Tangents and Normals



✏

✒

✑

13.1

Introduction In this block we see how the equations of the tangent line and the normal line at a particular point on the curve y = f (x) can be obtained. The equations of tangent and normal lines have form: y = mx + c, y = nx + d respectively. We shall show that the product mn is −1 if these lines are to be perpendicular. The constants c, d are then obtained by using the property that the both the normal and tangent lines and the curve pass through a common point.

✬

✩

① be able to differentiate standard functions

Prerequisites Before starting this Block you should . . . ✫

Learning Outcomes

② understand the geometrical interpretation of a derivative ③ understand the trigonometric expansions of sin(A + B), cos(A + B)

Learning Style

After completing this Block you should be able To achieve what is expected of you . . . to . . . ✓ obtain the condition that two given lines be perpendicular

☞ allocate sufficient study time

✓ obtain the equation of the tangent line to a curve

☞ briefly revise the prerequisite material

✓ obtain the equation of the normal line to a curve

☞ attempt every guided exercise and most of the other exercises

✪

1. Perpendicular Lines One form for the equation of a straight line is y = mx + c where m and c are constants. We remember that m is the gradient of the line and its value is the tangent of the angle θ that the line makes with the positive x−axis. The constant c is the value obtained where the line intersects the y−axis. See the following diagram: y y = mx + c m = tan θ

c θ

x

If we have a second line, with equation y = nx + d then, unless m = n, the two lines will intersect. These are drawn together on the following diagram. The second line makes an angle ψ with the positive x−axis. y y = mx + c c ψ

θ

x

y = nx + d n = tan ψ

A simple question to ask is “what is the relation between m and n if the lines are perpendicular?” If the lines are perpendicular, as shown in the next figure, the angles θ, ψ must satisfy the relation: ψ − θ = 90◦ y

c θ

ψ x

This is true since the angles in a triangle add up to 180◦ . According to the figure the three angles are 90◦ , θ and 180◦ − ψ. Therefore 180◦ = 90◦ + θ + (180◦ − ψ) Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

implying

ψ − θ = 90◦ 2

In this special case that the lines are perpendicular or normal to each other the relation between the gradients m and n is easily obtained. In this deduction we remind the reader of some basic trigonometric relations and identities: sin(A − B) = sin A cos B − cos A sin B tan A =

sin A cos A

cos(A − B) = cos A cos B + sin A sin B

sin 90◦ = 1

cos 90◦ = 0

Now m = tan θ = tan(ψ − 90o ) sin(ψ − 90o ) = cos(ψ − 90o ) 1 1 − cos ψ =− =− = sin ψ tan ψ n Key Point Two straight lines y = mx + c, y = nx + d are perpendicular if m=−

1 n

mn = −1

or equivalently

This result assumes that neither of the lines are parallel to the x−axis or to the y−axis. More exercises for you to try Which of the following pairs of lines are perpendicular: i. y = −x + 1, y = x + 1 ii. y + x − 1 = 0, y + x − 2 = 0 iii. 2y = 8x + 3, y = −0.25x − 1 Answer

2. Tangents and Normals to a curve As we know, the relationship between an independent variable x and a dependent variable y is denoted by y = f (x) As we also know, the geometrical interpretation of this relation takes the form of a curve in an xy plane as illustrated in the following diagram. y

y = f (x)

x

3

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

We know how to calculate a value of y given a value of x. We can either do this graphically (which is inaccurate) or else use the function itself. So, at an x value of x0 the corresponding y value is y0 where y0 = f (x0 ) Let us examine the curve in the neighbourhood of the point (x0 , y0 ). There are two important constructions of interest • the tangent line at (x0 , y0 ) • the normal line at (x0 , y0 ) These are shown in the following figure y tangent line y0 ψ

θ

x

x0 normal line

We note the geometrically obvious fact: the tangent and normal lines at any given point on a curve are perpendicular to each other. Try each part of this exercise Part (a) The curve y = x2 is drawn below. On a copy of this graph draw the tangent line and the normal line at the point (x0 = 1, y0 = 1) y

1 1

x

Answer Part (b) From your graph, estimate the values of θ and ψ. (You will need a protractor). Answer Returning to the curve y = f (x) : we know, from the geometrical interpretation of the derivative that  df  = tan θ dx x0 (in case you’ve forgotten, the notation on the left-hand side of this relation means evaluate at the value x0 )

df dx

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

4

Also, in this relation, θ is the angle the tangent line to the curve y = f (x) makes with the positive x−axis. This is highlighted in the following diagram: y y = f (x)

 df  = tan θ dx x0 θ

x

x0

3. The Tangent Line to a Curve Let the equation of the tangent line to the curve y = f (x) at the point (x0 , y0 ) be: y = mx + c where m and c are constants to be found. The line just touches the curve y = f (x) at the point (x0 , y0 ) so, at this point both must have the same value for the derivative. That is:  df  m= dx x0 Since we know (in any particular case) f (x) and the value x0 we can readily calculate the value for m. The value of c is found by using the fact that the tangent line and the curve pass through the same point (x0 , y0 ). y0 = mx0 + c and y0 = f (x0 ) Thus mx0 + c = f (x0 )

leading to

c = f (x0 ) = mx0

Key Point The equation of the tangent line to the curve y = f (x) at the point (x0 , y0 ) is  df  y = mx + c where m = and c = f (x0 ) − mx0 dx x0

Example Find the equation of the tangent line to the curve y = x2 at the point (1,1) Solution Here f (x) = x2 and x0 = 1.  df df  Thus =2 = 2x ∴ m = dx dx x0 Also c = f (x0 ) − mx0 = f (1) − m = 1 − 2 = −1. Therefore the equation of the tangent line is y = 2x − 1. (Check back to the previous guided exercise to see if this ‘looks right’).

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Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

Try each part of this exercise Find the equation of the tangent line to the curve y = ex at the point x = 0. The curve and the line are displayed in the following figure: y tangent line x

Part (a) Specify x0 and f Answer  df  , f (x0 ) − mx0 Part (b) Now obtain the values of dx x0 Answer Part (c) Now obtain the equation of the tangent line Answer Try each part of this exercise Find the equation of the tangent line to the curve y = sin 3x at the point x = the tangent line intersects the x−axis. See the following figure.

π 4

and find where

y x tangent line

Part (a) Specify x0 and f Answer  df  , f (x0 ) − mx0 Part (b) Now obtain the values of dx x0 Answer Part (c) Now obtain the equation of the tangent line Answer Part (d) Where does the line intersect the x−axis? Answer

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

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4. The Normal Line to a Curve We have already noted that, at any point (x0 , y0 ) on a curve y = f (x) the tangent and normal lines are perpendicular. Thus if the equations of the tangent and normal lines are, respectively y = mx + c then m = − n1 or, equivalently n = − m1 . We have also noted, for the tangent line

y = nx + d

 df  m= dx x0

so n can easily be obtained. To find d, we again use the fact that the normal line y = nx + d and the curve have a point in common: y0 = nx0 + d

and

y0 = f (x0 )

so nx0 + d = f (x0 ) leading to d = f (x0 ) − nx0 . Try each part of this exercise Part (a) Find the equation of the normal line to curve y = sin 3x at the point x = π4 . See the earlier guided exercise for the value of ‘m’. Answer Part (b) Hence find the value of n The equation of the normal line is y =

Answer

√

2 x 3

+ d.

Part (c) Now find the value of d. (Remember the normal line and the curve pass through the same point. Answer Part (d) Now obtain the equation of the normal line. Answer The curve and the normal line are shown in the following figure: y

normal line x

Try each part of this exercise Find the equation of the normal line to the curve y = x3 at x = 1.  df  Part (a) First find f (x), x0 , , m, n dx x0 Answer 7

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

Part (b) Now use the property that the normal line y = nx + d and the curve y = x3 pass through the point (1,1) Answer Thus the equation of the normal line is y = − 13 x + 43 . The curve and the normal line are shown below: y

normal line x

More exercises for you to try 1. Find the equations of the tangent and normal lines to the following curves at the points indicated (a) y = x4 + 2x2 (1, 3) √ √ √ (b) y = 1 − x2 ( 22 , 22 ) (What would be obtained if the point was (1, 0)?) 1

(c) y = x 2

(1, 1)

2. Find the value of a (a = −1) if the two curves y = e−x and y = eax are to intersect at right-angles. Answer

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

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5. Computer Exercise or Activity

For this exercise it will be necessary for you to access the computer package DERIVE.

You can check, visually, on the examples, guided exercises and exercises in this block by using DERIVE to draw curves, tangent lines and normal lines. Make sure that the scales on the x− and y−axes are the same otherwise the angle between tangent and normal lines will appear not to be 90o . DERIVE has two commands: TANGENT (y, x, x0) and PERPENDICULAR (y, x, x0) which returns, respectively, the tangent and normal lines to a curve y(x) at the point x = x0. To access these commands the user must first open the library DIF APPS.MTH. Use the command File:Open and double click on the icon Dif apps. For example to find the tangent and normal lines to the curve y = x4 + 2x2 at x = 1 you would key in y := x ∧ 4 + 2x ∧ 2. DERIVE responds with y := x4 + 2.x2 Then key in TANGENT (y, x, 1). DERIVE responds with TANGENT (y, x, 1) If you now choose Simplify:Basic DERIVE responds with 8.x − 5 (so the equation of the tangent line is y = 8x − 5). Similarly, keying PERPENDICULAR (y, x, 1) and DERIVE responds with PERPENDICULAR (y, x, 1) Then hit Simplify:Basic as before to get the response 25−x 8

implying the equation of the normal line is y = − x8 +

9

25 . 8

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

End of Block 13.1

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

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(i) perpendicular

(ii) not perpendicular

(iii) perpendicular

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

y

θ

1

ψ 1

tangent line

x normal line

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

12

θ = 63.4o

ψ = 153.4o

Back to the theory

13

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

x0 = 0

f (x) = ex

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

14

df dx

x

=e

∴

 

df  dx 

=1

and f (0) − 1(0) = e0 − 0 = 1

0

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

y =x+1 Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

16

x0 =

π 4

f (x) = sin 3x

Back to the theory

17

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

df dx

∴

= 3 cos 3x

and f

π 4

−

mπ 4

 

df  dx 

= sin 3π − 4

π 4



= 3 cos 3π = − √32 = −2.12 4

−3 √ 2



π 4

=

√1 2

+

√3 π 24

= 2.37

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

18

y=

−3 √ x 2

+

1 √ (4 4 2

+ 3π) = −2.12x + 2.37

Back to the theory

19

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

when y = 0

∴

−2.12x + 2.37 = 0

∴

x = 1.12

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

20

m=

−3 √ 2

Back to the theory

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Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

nm = −1

∴

n=

√

2 3

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

22

√

2 3

π 4

+ d = sin π4

∴

d=

√1 2

−

√

2π 3 4

 0.34

Back to the theory

23

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

y = 0.47x + 0.34 Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

24

   df  2 f (x) = x , x0 = 1, = 3x  = 3  dx 1 1 1 ∴ m = 3 and n = − 3 3

Back to the theory

25

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

1=n+d

∴

d=1−n=1+

1 3

=

4 3

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

26

4

1. (a) f (x) = x + 2x

df dx

2

3

= 4x + 4x,

 

df  dx 

=8 x=1

tangent line y = 8x + c. This passes through (1, 3) so 3=8+c

∴

c = −5

∴

y = 8x − 5

normal line y = − 18 x + d. This passes through (1, 3) so 3 = − 18 + d

y = − 18 x + 25 . 8   √ df  √ −x (b) f (x) = 1 − x2 df = = −1 √ dx dx  1−x2 x= 22 √ √  tangent line y = −x + c. This passes through 22 , 22 so √ √ √ √ 2 2 = − +c ∴ c= 2 ∴ y = −x + 2 2 2 √ √  normal line y = x + d. This passes through 22 , 22 so √

∴

d=

25 8

∴

√

= 22 + d ∴ d = 0 ∴ y = x. At (1,0) the tangent line is x = 1 and the normal line is y = 0.   1 1 df df 1 −  = 2x 2 = 12 (c) f (x) = x 2 dx dx  2 2

x=1

tangent line: y = 12 x + c. This passes through (1,1) so 1 = 12 + c ∴ c = 12 ∴ y = 12 x + 12 normal line: y = −2x + d. This passes through (1,1) so 1 = −2 + d

∴

d=3

∴

y = −2x + 3.

2. The curves will intersect at right-angles if their tangent lines, at the point of intersection, are perpendicular. Point of intersection:

e−x = eax

The tangent line to y = eax

i.e. −x = ax

∴ x=0  is y = mx + c where m = aeax x=0 = a

 The tangent line to y = e−x is y = kx + g where k = −e−x x=0 = −1 These two lines are perpendicular if a(−1) = −1 27

i.e.

a = 1.

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

y y = e−x

y = ex

x

Back to the theory

Engineering Mathematics: Open Learning Unit Level 1 13.1: Tangents and Normals

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