# Quiz Chapter 05.03: Newton’s Divided Difference Method

 MULTIPLE CHOICE TEST NEWTON’S DIVIDED DIFFERENCE INTERPOLATION INTERPOLATION

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

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

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

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

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

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