Question

Use a linear approximation (or differentials) to estimate the given number $\sqrt[3]{328}$

Answer 1

$y=\sqrt[3]{3x}=x^{\frac{1}{3}}⟹\left(1\right)$

$y+\delta y={(x+\delta x)}^{\frac{1}{3}}$

${(x+\delta x)}^{\frac{1}{3}}=y+\delta y\approx x^{\frac{1}{3}}+dx$

$\therefore y=x^{\frac{1}{3}},\delta y\approx dy⟹\left(2\right)$

Taking the differential of equation(1),we get

$dy=d{x}^{\frac{1}{3}}=\frac{1}{3}13x^{\frac{1}{3}-1}dx=13x^{\frac{-2}{3}}dx$

Putting this value in equation (2)

$(x+\delta x{)}^{\frac{1}{3}}\approx x^{\frac{1}{3}}+\frac{1}{3}{x}^{\frac{-2}{3}}dx$

$(27+1{)}^{\frac{1}{3}}\approx (27{)}^{\frac{1}{3}}+\frac{1}{3}\left(27{)}^{\frac{-2}{3}}\right(1)$

$\because x=27,dx=1=\delta x$

$(28{)}^{\frac{1}{3}}\approx (3^3{)}^{\frac{1}{3}}+\frac{1}{3}(3^3{)}^{\frac{-2}{3}}=3+\frac{1}{3}(3^{-2})$

$\sqrt[3]{328}\approx 3+\frac{1}{3}\times \frac{1}{9}=3+\frac{1}{27}=\frac{81+1}{27}=\frac{82}{27}=3.037$