Question

Determine the average of the functionf(x)-sin(4x)onthe intervalp.6.2.2jusing (a) right Riemann sum with 8 segments (b) midpoint rule with 8 segments (c) trapezoidal rule with 8 segments. (d) Simpson's rule with 8 segments 1. (e) Romberg rule for 1,2, 4 and 8 segments A new fuel for recreational boats being developed at the local university was tested at an area pond by a team of engineers. Their interest is to document the environmental impact of the fuel- how quickly does the slick spread? The table below shows the video camera record of the radius of the wave generated by a drop of the fuel that fell into the pond Time, (s 2. Radius, R (m) o 0.667 1.886. 3.464 4.365 Use second order polynomial interpolation to compute the rate at which the radius of (a) the drop was changing at 31 25 seconds. Use the forward divided difference approximation to compute the rate at which the area (b) of the drop was changing at 2.785 seconds.

Numerical Methods problems 1 and 2

Answer 1

Function : $f(x) = sin(\sqrt{4x})$

Interval :

0.6,2.2

Use : Right Riemann sum with 8 segments

Solution : The eight segments turn out to be

2.2- 0.6 0.2

For the right-riemann sum, we take all the right vertices of each interval to be the points on the graph,

This would give the following regions :

0.6.0.8] → f(z) = f(0.8) sin(V3.2)

10.8. 1.01 → f(z) = f(1,0) = sin(V4.0)

11.0. 1.2 → f(x)-f(12)-sin ( V4.8)

11.2. 1.41 → f(z) _ f(14)-sin (V5.6)

$[1.4,1.6] \rightarrow f(x) = f(1.6) = sin(\sqrt{6.4})$

$[1.6,1.8] \rightarrow f(x) = f(1.8) = sin(\sqrt{7.2})$

11.8.2.01 → f(z) f(2.0)-sin (V8.0)

2.0, 2.21 → f(z) f(2.2) sin(V8.8)

Each interval is of size

So, computing the average, we get,

$\Rightarrow 0.2\times[sin(\sqrt{3.2})+sin(\sqrt{4.0})+sin(\sqrt{4.8})+sin(\sqrt{5.6})+sin(\sqrt{6.4})+sin(\sqrt{7.2}) +sin(\sqrt{8.0})]$

$\Rightarrow 0.2\times (0.285935)$

So, the average is

The eight mid points of the given intervals are

Computing average over all these intervals, we get,

$\Rightarrow 0.2\times[sin(\sqrt{2.8})+sin(\sqrt{3.6})+sin(\sqrt{4.4})+sin(\sqrt{5.2})+sin(\sqrt{6.8})+sin(\sqrt{7.6}) +sin(\sqrt{8.4})]$

$\Rightarrow 0.2\times (0.282858)$

So, the average is

$\int_{0.6}^{2.2}sin(\sqrt{4x})dx$

This computes the average over the entire interval

For the segements,

$h = \frac{b-a}{n} = \frac{2.2-0.6}{8} = 0.2$

So, we have for trapezoidal rule,

$\int_{a}^{b}f(x)\approx \frac{b-a}{2n}[f(a)+2[\sum_{i=1}^{n-1}f(a+ih)]+f(b)]$

$\int_{0.6}^{2.2}f(x)\approx \frac{2.2-0.6}{16}[f(0.6)+2[\sum_{i=1}^{n-1}f(0.6+i0.2)]+f(2.2)]$

$\Rightarrow (0.1)[f(0.6)+2[\sum_{i=1}^{7}f(0.6+i0.2)]+f(2.2)]$

Computing and substituting the values of we get,

$\Rightarrow (0.1)[(0.027)+2[\sum_{i=1}^{7}f(0.6+i0.2)]+(0.0518)]$

$\Rightarrow (0.1)[(0.79)+2[\sum_{i=1}^{7}f(0.6+i0.2)]]$

Calculating the value of the summation we get,

$\Rightarrow (0.1)[(0.79)+2[0.286]]$

So, we get the value to be

Simpson's rule states that :

$\int_{a}^{b}f(x)dx = \frac{(b-a)/n}{3}[f(x_{0})+2f(x_{1})+4f(x_{2})+...+4f(x_{n-1})+f(x_{n})]$

2.2 0.2 0.6 2f (1.8) +4f (2.0)f(2.2)]

$\Rightarrow \frac{0.2}{3}[0.856]$

This gives a value of

Romberg rule is a sub-part of the trapezoidal rule in the trapezoidal rule,we substitute the values of and find the approximate value of the integral. Now, we approximate the value of the integral by taking an average of all the obtained values.

We need to compute the rate at which the radius of the drop was changing at

t-3.125seconds

For this, we need to use the second order polynomial interpolation. we use the Newton's divided difference method here

t	R	1st Order	2nd Order	3rd Order	4th order	5th order
0.0	0
1.0	0.667	$\frac{0.667-0}{1-0} = 0.667$
2.0	1.886	$\frac{1.866-0.667}{2-1} = 1.199$	$\frac{1.199-0.667}{2-0} = 0.266$
2.5	2.635	$\frac{2.635-1.886}{2.5-2} = 1.498$	1.498 - 1.199 0.199 2.5-1
3.0	3.464	$\frac{3.464-2.635}{3-2.5} = 1.658$	$\frac{1.658-1.498}{3-2} = 0.16$
3.5	4.365	$\frac{4.365-3.464}{3.5-3} = 1.802$	$\frac{1.802-1.658}{3.5-2.5} = 0.144$

3! (-D-2)A 4! 5! (2n -I)! (2n)! -n

Using this, we substitute the value of x to be 2.785 seconds and find out the value of radius

f(x)= f(x.) f(x) (x-x)(r-)(x-x1) + f(%) This is called Lagrange's interpolation formula and can be used both equal and unequal intervals.

In this formula, we take the values of x to be time and the values of f(x) to be the radius and substitute the values of and

and the value of to find the rate of change of the radius

The equation comes out to be :

Substituting , we get,

We get the value of radius to be