CHAPTER 12 TRANSPOSING FORMULAE
EXERCISE 48 Page 97
1. Transpose a + b = c – d – e for d.
Since
a + b = c – d – e then d = c – e – a – b
2. Transpose y = 7x
for x.
Since y = 7x then x =
y
7
3. Transpose pv = c for v.
Since pv = c then v =
c
p
4. Transpose v = u + at for a.
Since
then
and
v = u + at
v – u = at
a=
v −u
t
5. Transpose V = IR for R.
Since V = IR then R =
V
I
6. Transpose x + 3y = t for y.
Since x + 3y = t then 3y = t – x
and dividing both sides by 3 gives:
y=
1
t−x
or =
y
(t − x )
3
3
163
© 2014, John Bird
7. Transpose c = 2πr for r.
Dividing both sides of c = 2πr by 2π gives:
c
c
= r or r =
2π
2π
8. Transpose y = mx + c for x.
Since
y = mx + c
then
y – c = mx
and
x=
y−c
m
9. Transpose I = PRT for
T.
Since
I = PRT
then
T=
I
PR
10. Transpose X L = 2π f L for L.
Since
X L = 2π f L
then
L=
11. Transpose I =
E
R
XL
2π r
for R.
Multiplying both sides of I =
E
by R gives: IR = E
R
and dividing both sides by I gives:
R=
164
E
I
© 2014, John Bird
x
+ 3 for x.
a
12. Transpose y=
x
+3
a
Since
y=
then
y–3=
and
x = m( y − 3)
13. Transpose F =
Rearranging F =
9
C + 32
5
x
a
for C.
9
C + 32 gives:
5
Multiplying both sides by
9
C
5
5
5
 5  9 
gives:
( F − 32 ) =
  C 
9
9
 9  5 
5
5
C
C or =
( F − 32 )
( F − 32 ) =
9
9
i.e.
1
2π f C
for f.
Since
XC =
1
2π f C
then
f=
14. Transpose X C =
F – 32 =
1
2π C X C
165
© 2014, John Bird
EXERCISE 49 Page 99
1. Transpose S =
a
1− r
for r.
a
by (1 – r) gives: S(1 – r) = a
1− r
Multiplying both sides of S =
i.e.
S – Sr = a
from which,
S – a = Sr
and dividing both sides by S gives:
S −a
=r
S
2. Transpose y =
λ(x − d )
λ (x − d)
yd = λ(x – d)
by d gives:
d
yd
Dividing both sides by λ gives:
λ
d+
and
Alternatively, from the first step,
yd = λ(x – d)
i.e.
yd = λx – λd
from which,
3. Transpose A =
i.e.
yd
λ
= x−d
= x or
x= d +
yd
λ
yd + λd = λx
and
Since A =
S −a
a
or r = 1 –
S
S
for x.
d
Multiplying both sides of y =
i.e. r =
x=
3( F − f )
L
yd + λ d
λ
=
d ( y + λ)
λ
i.e. x =
d
λ
( y + λ)
for f.
3( F − f )
then AL = 3(F – f)
L
AL = 3F – 3f
166
© 2014, John Bird
and
3f = 3F – AL
from which,
Since
f=
3F − AL
3
AL
3F − AL 3F AL
= −
=F −
3
3
3
3
4. Transpose y =
Since y =
A B2
5 CD
then f =
3F − AL
3
may also be written as f = F –
AL
3
A B2
for D.
5 CD
then D =
5. Transpose R = R 0 (1 + αt)
A B2
5 Cy
by cross-multiplying
for t.
Removing the bracket in=
R R0 (1 + α t ) gives: R
= R0 + R0 α t
from which,
R − R0 =
R0 α t
and
R − R0
=t
R0α
6. Transpose
1
1
1
=
+
R1
R2
R
or
R − R0
R0α
for R 2
1 1
1
Rearranging =
gives:
+
R R1 R2
1 1
1
− =
R R1 R2
i.e.
1
1 1 R1 − R
= − =
R2 R R1
R R1
Turning both sides upside down gives:
R2 =
7. Transpose I =
t=
R R1
R1 − R
E −e
for R.
R+r
Multiplying both sides by (R + r) gives:
I(R + r) = E – e
i.e.
and
IR+Ir=E–e
IR=E–e–Ir
167
© 2014, John Bird
and dividing both sides by I gives:
R=
E −e− I r
I
or R =
E −e
−r
I
8. Transpose y = 4ab2c2 for b.
Dividing both sides by 4a c 2 gives:
y
= b2
4ac 2
or
Taking the square root of both sides gives:
9. Transpose
Rearranging
b2
a2
+
=1
y2
x2
b2 =
y
4ac 2
y
4ac 2
b=
for x.
a 2 b2
+
=
1 gives:
x2 y 2
Turning both sides upside down gives:
a2
b2 y 2 − b2
=−
1
=
x2
y2
y2
x2
y2
=
a 2 y 2 − b2
 y2 
a2 y2
2
Multiplying both sides by a 2 gives: =
x 2 a=
 2 2
 y − b  y 2 − b2
Taking the square root of both sides gives:
a2 y2
x= =
y 2 − b2
i.e.
x=
10. Transpose for t = 2π
L
g
Dividing both sides of t = 2π
a2 y2
=
y 2 − b2
ay
y 2 − b2
ay
y 2 − b2
for L.
l
by 2π gives:
g
t
=
2π
l
g
2
Squaring both sides gives:
l
 t 

 =
g
 2π 
or
 t 
l = g

 2π 
Multiplying both sides by g gives:
168
l  t 
=

g  2π 
2
or
l=
2
g t2
4π 2
© 2014, John Bird
11. Transpose v2 = u2 + 2as for u.
Since v2 = u2 + 2as then v2 – 2as = u2
from which,
12. Transpose A =
u=
π R 2θ
360
v 2 − 2as
for R.
Multiplying both sides of A =
π R 2θ
360
by 360 gives:
360A = π R 2θ
360A
and dividing both sides by πθ gives:
πθ
Taking the square root of both sides gives:
13. Transpose N =
a+x

 for
y


Squaring both sides of N =
R2 =
or
360A
πθ
360A
R=
πθ
a.
a+x
gives:
y
N2 =
a+x
y
N 2 y= a + x
Multiplying both sides by y gives:
from which,
14. Transpose Z =
= R2
or a + x = N 2 y
a = N2y – x
R 2 + (2π fL) 2 for L, and evaluate L when Z = 27.82, R = 11.76 and f = 50
Squaring both sides of Z =
R 2 + ( 2π fL )
2
gives:
2
Z=
R 2 + ( 2π fL )
from which,
Z 2 − R2 =
( 2π fL )
Taking the square root of both sides gives:
Z 2 − R2 =
2π fL
Dividing both sides by 2πf gives:
L=
169
2
2
or
2π=
fL
Z 2 − R2
Z 2 − R2
2π f
© 2014, John Bird
15. The lift force, L, on an aircraft is given by:
L=
1 2
ρ v ac where ρ is the density, v is the
2
velocity, a is the area and c is the lift coefficient. Transpose the equation to make the velocity
the subject.
Since
L=
1 2
ρ v ac
2
from which,
then
2L
= v2
ρ ac
velocity, v =
2L
ρ ac
16. The angular deflection θ of a beam of electrons due to a magnetic field is given by:
 HL 
 . Transpose the equation for V.
 1 
2
V 
θ =k
 HL 
Since θ = k  1 


V2 
from which,
1
then
V2 =
kHL
θ
kH L
V= 

 θ 
2
or V =
170
k 2 H 2 L2
θ2
© 2014, John Bird
EXERCISE 50 Page 101
1. Transpose y =
a2m − a2n
x
Multiplying both sides of y =
for a.
a2m − a2n
by x gives:
x
xy = a 2 m − a 2 n
xy = a 2 ( m − n )
and factorizing gives:
xy
= a2
m−n
Dividing both sides by (m – n) gives:
2. Transpose M = π(R4 – r4)
for R.
Dividing both sides of M = π ( R 4 − r 4 ) by π gives:
M
and rearranging gives:
π
Taking the fourth root of both sides gives:
3. Transpose x + y =
r
3+ r
xy
m−n
a=
Taking the square root of both sides gives:
xy
m−n
a2 =
or
M
= R4 − r 4
π
+ r4 =
R4
R=
4
or
M
π
4
R=
M
π
+ r4
+ r4
for r.
Multiplying both sides of x + y =
r
by (3 + r) gives:
3+ r
Multiplying the brackets gives:
(x + y)(3 + r) = r
3x + xr + 3y + yr = r
and rearranging gives:
xr + yr – r = –3x – 3y
Factorizing gives:
r(x + y – 1) = –3(x + y)
Dividing both sides by (x + y – 1) gives:
r=
Multiplying numerator and denominator by –1 gives:
171
r=
−3( x + y )
x + y −1
3( x + y )
1− x − y
© 2014, John Bird
4. Transpose m =
µL
L + rCR
for L.
Multiplying both sides of m =
µL
L+rC R
by (L + rCR) gives:
m(L + rCR) = µL
mL + mrCR = µL
Removing brackets gives:
and rearranging gives:
mrCR = µL – mL
Factorizing gives:
mrCR = L(µ – m)
Dividing both sides by (µ – m) gives:
5. Transpose a2 =
b2 − c2
b2
L=
for b.
Multiplying both sides by b 2 gives:
2
a 2b=
b2 − c2
2
c=
b 2 − a 2b 2
and rearranging gives:
or
b 2 − a 2b 2 =
c2
b 2 (1 − a 2 ) =
c2
Factorizing gives:
Dividing both sides by (1 − a 2 ) gives:
b2 =
c2
1 − a2
Taking the square root of both sides gives:
b=
c2
c2
=
1 − a2
1 − a2
Hence,
b=
c
1 − a2
6. Transpose
x 1+ r2
=
y 1− r2
for r.
Rearranging by cross-multiplying gives:
Removing brackets gives:
and rearranging gives:
mrCR
µ −m
x (1 − r 2 ) = y (1 + r 2 )
x − xr 2 =y + yr 2
x − y= yr 2 + xr 2 or
172
yr 2 + xr 2 =−
x y
© 2014, John Bird
r 2 ( x + y ) =x − y
Factorizing gives:
Dividing both sides by (x + y) gives:
r2 =
Taking the square root of both sides gives:
r=
7. Transpose
p
=
q
 a + 2b 


 a − 2b 
x− y
x+ y
for b.
2
Squaring both sides gives:
x− y
x+ y
 p
a + 2b
  =
a − 2b
q
i.e.
p 2 a + 2b
=
q 2 a − 2b
Rearranging by cross-multiplying gives:
p 2 ( a − 2b ) = q 2 ( a + 2b )
Removing brackets gives:
ap 2 − 2bp 2 =aq 2 + 2bq 2
ap 2 − aq 2 = 2bq 2 + 2bp 2
Rearranging gives:
2bq 2 + 2bp 2 =ap 2 − aq 2
or
2b(q 2 + p 2 ) = a ( p 2 − q 2 )
Factorizing gives:
Dividing both sides by 2 ( p 2 + q 2 ) gives:
b=
8. A formula for the focal length, f, of a convex lens is:
a ( p2 − q2 )
2 ( p2 + q2 )
1
1
1
= + . Transpose the formula to make
f
u
v
v the subject and evaluate v when f = 5 and u = 6
Rearranging
1 1 1 u− f
= − =
v f u
uf
1 1 1
=
+ gives:
f u v
Turning each side upside down gives:
When f = 5 and u = 6, then
v=
uf
u− f
uf
(6)(5) 30
= 30
v= = =
u− f
6−5
1
173
© 2014, John Bird
9. The quantity of heat Q is given by the formula Q = mc(t 2 – t1 ). Make t 2 the subject of the formula
and evaluate t 2 when m = 10, t1 = 15, c = 4 and Q = 1600
Removing the brackets in Q = mc ( t2 − t1 ) gives:
Q = mct2 − mct1
Q + mct1 = mct2
and rearranging gives:
mct2= Q + mvt1
or
Dividing both sides by mc gives:
t2 =
Q + mvt1
mc
or =
t2
Q
+ t1
mc
or
t2= t1 +
Q
mc
When m = 10, t1 = 15, c = 4 and Q = 1600,
t2 = t1 +
Q
1600
1600
=15 +
=15 +
=15 + 40 = 55
mc
(10)(4)
40
10. The velocity v of water in a pipe appears in the formula h =
0.03Lv 2
. Express v as the subject
2dg
of the formula and evaluate v when h = 0.712, L = 150, d = 0.30 and g = 9.81
Multiplying both sides of h =
0.03Lv 2
by 2dg gives: 2dgh = 0.03L v 2
2dg
2dgh
= v2
0.03L
Dividing both sides by 0.03L gives:
Taking the square root of each side gives:
v=
or
v2 =
2dgh
0.03L
2dgh
0.03L
When h = 0.712, L = 150, d = 0.30 and g = 9.81,
2dgh
v=
=
0.03L
2(0.30)(9.81)(0.712)
=
0.03(150)
0.931296 = 0.965
 3d (l − d ) 

.
8


11. The sag S at the centre of a wire is given by the formula S =
Make l the subject of the formula and evaluate l when d = 1.75 and S = 0.80
Squaring both sides of S =
3 d (l − d )
gives:
8
S2 =
174
3d ( l − d )
8
© 2014, John Bird
Multiplying both sides by 8 gives:
8=
S 2 3d ( l − d )
Removing the bracket gives:
2
8 S=
3dl − 3d 2
8 S 2 + 3d 2 =
3dl
Rearranging gives:
dl 8 S 2 + 3d 2
3=
or
Dividing both sides by 3d gives:
l=
8S 2 + 3d 2 8S 2 3d 2
=
+
3d
3d
3d
i.e.
l=
8S 2
+d
3d
When d = 1.75 and S = 0.80,
l=
8S 2
8(0.80) 2
+=
d
+ 1.75
= 0.975 + 1.75 = 2.725
3d
3(1.75)
12. In an electrical alternating current circuit the impedance Z is given by:
Z=
2

1  

2 + ωL −
R


  . Transpose the formula to make C the subject and hence evaluate C
C
ω

 

when Z = 130, R = 120, ω = 314 and L = 0.32
1 

Squaring both sides of Z = R +  ω L −
ωC 

2
2
1 

Z =R +  ω L −
ωC 

gives:
2
Rearranging gives:
1 

Z − R = ωL −
ωC 

Taking the square root of both sides gives:
Z 2 − R2 = ω L −
2
2
2
2
2
1
ωC
1
=ω L − Z 2 − R 2
ωC
Rearranging gives:
{
Multiplying both sides by ω gives:
1
= ω ω L − Z 2 − R2
C
Turning both sides upside down gives:
C=
{
}
1
ω ω L − Z 2 − R2
}
When Z = 130, R = 120, ω = 314 and L = 0.32,
175
© 2014, John Bird
C=
1
1
1
1
=
= =
314 {100.48 − 50} 15 850.72
ω ω L − Z 2 − R2
314 (314)(0.32) − 1302 − 1202
{
}
{
}
= 0.0000631 = 63.1×10−6
13. An approximate relationship between the number of teeth T on a milling cutter, the diameter
of cutter D and the depth of cut d is given by T =
12.5 D
. Determine the value of D when T =
D + 4d
10 and d = 4 mm.
Multiplying both sides of T =
12.5 D
by D + 4d gives:
D + 4d
Removing brackets gives:
T(D + 4d) = 12.5D
TD + 4dT = 12.5D
Rearranging gives:
4dT = 12.5D – TD
or
12.5D – TD = 4dT
Factorizing gives:
D(12.5 – T) = 4dT
Dividing both sides by (12.5 – T) gives:
D=
4dT
12.5 − T
4dT
4(4)(10) 160
When T = 10 and d = 4 mm, then D = = =
= 64 mm
12.5 − T 12.5 − 10 2.5
14. Make λ, the wavelength of X-rays, the subject of the following formula:
µ C Z 4 λ5 n
=
ρ
a
Rearranging (by cross-multiplying) gives:
a µ = ρ CZ 4 λ 5 n
and
λ5 =
i.e.
λ2 =
5
176
aµ
ρ CZ 4 n
aµ
ρ CZ 4 n
© 2014, John Bird
2
2
5
 aµ  5
 
λ  =  ρ CZ 4 n 


5
2
and
 aµ 
λ= 

 ρ CZ 4 n 
from which,
2
5
15. A simply supported beam of length L has a centrally applied load F and a uniformly distributed
load of w per metre length of the beam. The reaction at the beam support is given by:
R=
1
( F + wL )
2
Rearrange the equation to make w the subject. Hence determine the value of w when L = 4 m,
F = 8 kN and R = 10 kN.
Since
R=
1
( F + wL ) then
2
2R = F + wL
and
2R – F = wL
from which,
w=
2R − F
L
When L = 4 m, F = 8 kN and R = 10 kN,
w=
2(10) − 8 12
= = 3 kN/m
4
4
16. The rate of heat conduction through a slab of material, Q, is given by the formula
Q=
kA(t1 − t2 )
where t1 and t2 are the temperatures of each side of the material, A is the area
d
of the slab, d is the thickness of the slab, and k is the thermal conductivity of the material.
Rearrange the formula to obtain an expression for t2
Since
i.e.
Q=
kA(t1 − t2 )
then
d
Qd = kA ( t1 − t2 )
Qd
= t1 − t2
kA
177
© 2014, John Bird
from which,
t2 = t1 –
Qd
kA
 rω 
17. The slip s of a vehicle is given by: s = 1 −
 ×100% where r is the tyre radius, ω is the
v 

angular velocity and v the velocity. Transpose to make r the subject of the formula.
Since
 rω 
s = 1 −
 ×100%
v 

then
s
rω
=1–
100
v
rω
s
=1–
v
100
and
from which,
r=
v
s 
1 −
ω  100 
18. The critical load, F newtons, of a steel column may be determined from the formula
L
F
= nπ where L is the length, EI is the flexural rigidity, and n is a positive integer.
EI
Transpose for F and hence determine the value of F when n = 1, E = 0.25 ×1012 N/m 2 ,
I = 6.92 ×10−6 m 4 and L = 1.12 m.
Since
L
F
= nπ
EI
F
nπ
=
EI
L
then
F  nπ 
=

EI  L 
and
2
 nπ 
F = EI 

 L 
i.e.
2
When n = 1, E = 0.25 ×1012 N/m 2 , I = 6.92 ×10−6 m 4 and L = 1.12 m,
1× π 
 nπ 
6
load, F = EI 
( 0.25 ×1012 )( 6.92 ×10−6 ) 
 =
 = 13.61 ×10 N = 13.61 MN
 L 
 1.12 
2
2
19. The flow of slurry along a pipe on a coal processing plant is given by: V =
π pr 4
8η 
Transpose the equation for r.
178
© 2014, John Bird
Since V =
π pr 4
8η 
then
8η V = π pr 4
8η V
= r4
πp
and
from which,
r=
4
 8η  V 


 π p 
3
20. The deflection head H of a metal structure is given by: H =
I ρ 4 D 2 2
20 g
Transpose the formula for length  .
3
3
Since
H=
I ρ 4 D 2 2
20 g
then
I ρ 4 D2  2
H2 =
20 g
3
and
20 g H 2 = I ρ 4 D 2  2
i.e.
3
20 g H 2
= 2
I ρ4 D
from which,
 20 g H 2 
= 3

 I ρ4 D 
179
2
© 2014, John Bird