Higher Engineering Mathematics

310 DIFFERENTIAL CALCULUS

10. An electrical voltageEis given by
    E=(15 sin 50πt+40 cos 50πt) volts,
    wheretis the time in seconds. Determine
    the maximum value of voltage.
    [42.72 volts]


11. The fuel economyEof a car, in miles per
    gallon, is given by:

E= 21 + 2. 10 × 10 −^2 v^2

− 3. 80 × 10 −^6 v^4

wherevis the speed of the car in miles per

hour.

Determine, correct to 3 significant figures,

the most economical fuel consumption, and

the speed at which it is achieved.

[50.0 miles/gallon, 52.6 miles/hour]

28.5 Tangents and normals

Tangents

The equation of the tangent to a curvey=f(x)atthe
point (x 1 ,y 1 ) is given by:

y−y 1 =m(x−x 1 )

wherem=

dy

dx

=gradient of the curve at (x 1 ,y 1 ).

Problem 21. Find the equation of the tangent

to the curvey=x^2 −x−2 at the point (1,−2).

Gradient,m

=

dy

dx

= 2 x− 1

At the point (1,−2),x=1 andm=2(1)− 1 =1.
Hence the equation of the tangent is:

y−y 1 =m(x−x 1 )

i.e. y−(−2)=1(x−1)

i.e. y+ 2 =x− 1

or y=x− 3

The graph ofy=x^2 −x−2 is shown in Fig. 28.12.
The lineABis the tangent to the curve at the pointC,
i.e. (1,−2), and the equation of this line isy=x−3.

Figure 28.12

Normals

The normal at any point on a curve is the line which

passes through the point and is at right angles to

the tangent. Hence, in Fig. 28.12, the lineCDis the

normal.

It may be shown that if two lines are at right angles

then the product of their gradients is−1. Thus ifm

is the gradient of the tangent, then the gradient of the

normal is−

1

m

Hence the equation of the normal at the point (x 1 ,y 1 )

is given by:

y−y 1 =−

1

m

(x−x 1 )

Problem 22. Find the equation of the normal

to the curvey=x^2 −x−2 at the point (1,−2).

m=1 from Problem 21, hence the equation of the

normal is

y−y 1 =−

1

m

(x−x 1 )

i.e. y−(−2)=−

1

1

(x−1)

i.e. y+ 2 =−x+ 1

or y=−x− 1

Thus the lineCDin Fig. 28.12 has the equation

y=−x−1.