Question

The following table gives the viscosity of an oil as a function of temperature. Use Lagrange’s formula to find the viscosity of oil at a temperature of 140°.

Temp° : 110 130 160 190

Viscosity: 10.8 8.1 5.5 4.8

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} x & 110 & 130 & 160 & 190 \\ \hline y & 10.8 & 8.1 & 5.5 & 4.8 \\ \end{array}$ , where $x$ is temperature, $y$ is viscosity.

By Lagrange’s interpolation formula we have:

$y=f(x)= \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}y_0+ \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} y_1+ \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}y_2+ \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} y_3$

We put $x=140$ :

$y(140)=f(140)= \frac{(140-130)(140-160)(140-190)}{(110-130)(110-160)(110-190)}10.8+ \frac{(140-110)(140-160)(140-190)}{(130-110)(130-160)(130-190)} 8.1+ \frac{(140-110)(140-130)(140-190)}{(160-110)(160-130)(160-190)}5.5+ \frac{(140-110)(140-130)(140-160)}{(190-110)(190-130)(190-160)} 4.8=-\frac{1}{8}\cdot 10.8+\frac{5}{6}\cdot8.1+\frac{1}{3}\cdot 5.5-\frac{1}{24}\cdot4.8=7.033$

Answer: the viscosity of oil at a temperature of $140^\circ$ is $7.033$ .