Quiz Chapter 05.03: Newton’s Divided Difference Method

1. If a polynomial of degree $n$ has more than $n$ zeros, then the polynomial is

oscillatory

zero everywhere

quadratic

not defined

2. The following $x-y$ data is given

$x$	$15$	$18$	$22$
$y$	$24$	$37$	$25$

The Newton’s divided difference second order polynomial for the above data is given by

$f_{2} (x) = b_{0} + b_{1} \left( x-15 \right) + b_{2} \left( x-15 \right) \left( x-22 \right)$

The value of $b_{1}$ is

-1.048

0.1433

4.333

24.00

3. The polynomial that passes through the following $x-y$ data

$x$	$18$	$22$	$24$
$y$	?	$25$	$123$

is given by

- - $8.125x^{2} - 324.75x + 3237, \, 18 \leq x \leq 24$

The corresponding polynomial using Newton’s divided difference polynomial is given by

- - $f_{2}(x) = b_{0} + b_{1} \left( x-18 \right) + b_{2} \left( x-18 \right) \left( x-22 \right)$

The value of $b_{2}$ is

0.2500

8.125

24.00

not obtainable with the information given

4. Velocity vs. time data for a body is approximated by a second order Newton’s divided difference polynomial as

- - $v (t) = b_{0} + 39.622 \left( t - 20 \right) + 0.5540 \left( t - 20 \right) \left( t - 15 \right), \, 10 \leq t \leq 20$

The acceleration in m/s² at $t=15$ seconds is

0.5540

m/s²

39.622

m/s²

36.852

m/s²

not obtainable with the given information

5. The path that a robot is following on an $x-y$ plane is found by interpolating the following four data points

$x$	$2$	$45$	$5.5$	$7$
$y$	$7.5$	$7.5$	$6$	$5$

- - $0.1524x^{3} - 2.257x^{2} + 9.605x - 3.900$

The length of the path from $x = 2$ to $x = 7$ is

\, \sqrt{ \left( 7.5 - 7.5 \right)^{2} + \left( 4.5 - 2 \right)^{2}} + \sqrt{ \left( 6 - 7.5 \right)^{2} + \left( 5.5 - 4.5 \right)^{2}} + \sqrt{ \left( 5 - 6 \right)^{2} + \left( 7 - 5.5 \right)^{2}}

\, \displaystyle\int_{2}^{7} \sqrt{ 1 + \left( 0.1524x^{3} - 2.257x^{2} + 9.605x - 3.900 \right)^{2}} \, dx

\, \displaystyle\int_{2}^{7} \sqrt{ 1 + \left( 0.4572x^{2} - 4.514x + 9.605 \right)^{2}} \, dx

\, \displaystyle\int_{2}^{7} \left( 0.1524x^{3} - 2.257x^{2} + 9.605x - 3.900 \right)^{2} \, dx

6. The following data of the velocity of a body is given as a function of time.

Time (s)	$0$	$15$	$18$	$22$	$24$
Velocity (m/s)	$22$	$24$	$37$	$25$	$123$

If you were going to use quadratic interpolation to find the value of the velocity at ?? seconds, the three data points of time you would choose for interpolation are

0, \, 15, \, 18

15, \, 18, \, 22

0, \, 15, \, 22

0, \, 18, \, 24