Trigonometry

**headbazza** · March 23rd 2010, 11:17 AM

Can anyone please help
An engineering process is tested over a period of 60secs. the equation can be given by y=e(to the power)0.06t.
a) Use numerical integration with 10 second itervals to find the area.
b) use integral calculus to find the area.
c) Compare your answers to part (a) and (b), stating which is more accurate and giving your reasons.

Thanks
Headba

zza

**Grandad** · March 23rd 2010, 11:45 AM

Hello headbazza

Originally Posted by headbazza

Can anyone please help
An engineering process is tested over a period of 60secs. the equation can be given by y=e(to the power)0.06t.
a) Use numerical integration with 10 second itervals to find the area.
b) use integral calculus to find the area.
c) Compare your answers to part (a) and (b), stating which is more accurate and giving your reasons.

Thanks
Headba

zza

What attempt have you made? Have you worked out the values of $y$ for $t = 0, 10, ..., 60$ ?

Do you know how to integrate $e^{0.06t}$ ?

Please give us some indication of where you're finding this question difficult.

Grandad

**headbazza** · March 23rd 2010, 12:07 PM

Hi Grandad
to be totally honest with you i am completly stuck with this and could do with a complete answer, i will be doing an exam shortly with this question as an example of what i might encounter.

Thanks
headbazza

**Grandad** · March 23rd 2010, 01:48 PM

Hello headbazza

The simplest method of doing a numerical integration is to use the Trapezium Rule.

Imagine the area under the curve $y = f(x)$ divided vertically into $n$ strips of equal width $d$ , with the curve replaced by straight line segments. Then the area is approximately the same as the sum of the areas of the trapezia; which is:

$\Big(\frac{f(x_0)+f(x_n)}{2} + f(x_1) + f(x_2) + ... + f(x_{n-1})\Big)d$ , where $x_0, x_1, ...,x_n$ are the $n+1$ equally spaced values of $x$ .

So, to use this formula in your question, you'll work out $y(0), y(10), y(20), ... ,y(60)$ , and then:

Area = $10\Big(\frac{y(0)+y(60)}{2} + y(10) + y(20) + ... + y(50)\Big)$

Take care with the calculation, and it should be very straightforward.

For (b):

Area $= \int_0^{60}e^{0.06t}dt$

$= \frac{1}{0.06}\Big[e^{0.06t}\Big]_0^{60}$

$= \frac{1}{0.06}(e^{3.6}-e^0)$

$\approx 593$

For (c), the answer to (b) will be more accurate because in (a) the curve has been replaced by straight line segments joining the ordinates at $x = 0, 10, ..., 60$ .

Grandad

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