Transposing formulae 87
Taking the cube root of both sides gives
3
√
r 3 = 3
3V
2π
i.e.
r = 3
3V
2π
(b) When V = 32cm 3 ,
radius r = 3
3V
2π
= 3
3 × 32
2π
= 2.48 cm.
Now try the following Practice Exercise
Practice Exercise 47 Further transposing
formulae (answers on page 345)
Make the symbol indicated the subject of each of
the formulae shown in problems 1 to 13 and express
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
5C D
(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 = 4ab 2 c 2
(b)
9.
a 2
x 2 +
b 2
y 2 = 1
(x)
10. t = 2π
L
g
(L)
11. v 2 = u 2 + 2as
(u)
12. A =
π R 2 θ
360
(R)
13. N =
a + x
y
(a)
14. Transpose Z =
R 2 + (2π f L) 2 for L and
evaluate L when Z = 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) Transpose S =
3d (L − d)
8
to
make l the subject. (b) Evaluate L when d = 1.65
and S = 0.82
The formula S =
3d (L − d)
8
represents the sag S at
the centre of a wire.
(a) Squaring both sides gives
S
2
=
3d(L − d)
8
Multiplying both sides by 8 gives
8S
2
= 3d(L − d)
Dividing both sides by 3d gives
8S 2
3d
= L − d
Rearranging gives
L = d +
8S 2
3d
(b) When d = 1.65 and S = 0.82,
L = d +
8 S 2
3d
= 1.65 +
8 × 0.82 2
3 × 1.65
= 2.737
Problem 19. Transpose the formula
p =
a
2 x
2
+ a
2 y
r
to make a the subject
Rearranging gives
a 2 x 2 + a 2 y
r
= p
Multiplying both sides by r gives
a
2 x + a
2 y = r p
Factorizing the LHS gives
a
2
(x + y) = r p
Dividing both sides by (x + y) gives
a 2 (x + y)
(x + y)
=
r p
(x + y)