Answer to Question #269315 in Calculus for Angel

Question

Find:Differential to approximate value of cube root of 3.3

Answer 1

Let $f(x)=x^{\frac 1 3}$ . Formula for approximate calculation using differential is presented below

$f(x_0+Δx)\approx f(x_0)+df(x_0)$ , where $df(x_0)=f'(x_0)*Δx$

$f'(x)={\frac 1 3}x^{-{\frac {2} 3}} = {\frac 1 3}*{\frac 1 {(x^2)^{\frac 1 3}}}$

Now we have to present number 3.3 as $x_0+Δx$ such that $f(x_0)$ is a perfect cube and $Δx$ is relativately small. We can present it as 3.3 = 1 + 2.3, but 2.3 is a relativately big number, so the approximation error will be significant. Since $3.3^{\frac 1 3}={\frac {33^{\frac 1 3}} {10^{\frac 1 3}}}$ , then we can present 3.3 as ${\frac {33} {10}}$

and calculate both f(33) and f(10) using differentiation

$f(33)=f(27+6)=f(27)+f'(27)*6=27^{\frac 1 3}+{\frac 1 3}*{\frac 1 {(27^2)^{\frac 1 3}}}*6=3+2*{\frac 1 9}={\frac {29} 9}$

$f(10)=f(8+2)=f(8)+f'(8)*2=8^{\frac 1 3}+{\frac 1 3}*{\frac 1 {(8^2)^{\frac 1 3}}}*2=2+2*{\frac 1 3}*{\frac 1 4}=2+{\frac 1 6}={\frac {13} 6}$

$f(3.3)={\frac {f(33)} {f(10)}}={\frac {29*6} {9*13}}=1{\frac {19} {39}}\approx1.49$

Lets check the accuracy

$1.49^3=3.307959$ . The obtained number is close to 3.3, so we answer is very accurate