Maths bridge › Rearranging equations
Rearranging equations
Changing the subject of an equation, reliably, including through squares and roots.
One rule, applied in reverse order
An equation is a stack of operations applied to the quantity you want. Undo them to get it out. Apply the inverse of each operation to both sides, in the reverse of the order they went on. Adding undoes subtracting, dividing undoes multiplying, and a square root undoes a square.
Formula triangles stop at three symbols and cannot survive A-level equations. The both-sides method never runs out, so practise it on easy cases until your hand does it without asking.
Through brackets, squares and roots
WORKED EXAMPLE
Freeing a buried symbol
Solve ε = I(R + r) for r.
The r sits inside a bracket that is multiplied by I. Undo the outermost layer first, so divide both sides by I, giving ε/I = R + r.
Now subtract R and you have r = ε/I − R. Two moves, each applied to both sides, outermost first.
WORKED EXAMPLE
When the subject is under a root
The pendulum equation is T = 2π√(l/g). Solve it for g.
The g hides inside a root, and a root is undone by squaring, so square both sides before any dividing at all. That gives T2 = 4π2l/g.
Now g is a simple divisor: multiply both sides by g, divide by T2, and g = 4π2l/T2.
A pendulum of length 0.99 m with a period of 2.0 s gives g = 4π2 × 0.99/2.02 = 9.77 m s−2, which is the sanity check the algebra deserves.
GUIDED PRACTICE
A rearrangement under pressure
The time constant relation Q = Q0e−t/RC will meet you in the capacitance unit. Without touching the exponential yet, solve the simpler v2 = u2 + 2as for a, then check your result's units make sense.
Show the working
Subtract u2 from both sides, then divide by 2s: a = (v2 − u2)/2s.
Now check the units. (m s−1)2 divided by metres is m2 s−2/m = m s−2, an acceleration. Algebra and units agree, so the rearrangement stands.
Where physics leans on this: EMF and internal resistance · Motion graphs and SUVAT · Pendulums and springs. All eight skills: the maths bridge.