### Substitution

#### Definition

The process known as substitution is the way in which symbols in formulas are replaced by actual numerical values, other symbols or even other formulas.

#### Worked Examples

###### Example 1

Find the value of $I=x P + n$ when $x=0.05,\,P=100,\,n=3$.

###### Solution

$I = x P + n = (0.05)(100) + (3) = 5+3 = 8$

###### Example 2

Find the value of $\dfrac{i}{(1+i)^n-1}$ when $n=10$ and $i=0.11$.

###### Solution

\begin{align} \frac{i}{(1+i)^n-1}&=\frac{0.11}{(1+0.11)^{10}-1}\\\\ &=\frac{0.11}{1.11^{10}-1}\\\\ &=0.060 \text{ (to 3 d.p.)} \end{align}

###### Example 3

This is an example of substituting a formula for a variable in an expression.

The kinetic energy of a body of mass $m$ and velocity $v$ is given by $K=\frac{1}{2}mv^2$.

Suppose that the body starts at time $t=0$ with velocity $u$ and constant acceleration $a$ then at time $t$ the velocity is $v=u+at$.

Express the kinetic energy of the body as a function of time $t$.

###### Solution

In $K=\frac{1}{2}mv^2$ substitute $v=u+at$ to obtain: $K=\frac{1}{2}m(u+at)^2$ This gives the kinetic energy in terms of the time travelled $t$, the initial velocity $u$ and the acceleration $a$.

#### Video Example

Prof. Robin Johnson evaluates $(1+x)^5-(1-x)^7$ at $x=0.1$.

#### Workbook

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

#### External Resources #### More Support

You can get one-to-one support from Maths-Aid.