Basic Engineering Mathematics, Fifth Edition

86 Basic Engineering Mathematics

Whenever the prospective new subject is a squared

term,thattermisisolatedontheLHSandthenthesquare

root of both sides of the equation is taken.

Multiplying both sides by 2 gives mv^2 = 2 k

Dividing both sides bymgives mv

2

m

=

2 k

m

Cancelling gives v^2 =

2 k

m

Taking the square root of both sides gives

√

v^2 =

√(

2 k

m

)

i.e. v=

√(

2 k

m

)

Problem 13. In a right-angled triangle having

sidesx,yand hypotenusez, Pythagoras’ theorem

statesz^2 =x^2 +y^2. Transpose the formula to findx

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. Transposey=

ML^2

8 EI

to makeLthe

subject

Multiplying both sides by 8EIgives 8 EIy=ML^2

Dividing both sides byMgives

8 EIy

M

=L^2

or L^2 =

8 EIy

M

Taking the square root of both sides gives

√

L^2 =

√

8 EIy

M

i.e.

L=

√

8 EIy

M

Problem 15. Givent= 2 π

√

l

g

,findgin terms of

t,landπ

Whenever the prospective new subject is withina square

root sign, it is best to isolate that term on the LHS and

then to square both sides of the equation.

Rearranging gives 2 π

√

l

g

=t

Dividing both sides by 2πgives

√

l

g

=

t

2 π

Squaring both sides gives

l

g

=

(

t

2 π

) 2

=

t^2

4 π^2

Cross-multiplying, (i.e. multiplying

each term by 4π^2 g), gives 4 π^2 l=gt^2

or gt^2 = 4 π^2 l

Dividing both sides byt^2 gives

gt^2

t^2

=

4 π^2 l

t^2

Cancelling gives g=

4 π^2 l

t^2

Problem 16. The impedanceZof an a.c. circuit

is given byZ=

√

R^2 +X^2 whereRis 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 volumeVof a hemisphere of

radiusris given byV=

2

3

πr^3 .(a)Findrin terms

ofV. (b) Evaluate the radius whenV=32cm^3

(a) Rearranging gives

2

3

πr^3 =V

Multiplying both sides by 3 gives 2πr^3 = 3 V

Dividing both sides by 2πgives

2 πr^3

2 π

=

3 V

2 π

Cancelling gives r^3 =

3 V

2 π