Basic Engineering Mathematics, Fifth Edition

Transposing formulae 87

Taking the cube root of both sides gives

√ 3

r^3 =^3

√(

3 V

2 π

)

i.e. r=^3

√(

3 V

2 π

)

(b) WhenV=32cm^3 ,

radiusr=^3

√(

3 V

2 π

)

=^3

√(

3 × 32

2 π

)

=2.48cm.

Now try the following Practice Exercise

PracticeExercise 47 Further transposing

formulae (answers on page 345)

Make the symbol indicated the subject of each of

theformulaeshowninproblems1to13andexpress

each in its simplest form.

1. S=

a

1 −r

(r)

2. y=

λ(x−d)

d

(x)

3. A=

3 (F−f)

L

(f)

4. y=

AB^2

5 CD

(D)

5. R=R 0 ( 1 +αt)(t)

6.

1

R

=

1

R 1

+

1

R 2

(R 2 )

7. I=

E−e

R+r

(R)

8. y= 4 ab^2 c^2 (b)

9.

a^2

x^2

+

b^2

y^2

= 1 (x)

10. t= 2 π

√

L

g

(L)

11. v^2 =u^2 + 2 as (u)


12. A=

πR^2 θ

360

(R)

13. N=

√(

a+x

y

)

(a)

14. TransposeZ=

√

R^2 +( 2 πfL)^2 forLand

evaluateLwhenZ= 27. 82 ,R= 11 .76 and

f=50.

12.4 More difficult transposing of formulae

Here are some more transposition examples to help

us further understand how more difficult formulae are

transposed.

Problem 18. (a) TransposeS=

√

3 d(L−d)

8

to

makelthe subject. (b) EvaluateLwhend= 1. 65

andS= 0. 82

The formulaS=

√

3 d(L−d)

8

represents the sagSat

the centre of a wire.

(a) Squaring both sides gives S^2 =

3 d(L−d)

8

Multiplying both sides by 8 gives

8 S^2 = 3 d(L−d)

Dividing both sides by 3dgives

8 S^2

3 d

=L−d

Rearranging gives L=d+

8 S^2

3 d

(b) Whend= 1 .65 andS= 0 .82,

L=d+

8 S^2

3 d

= 1. 65 +

8 × 0. 822

3 × 1. 65

= 2. 737

Problem 19. Transpose the formula

p=

a^2 x^2 +a^2 y

r

to makeathe subject

Rearranging gives

a^2 x^2 +a^2 y

r

=p

Multiplying both sides byrgives

a^2 x+a^2 y=rp

Factorizing the LHS gives a^2 (x+y)=rp

Dividing both sides by(x+y)gives

a^2 (x+y)

(x+y)

=

rp

(x+y)