G
Guest
Guest
By using y^3 = x and small error theory, evaluate the cube root of 8.2
G
Guest
Guest
okay...
y^3 = x
3y^2dy/dx = 1
thus
where do i go from here to evaluate the cube root of 8.2?
y^3 = x
3y^2dy/dx = 1
thus
where do i go from here to evaluate the cube root of 8.2?
Are you talking differentials, as Stapel said?.
I've not heard it called small number theory.
Approximate \(\displaystyle \L\\f(x)=\sqrt[3]{8.2}\)
\(\displaystyle \L\\f(x)=\sqrt[3]{x}\)
\(\displaystyle \L\\f'(x)=\frac{1}{3x^{\frac{2}{3}}\)
\(\displaystyle \L\\f(x+{\Delta}x)=f(x)+f'(x){\Delta}x\)
\(\displaystyle \L\\x=8 \;\ and \;\ x+{\Delta}x=8.2\)
Therefore, \(\displaystyle \L\\{\Delta}x=0.2\)
\(\displaystyle \L\\\sqrt[3]{8}+\frac{1}{3(8)^{\frac{2}{3}}}(0.2)\)
I've not heard it called small number theory.
Approximate \(\displaystyle \L\\f(x)=\sqrt[3]{8.2}\)
\(\displaystyle \L\\f(x)=\sqrt[3]{x}\)
\(\displaystyle \L\\f'(x)=\frac{1}{3x^{\frac{2}{3}}\)
\(\displaystyle \L\\f(x+{\Delta}x)=f(x)+f'(x){\Delta}x\)
\(\displaystyle \L\\x=8 \;\ and \;\ x+{\Delta}x=8.2\)
Therefore, \(\displaystyle \L\\{\Delta}x=0.2\)
\(\displaystyle \L\\\sqrt[3]{8}+\frac{1}{3(8)^{\frac{2}{3}}}(0.2)\)