Question

Use differentials to a approximate the values of ; (i) $\sqrt{36.6}$ and (ii) $\sqrt[3]{326}$

• A. (i) $6.02$, (ii) $\frac{80}{9}$
• B. (i) $6.03$, (ii) $\frac{80}{18}$
• C. (i) $6.25$, (ii) $\frac{80}{27}$
• D. (i) $6.05$, (ii) $\frac{80}{27}$

Answer 1

The correct option is D (i) $6.05$,

(ii) $\frac{80}{27}$

Using first order approximation,
$f(x+\Delta x)=f\left(x\right)+f^{\prime }\left(x\right)\Delta x$
(i) Here assume $f\left(x\right)=\sqrt{x}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}$
$\therefore \sqrt{36.6}=\sqrt{36+.6}=\sqrt{36}+\frac{1}{2\sqrt{36}}\times .6=6.05$
(ii) Here assume $f\left(x\right)=x^{1/3}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{3}{x}^{-2/3}$
$\therefore (26{)}^{1/3}=(27-1{)}^{1/3}=(27{)}^{1/3}-\frac{1}{3}(27{)}^{-2/3}\times 1=\frac{80}{27}$