The following table gives you the viscosity of a liquid as the function of temperature. Apply interpolation to compute the temperature of the liquid if the viscosity is 13.6 . Temperature(x) 30 35 40 45 50 Viscosity(y) 15.9 14.9 14.1 13.3 12.5
Answers
Answer:
\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}
x
y
110
10.8
130
8.1
160
5.5
190
4.8
, where xx is temperature, yy 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_3y=f(x)=
(x
0
−x
1
)(x
0
−x
2
)(x
0
−x
3
)
(x−x
1
)(x−x
2
)(x−x
3
)
y
0
+
(x
1
−x
0
)(x
1
−x
2
)(x
1
−x
3
)
(x−x
0
)(x−x
2
)(x−x
3
)
y
1
+
(x
2
−x
0
)(x
2
−x
1
)(x
2
−x
3
)
(x−x
0
)(x−x
1
)(x−x
3
)
y
2
+
(x
3
−x
0
)(x
3
−x
1
)(x
3
−x
2
)
(x−x
0
)(x−x
1
)(x−x
2
)
y
3
We put x=140x=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=7.033y(140)=f(140)=
(110−130)(110−160)(110−190)
(140−130)(140−160)(140−190)
10.8+
(130−110)(130−160)(130−190)
(140−110)(140−160)(140−190)
8.1+
(160−110)(160−130)(160−190)
(140−110)(140−130)(140−190)
5.5+
(190−110)(190−130)(190−160)
(140−110)(140−130)(140−160)
4.8=−
8
1
⋅10.8+
6
5
⋅8.1+
3
1
⋅5.5−
24
1
⋅4.8=7.033
Answer: the viscosity of oil at a temperature of 140^\circ140
∘
is 7.0337.033 .
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