Higher Engineering Mathematics

SOME APPLICATIONS OF DIFFERENTIATION 311

G

Problem 23. Determine the equations of the

tangent and normal to the curvey=

x^3

5

at the

point

(

−1,−

1

5

)

Gradientmof curvey=

x^3

5

is given by

m=

dy

dx

=

3 x^2

5

At the point

(

−1,−^15

)

,x=−1 andm=

3(−1)^2

5

=

3
5
Equation of the tangentis:

y−y 1 =m(x−x 1 )

i.e. y−

(

−

1

5

)

=

3

5

(x−(−1))

i.e. y+

1

5

=

3

5

(x+1)

or 5 y+ 1 = 3 x+ 3

or 5 y− 3 x= 2

Equation of the normal is:

y−y 1 =−

1

m

(x−x 1 )

i.e. y−

(

−

1

5

)

=

− 1

(3/5)

(x−(−1))

i.e. y+

1

5

=−

5

3

(x+1)

i.e. y+

1

5

=−

5

3

x−

5

3

Multiplying each term by 15 gives:

15 y+ 3 =− 25 x− 25

Henceequation of the normalis:

15 y+ 25 x+ 28 = 0

Now try the following exercise.

Exercise 126 Further problems on tangents

and normals

For the curves in problems 1 to 5, at the points

given, find (a) the equation of the tangent, and

(b) the equation of the normal.

1.y= 2 x^2 at the point (1, 2)

[

(a)y= 4 x− 2

(b) 4y+x= 9

]

2.y= 3 x^2 − 2 xat the point (2, 8)

[

(a)y= 10 x− 12

(b) 10y+x= 82

]

3.y=

x^3

2

at the point

(

−1,−

1

2

)

[

(a) y=^32 x+ 1

(b) 6y+ 4 x+ 7 = 0

]

4.y= 1 +x−x^2 at the point (−2,−5)

[

(a) y= 5 x+ 5

(b) 5y+x+ 27 = 0

]

5.θ=

1

t

at the point

(

3,

1

3

)

[

(a) 9θ+t= 6

(b)θ= 9 t− 2623 or 3θ= 27 t− 80

]

28.6 Small changes

Ifyis a function ofx, i.e.y=f(x), and the approxi-

mate change inycorresponding to a small change

δxinxis required, then:

δy

δx

≈

dy

dx

andδy≈

dy

dx

·δx or δy≈f′(x)·δx

Problem 24. Giveny= 4 x^2 −x, determine the

approximate change inyifxchanges from 1 to

1.02.

Sincey= 4 x^2 −x, then

dy

dx

= 8 x− 1

Approximate change iny,

δy≈

dy

dx

·δx≈(8x−1)δx

Whenx=1 andδx= 0 .02,δy≈[8(1)−1](0.02)

≈ 0. 14