Browse Questions

# Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(0.009)^{\Large\frac{1}{3}}$

$\begin{array}{1 1} 0.208 \\ 0.308 \\ 0.408 \\ 0.228 \end{array}$

Toolbox:
• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
Step 1:
Let $y=x^{\Large\frac{1}{3}}$
Let $x=0.008$
$dx=0.001$
So that $x+dx=0.009$
Now $(x+\Delta x)^{\Large\frac{1}{3}}-x^{\Large\frac{1}{3}}$
$\Rightarrow (0.009)^{\Large\frac{1}{3}}-(0.008)^{\Large\frac{1}{3}}$
$\Rightarrow (0.009)^{\Large\frac{1}{3}}-0.2$
$\therefore (0.009)^{\Large\frac{1}{3}}-0.2+\Delta y$-----(1)
Step 2:
Also $\large\frac{dy}{dx}$$\Delta x is approximately equal to dy dy=\big(\large\frac{dy}{dx}\big)$$\Delta x$
$\quad=\large\frac{1}{3x^{\Large\frac{2}{3}}}$$\Delta x \quad=\large\frac{1}{3(0.008)^{\Large\frac{2}{3}}}$$\times 0.001$
$\quad=\large\frac{1}{3(0.2)^2}$$\times 0.001$
$\quad=\large\frac{0.001}{3\times 0.04}$
$\quad=\large\frac{0.001}{0.12}$
$\quad=\large\frac{1}{120}$
$\quad=0.008$
Step 3:
Approximate value of $\Delta y=dy=0.008$
Hence from Equation(1) we have
$(0.009)^{\Large\frac{1}{3}}=0.2+0.008$
$\qquad\quad\;\;\;=0.208$