From Zero to Engineer

Rearranging formulae and evaluating functions

Making a quantity the subject when it hides under a root or appears on both sides of a formula, and evaluating f(x) for whole, fractional and negative arguments.

The hardest case of rearranging is the one where the letter you want appears on both sides. The recipe is always the same: gather every term containing it on one side, everything else on the other, then take it outside a bracket and divide. The second half of the lesson is the notation f(x): it is nothing but a formula with a name, and f(3) means "put 3 in place of x".

Rearranging formulae

Rearrange the formula to make r the dependent variable.

\begin{aligned}d &= 2\sqrt{h(2r - h)} \;\Big/ : 2 \\\frac{d}{2} &= \sqrt{h(2r - h)} \;\Big/ ^{2} \\\frac{d^{2}}{4} &= h(2r - h) \;\Big/ \cdot \frac{1}{h} \\\frac{d^{2}}{4h} &= 2r - h \\\frac{d^{2}}{4h} + h &= 2r \;\Big/ \cdot \frac{1}{2} \\r &= \frac{1}{2}\left(\frac{d^{2}}{4h} + h\right)\end{aligned}

Make R the dependent variable.

\begin{aligned}V &= \frac{\pi h(3R^{2} + h^{2})}{6} \;\Big/ \cdot 6 \\6V &= \pi h(3R^{2} + h^{2}) \;\Big/ : \pi h \\\frac{6V}{\pi h} &= 3R^{2} + h^{2} \\\frac{6V}{\pi h} - h^{2} &= 3R^{2} \;\Big/ \cdot \frac{1}{3} \\\frac{2V}{\pi h} - \frac{1}{3}h^{2} &= R^{2} \;\Big/ \sqrt{\phantom{x}} \\R &= \sqrt{\frac{2V}{\pi h} - \frac{1}{3}h^{2}}\end{aligned}

Rearrange the formula to make J the dependent variable.

\begin{aligned}n &= \frac{JR}{E - Jr} \;\Big/ \cdot (E - Jr) \\n(E - Jr) &= JR \\nE - nJr &= JR \\nE &= JR + nJr = J(R + nr) \;\Big/ : (R + nr) \\J &= \frac{nE}{R + nr}\end{aligned}

Make f the dependent variable.

\begin{aligned}\frac{R}{r} &= \sqrt{\frac{f + P}{f - P}} \;\Big/ ^{2} \\\frac{R^{2}}{r^{2}} &= \frac{f + P}{f - P} \\R^{2}(f - P) &= r^{2}(f + P) \\fR^{2} - PR^{2} &= fr^{2} + Pr^{2} \\fR^{2} - fr^{2} &= PR^{2} + Pr^{2} \\f(R^{2} - r^{2}) &= P(R^{2} + r^{2}) \;\Big/ : (R^{2} - r^{2}) \\f &= \frac{P(R^{2} + r^{2})}{R^{2} - r^{2}}\end{aligned}

Rearrange the formula to make m the dependent variable.

\begin{aligned}f &= \frac{S(M - m)}{M + m} \;\Big/ \cdot (M + m) \\f(M + m) &= S(M - m) \\fM + fm &= SM - Sm \\fm + Sm &= SM - fM \\m(f + S) &= M(S - f) \;\Big/ : (f + S) \\m &= \frac{M(S - f)}{f + S}\end{aligned}

Evaluating functions

The notation f(x) is a formula with a name, and f(3) means the result of putting 3 in place of every x. It is worth writing the argument in brackets, because for negative numbers the brackets decide the sign of the power.

f(x) = 4x^{3} - \frac{6}{2x}

\text{a)}\quad f(3) = 4\cdot 3^{3} - \frac{6}{2\cdot 3} = 4\cdot 27 - 1 = 107

\text{b)}\quad f(-4) = 4\cdot(-4)^{3} - \frac{6}{2\cdot(-4)} = 4\cdot(-64) + \frac{6}{8} = -256 + \frac{3}{4} = -255\tfrac{1}{4} = -255.25

\begin{aligned}\text{c)}\quad f\left(\tfrac{2}{5}\right) &= 4\cdot\left(\tfrac{2}{5}\right)^{3} - \frac{6}{2\cdot\tfrac{2}{5}} = 4\cdot\frac{8}{125} - \frac{6}{\tfrac{4}{5}} = \frac{32}{125} - 6\cdot\frac{5}{4} = \frac{32}{125} - \frac{15}{2} = \\&= \frac{64}{250} - \frac{1875}{250} = -\frac{1811}{250} = -7\tfrac{61}{250} = -7.244\end{aligned}

\begin{aligned}\text{d)}\quad f(-3.24) &= 4\cdot(-3.24)^{3} - \frac{6}{2\cdot(-3.24)} = -136.0489 + 0.926 = \\&= -135.12\end{aligned}

correct to five significant figures

Review exercises

Evaluate each of the following expressions, correct to three significant figures.

\text{a)}\quad J = \frac{nE}{R + nr}

n = 4,\quad E = 1.08,\quad R = 5,\quad r = 0.04

J = \frac{4\cdot 1.08}{5 + 4\cdot 0.04} = \frac{4.32}{5 + 0.16} = \frac{4.32}{5.16} = 0.837

4\cdot 1.08 = 4.32

4\cdot 0.04 = 0.16

5 + 0.16 = 5.16

\text{b)}\quad A = P\left(1 + \frac{r}{100}\right)^{n}

P = 285.79,\quad r = 5.25,\quad n = 12

\begin{aligned}A &= 285.79\left(1 + \frac{5.25}{100}\right)^{12} = 285.79\left(1 + 0.0525\right)^{12} = \\&= 285.79\cdot 1.0525^{12} = 285.79\cdot 1.848 = 528\end{aligned}

\text{c)}\quad P = A\,\frac{(nv/u)^{\frac{3}{2}}}{1 + (nv/u)^{3}}

A = 40,\quad u = 30,\quad n = 2.5,\quad v = 42.75

\begin{aligned}P &= 40\cdot\frac{(2.5\cdot 42.75 : 30)^{\frac{3}{2}}}{1 + (2.5\cdot 42.75 : 30)^{3}} = 40\cdot\frac{3.5625^{\frac{3}{2}}}{1 + 3.5625^{3}} = 40\cdot\frac{6.724}{1 + 45.213} = \\&= \frac{268.96}{46.213} = 5.82\end{aligned}

2.5\cdot 42.75 : 30 = 3.5625

40\cdot 6.724 = 268.96

Given that f(x) = 3x² − 4x, find f(2) and f(−3).

f(2) = 3\cdot 2^{2} - 4\cdot 2 = 12 - 8 = 4

f(-3) = 3\cdot(-3)^{2} - 4\cdot(-3) = 27 + 12 = 39

Frequently asked questions

What do you do when the letter you want appears on both sides?

Move every term containing it to one side and the rest to the other, then take it outside a bracket. From fM + fm = SM − Sm you get m(f + S) = M(S − f), that is m = M(S − f)/(f + S).

What does f(3) mean?

The value of the function at 3, that is the result of putting 3 in place of every x. For f(x) = 3x² − 4x we have f(2) = 3·4 − 8 = 4 and f(−3) = 3·9 + 12 = 39.

What should you watch out for with a negative argument?

The signs of the powers: (−4)³ = −64, but (−3)² = 9. It is always worth writing the argument in brackets, because without them −4³ means something quite different from (−4)³.

When do you turn division by a fraction into multiplication?

Whenever the denominator is itself a fraction. In f(2/5) we get 6 : (4/5), that is 6 · 5/4 = 15/2 - trying to divide it in your head is the commonest source of mistakes here.

← Previous

Evaluating expressions and rearranging formulae

Horner's method and the remainder theorem