Summer 2017 - Q.11 - Grad Plus

# Summer 2017 – Q.11

11. a) Fit a curve $latex y=a+bx^2$ for the following data : [7M]

 x 0 1 2 3 y 2 4 10 15

Given curve is $latex y=a+bx^2—–\left(i\right)$

∴ normal equations are

$latex \begin{array}{l}\Sigma y=na+b\Sigma x^2—–\left(ii\right)\\\\\Sigma x^2y=a\Sigma x^2+b\Sigma x^4—–\left(iii\right)\end{array}$

Here n = 4

The various calculations are as following table

 n x y x2 x4 x2y 1 0 2 0 0 0 2 1 4 1 1 4 3 2 10 4 16 40 4 3 15 9 81 135 Total 31 14 98 179

from equation (ii)

$latex \begin{array}{l}31=4a+b(14)\\\\\therefore4a+14b=31—–\left(iv\right)\end{array}$

form equation (iii)

$latex \begin{array}{l}179=a\left(14\right)+b\left(98\right)\\\\\therefore14a+98b=179—–\left(v\right)\end{array}$

on solving the equation (iv) and (v) we get

$latex \begin{array}{l}a=2.7143\;and\\b=1.4388\end{array}$

∴ The required curve is

$latex y=2.7143+1.4388\;x^2$

b) Using Lagrange’s interpolation formula, find the value of y when x = 10 from the
following table. [6M]

 x 5 6 9 11 y 12 13 14 16

By lagrange’s interpolation formula.

we get

$latex \begin{array}{l}y\left(x\right)=\frac{\left(x-6\right)\left(x-9\right)\left(x-11\right)}{\left(5-6\right)\left(5-9\right)\left(5-11\right)}\left(12\right)+\frac{\left(x-5\right)\left(x-9\right)\left(x-11\right)}{\left(6-5\right)\left(6-9\right)\left(6-11\right)}\\\\\;\;\;\;\;x\left(13\right)+\frac{\left(x-5\right)\left(x-6\right)\left(x-11\right)}{\left(9-5\right)\left(9-6\right)\left(9-11\right)}\times\left(14\right)+\frac{\left(x-5\right)\left(x-6\right)\left(x-9\right)}{\left(11-5\right)\left(11-6\right)\left(11-9\right)}\times16\\\\\;\;\;\;\;\;\;\;\;=\frac{-1}2\left(x^3-26x^2+219x-594\right)+\frac{13}{15}\left(x^3-25x^2-199x-495\right)\\\\\;\;\;\;\;\;\;\;\;\;-\frac7{12}\left(x^3-22x^2+151x-330\right)+\frac4{15}\left(x^3-20x^2+129x-270\right)\\\\\;\;\;\;\;\;\;\;\;=\frac3{60}x^3-\frac{70}{60}x^2+\frac{557}{60}x-\frac{690}{60}\end{array}$

$latex y\left(x\right)=\frac1{60}\left(3x^3-70x^2+557x-690\right)$

putting x = 0 —–(given)

$latex \begin{array}{l}y\left(10\right)=\frac1{60}\left(3\times10^3-70\times10^2+557\times10-690\right)\\\\\;\;\;\;\;\;\;\;\;\;=\frac1{60}\left(3\times1000-70\times100+557\times10-690\right)\\\\\;\;\;\;\;\;\;\;\;\;=\frac1{60}\left(3000-7000+5570-690\right)\\\\\;\;\;\;\;\;\;\;\;\;=\frac{880}{60}\end{array}$

$latex y\left(10\right)=14.66$

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