86 Basic Engineering Mathematics
Whenever the prospective new subject is a squared
term, that term is isolated on the LHS and then the square
root of both sides of the equation is taken.
Multiplying both sides by 2 gives
mv
2
= 2k
Dividing both sides by m gives
mv 2
m
=
2k
m
Cancelling gives
v
2
=
2k
m
Taking the square root of both sides gives
√
v 2 =
2k
m
i.e.
v =
2k
m
Problem 13. In a right-angled triangle having
sides x, y and hypotenuse z, Pythagoras’ theorem
states z 2 = x 2 + y 2 . Transpose the formula to find x
Rearranging gives
x
2
+ y
2
= z
2
and
x
2
= z
2
− y
2
Taking the square root of both sides gives
x =
z 2 − y 2
Problem 14. Transpose y =
ML 2
8EI
to make L the
subject
Multiplying both sides by 8E I gives 8EIy = ML
2
Dividing both sides by M gives
8EIy
M
= L
2
or
L
2
=
8EIy
M
Taking the square root of both sides gives
√
L 2 =
8EIy
M
i.e.
L =
8EIy
M
Problem 15. Given t = 2π
l
g
, find g in terms of
t, l and π
Whenever the prospective new subject is within a square
root sign, it is best to isolate that term on the LHS and
then to square both sides of the equation.
Rearranging gives
2π
l
g
= t
Dividing both sides by 2π gives
l
g
=
t
2π
Squaring both sides gives
l
g
=
t
2π
2
=
t 2
4π 2
Cross-multiplying, (i.e. multiplying
each term by 4π 2 g), gives
4π
2 l = gt
2
or
gt
2
= 4π
2 l
Dividing both sides by t 2 gives
gt 2
t 2 =
4π 2 l
t 2
Cancelling gives
g =
4π 2 l
t 2
Problem 16. The impedance Z of an a.c. circuit
is given by Z =
√
R 2 + X 2 where R is the
resistance. Make the reactance, X , the subject
Rearranging gives
R 2 + X 2 = Z
Squaring both sides gives
R
2
+ X
2
= Z
2
Rearranging gives
X
2
= Z
2
− R
2
Taking the square root of both sides gives
X =
Z 2 − R 2
Problem 17. The volume V of a hemisphere of
radius r is given by V =
2
3
πr 3 . (a) Find r in terms
of V. (b) Evaluate the radius when V = 32 cm 3
(a) Rearranging gives
2
3
πr
3
= V
Multiplying both sides by 3 gives 2πr
3
= 3V
Dividing both sides by 2π gives
2πr 3
2π
=
3V
2π
Cancelling gives
r
3
=
3V
2π