Quiz Chapter 05.04: Lagrangian Interpolation

1. Given $n + 1$ data pairs, a unique polynomial of degree _____ passes through the $n + 1$ data points.

n + 1

n

n

or less

greater than

n

2. Given the two points $\left[ a,f(a) \right], \left[ b, f (b) \right]$ , the linear Lagrange polynomial $f_{1}(x)$ that passes through these two points is given by

f_{1}(x) = \dfrac{x-b}{a-b} f(a) + \dfrac{x - a}{a - b} f(b)

f_{1}(x) = \dfrac{x}{b - a} f(a) + \dfrac{x}{b-a} f (b)

f_{1}(x) = f(a) + \dfrac{ f(b) - f \left( a \right)}{b - a} \left( b - a \right)

f_{1}(x) = \dfrac{x - b}{a - b} f(a) + \dfrac{x - a}{b - a} f(b)

3. The Lagrange polynomial that passes through three data points given by

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

- - $f_{2} \left( x \right) = L_{0}(x) \left( 24 \right) + L_{1}(x) \left( 37 \right) + L_{2}(x) \left( 25 \right)$

The value of $L_{1}(x)$ at $x = 16$ is

-0.071430

0.50000

0.57143

4.3333

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

Time (s)	$10$	$15$	$18$	$22$	$24$
Velocity (m/s)	$22$	$24$	$37$	$25$	$123$

A quadratic Lagrange interpolant is found using three data points, $t=15, \, 18,$ and $22$ . From this information, at what of the times given in seconds is the velocity of the body $26$ m/s during the time interval of $t = 15$ to $22$ seconds.

20.173

21.858

21.667

22.020

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

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

- - $y \left( x \right) = 0.15238x^{3} - 2.2571x^{2} + 9.6048x - 3.9000$

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( 0.15238x^{3} - 2.2571x^{2} + 9.6048x - 3.9000 \right)^{2}} \, dx

\, \displaystyle\int_{2}^{7} \sqrt{ 1 + \left( 0.45714x^{2} - 4.5142x + 9.6048 \right)^{2}} \, dx

\, \displaystyle\int_{2}^{7} \left( 0.15238x^{3} - 2.2571x^{2} + 9.6048x - 3.9000 \right) \, 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 $t = 14.9$ seconds, what three data points of time would you choose for interpolation?

0, \, 15, \, 18

15, \, 18, \, 22

0, \, 15, \, 22

0, \, 18, \, 24