From Zero to Engineer

Evaluating expressions and rearranging formulae

Substituting data into a formula and rounding the result, dependent and independent variables, and rearranging a formula to make any of its quantities the subject.

A formula on its own computes nothing - it computes only once numbers are put in place of the letters. This lesson shows both sides of that. First, straightforward substitution: we have the formula and every value except the one it gives directly. Then the situation that is more common in engineering practice: the quantity we want sits inside the formula, under a root or in a denominator, so the formula has to be rearranged first and only then can the numbers go in.

Evaluating expressions

Substitute the data for the letters and work through in the usual order of operations. Round only at the end, to the accuracy the exercise asks for.

2\pi\sqrt{\frac{l}{g}}

\pi = 3.1415,\quad l = 2,\quad g = 9.81

2\cdot 3.1415\sqrt{\frac{2}{9.81}} = 6.2830\sqrt{0.2039} = 6.2830\cdot 0.4516 = 2.84

correct to two decimal places

V = \frac{\pi h}{6}(3R^{2} + h^{2})

h = 2.85,\quad R = 6.24,\quad \pi = 3.142

\begin{aligned}V &= \frac{3.142\cdot 2.85}{6}\left(3\cdot 6.24^{2} + 2.85^{2}\right) = 1.492\left(116.8128 + 8.1225\right) = \\&= 1.492\cdot 124.9353 = 186.40\end{aligned}

3\cdot 6.24^{2} = 116.8128

2.85^{2} = 8.1225

116.8128 + 8.1225 = 124.9353

correct to two decimal places

R = \frac{R_{1}R_{2}}{R_{1} + R_{2}}

R_{1} = 276,\quad R_{2} = 145

R = \frac{276\cdot 145}{276 + 145} = \frac{40\,020}{421} = 95.06

\frac{276\cdot 145}{276 + 145} = \frac{40020}{421}

correct to two decimal places

V = \frac{\pi b}{12}(D^{2} + Dd + d^{2})

b = 1.46,\quad D = 0.864,\quad d = 0.517,\quad \pi = 3.142

\begin{aligned}V &= \frac{3.142\cdot 1.46}{12}\left(0.864^{2} + 0.864\cdot 0.517 + 0.517^{2}\right) \\V &= 0.382\left(0.746 + 0.447 + 0.267\right) \\V &= 0.382\cdot 1.46 = 0.558\end{aligned}

correct to three significant figures

Dependent and independent variables

In every formula one letter stands alone on the left and its value follows from the rest. That is the dependent variable. The letters on the right, the ones we substitute values for, are the independent variables.

r = 2s^{3} + 3t

r dependent, s and t independent

Evaluating independent variables

When the quantity we want is not alone on the left, we rearrange the formula first and substitute only afterwards. Each operation is applied to both sides - the note after the slash says what is being done to the whole equation.

Calculate l.

T = 2\pi\sqrt{\frac{l}{g}}

\pi = 3.14,\quad g = 9.81,\quad T = 1.03

\begin{aligned}T &= 2\pi\sqrt{\frac{l}{g}} \;\Big/ : 2\pi \\\frac{T}{2\pi} &= \sqrt{\frac{l}{g}} \;\Big/ ^{2} \\\frac{T^{2}}{4\pi^{2}} &= \frac{l}{g} \;\Big/ \cdot g \\l &= \frac{gT^{2}}{4\pi^{2}}\end{aligned}

\begin{aligned}l &= \frac{9.81\cdot 1.03^{2}}{4\cdot 3.14^{2}} = \frac{9.81\cdot 1.0609}{4\cdot 9.8596} \\l &= 0.264\end{aligned}

1.03^{2} = 1.0609

3.14^{2} = 9.8596

correct to three significant figures

Calculate r.

J = \frac{nE}{R + nr}

n = 6,\quad E = 2.01,\quad R = 12,\quad J = 0.98

\begin{aligned}J &= \frac{nE}{R + nr} \;\Big/ \cdot (R + nr) \\J(R + nr) &= nE \\JR + Jnr &= nE \\Jnr &= nE - JR \;\Big/ : Jn \\r &= \frac{nE - JR}{Jn}\end{aligned}

r = \frac{6\cdot 2.01 - 0.98\cdot 12}{0.98\cdot 6} = \frac{12.06 - 11.76}{5.88} = 0.051

6\cdot 2.01 = 12.06

0.98\cdot 12 = 11.76

0.98\cdot 6 = 5.88

correct to two significant figures

Rearranging formulae

Rearrange the formula to make a the dependent variable.

\begin{aligned}v &= u + at \\v - u &= at \;\Big/ : t \\a &= \frac{v - u}{t}\end{aligned}

Make u the dependent variable.

\begin{aligned}a &= \frac{2(ut - s)}{t^{2}} \;\Big/ \cdot t^{2} \\at^{2} &= 2ut - 2s \\at^{2} + 2s &= 2ut \;\Big/ : 2t \\u &= \frac{at^{2} + 2s}{2t}\end{aligned}

Frequently asked questions

What is the difference between a dependent and an independent variable?

In r = 2s³ + 3t the letter r is the dependent variable, because its value follows from the others, while s and t are the independent ones - those are what we substitute. Rearranging the formula swaps the roles.

How do you make a quantity under a root the subject?

First get the root on its own, then square both sides. From T = 2π√(l/g) divide by 2π, square, and multiply by g, which gives l = gT²/(4π²).

How do you make a letter that sits in a denominator the subject?

Multiply both sides by the whole denominator, expand, and collect the terms containing the letter on one side. From J = nE/(R + nr) you get Jnr = nE − JR, that is r = (nE − JR)/(Jn).

How many places should the result be rounded to?

To as many as the data have. If the exercise gives R₁ = 276 and R₂ = 145, the result 95.059382... is written as 95.06 - further digits are false precision, because the data themselves do not carry it.

← Previous

Algebra - review exercises

Rearranging formulae and evaluating functions