differentiation

310 DIFFERENTIAL CALCULUS
10. An electrical voltage E is given by
E = (15 sin 50πt + 40 cos 50πt) volts,
where t is the time in seconds. Determine
the maximum value of voltage.
[42.72 volts]
11. The fuel economy E of a car, in miles per
gallon, is given by:
E = 21 + 2.10 × 10 −2 v 2
− 3.80 × 10 −6 v 4
where v is 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 curve y = f (x)atthe
point (x 1 , y 1 ) is given by:
y − y 1 = m(x − x 1 )
where m = dy
dx = gradient of the curve at (x 1, y 1 ).
Problem 21. Find the equation of the tangent
to the curve y = x 2 − x − 2 at the point (1, −2).
Gradient, m
= dy
dx = 2x − 1
At the point (1, −2), x = 1 and m = 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 of y = x 2 − x − 2 is shown in Fig. 28.12.
The line AB is the tangent to the curve at the point C,
i.e. (1, −2), and the equation of this line is y = 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 line CD is the
normal.
It may be shown that if two lines are at right angles
then the product of their gradients is −1. Thus if m
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 curve y = 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 (x − 1)
1
i.e. y + 2 =−x + 1
or y =−x − 1
Thus the line CD in Fig. 28.12 has the equation
y =−x − 1.