝐿(𝑥) = 0 gives 𝑥 = −2 and 𝑥 =5

4 . In order

to determine the maximum length of QP all we need to do

is determine the turning point of the parabola represented

by 𝐿(𝑥).

We can do this by (i) completing the square and writing 𝐿(𝑥) in the form 𝑎(𝑥 − 𝑝)2 + 𝑞 from which the

turning point can be read, (ii) determining the axis of symmetry of 𝐿 and using it to find the corresponding

𝑦-value, or (iii) setting 𝐿′(𝑥) = 0 to determine the point at which the gradient is zero and then finding the

corresponding 𝑦-value. Using any of these methods gives the turning point of 𝐿 to be (−3

8;

169

32). The

maximum length of QP is thus 5,28 to two decimal places.

Irrespective of the approach taken, the most critical aspect is for pupils to appreciate that the expression

𝐿(𝑥) = 𝑔(𝑥) − 𝑝(𝑥) represents the length of the line segment QP. The use of dynamic geometry software

such as GeoGebra is a useful way of adding a more visual element to reinforce this. We began by drawing the

two original parabolas, 𝑝(𝑥) = 𝑥2 +3

2𝑥 − 1 and 𝑔(𝑥) = −𝑥2 + 4. The distance PQ was then defined by the

function 𝑠(𝑥) = −2𝑥2 −3

2𝑥 + 5. We then set E to be a point on 𝑠 between 𝑥 = −2 and 𝑥 =

5

4 , i.e. between

the points of intersection of 𝑝 and 𝑔. A slider was then created so that the 𝑥-value of point E could be

adjusted for 𝑥 ∈ [−2;5

4]. The file was set up so that as the slider is moved it not only traces the curve of

𝑔(𝑥) − 𝑝(𝑥) but also the line segment QP.

Page 2: Finding the Maximum Distance between Two Curves2 + v. The distance PQ was then defined by the function ( ) =− t 2 − 3 2 + w. We then set E to be a point on between and 5 4, i.e

Page 17

Learning and Teaching Mathematics, No. 20, 2016, pp. 16-17

As pupils engage with the dynamic geometry sketch their attention should be drawn to a number of

important visual elements: (i) the traced function represented by 𝑔(𝑥) − 𝑝(𝑥) has 𝑥-intercepts at the 𝑥-values

where 𝑔(𝑥) and 𝑝(𝑥) intersect, (ii) as the slider is moved from 𝑥 = −2 to 𝑥 =5

4 the length of QP increases

to a maximum and then decreases again, (iii) the maximum length of QP coincides with the point at which

the function represented by 𝑔(𝑥) − 𝑝(𝑥) has a local maximum.

The use of a dynamic geometry environment to reinforce visual elements of what is otherwise a fairly

abstract algebraic process can be incredibly powerful. For those interested, we have uploaded the GeoGebra

file to http://tube.geogebra.org/ under the title “Distance Between Parabolas”. Additional sliders have been

added so that the equation of 𝑝(𝑥) can also be adjusted.