Question

The area A of circle of diameter ‘d’ is given for the following values

D	80	85	90	95	100
A	5026	5674	6362	7088	7854

Find the approximate values for the areas of circles of diameter 82 and 91 respectively

Answer 1

To find A at D = 82

Since the value of A is required near the beginning of the table.

We use the Newton’s forward interpolation formula.

`"A"("D" = "D"_0 + "nh") = "A"_0 + "n"/(1!) Delta"A"_0 + ("n"("n" - 1))/(2!) Delta^2"A"_0 + ("n"("n" - 1)("n" - 2))/(3!) Delta^3"A"_0 + .....`

`"D"_0 + "nh" = "D" => 80 + "n"(5)` = 82

5n = 82 – 80 = 2

n = `2/5`

n = 0.4

D	A	`Delta"A"`	`Delta^2"A"`	`Delta^3"A"`	`Delta^4"A"`
80	5026
		648
85	5674		40
		688		– 2
90	6362		38		4
		726		2
95	7088		40
		766
100	7854

`"A"_(("at"  "D" = 82)) = 5026 + 0.4/(1!) (648) + ((0.4)(0.4 - 1))/(2!) (40) + ((0.4)(0.4 - 1)(0.4 - 2))/(3!) (- 2) + ((0.4)(0.4 - 1)(0.4 - 2)(0.4 - 3))/(4!) (4)`

= `5026 + 0.4(648) + ((0.4)(-0.6))/ (40) + ((0.4)(-0.6)(-1.6))/6 (-2) + ((0.4)(-0.6)(-1.6)(-2.6))/24 (4)`

= 5026 + 259.2 – 4.8 – 0.128 – 0.1664

= 5285.2 – 5.0944

= 5280.1056

A = 5280.11

To find Δ at D = 91

Since the value of A is required near the beginning of the table.

We use the Newton’s forward interpolation formula.

`"A"("D" = "D"_"n"  "nh") = "A"_"n" + "n"/(1!) ∇"A"_"n" + ("n"("n" - 1))/(2!) ∇^2"A"_"n" + ("n"("n" - 1)("n" - 2))/(3!) ∇^"A"_"n" + ......`

`Delta"n" + "n"` = D

100 + n(5) = 91

5n = 91 – 100

⇒ 5n = – 9

n = `(-9)/5`

n = – 1.8

D	A	`Delta"A"`	`Delta^2"A"`	`Delta^3"A"`
80	5026
		648
85	5674		40
		688		– 2
90	6362		38
		726		2
95	7088		40
		766
100	7854

`"A"_(("at"  "D" = 91)) = 7854 + ((-1.8))/(1) (766) + ((-1.8)(-1.8 + 1))/(2!) (40) + ((1.8)(-1.8 + 1)(-1.8 + 2))/(3!) (2) ((-1.8)(-1.8 + 1)(-1.8 + 2)(-1.8 + 3))/(4!) (4)`

= `7854 - 1378.8 + ((-1.8)(-0.8))/2 (40) + ((-1.8)(-0.8)(0.2))/6 (2) + ((-1.8)(-0.8)(0.2)(1.2))/24 (4)`

= 7854 – 1378.8 + 28.8 + 0.096 + 0.0576

= 7882.9536 – 1378.8

= 6504.1536

= 6504.15