First Order Error Analysis for O.D.E.

• Jan 14th 2011, 03:42 PM
jegues
First Order Error Analysis
Given that the velocity of the falling parachutist can be computed by,

$\displaystyle v(t) = \frac{gm}{c}(1-e^{(\frac{c}{m})t})$

Use a first-order error analysis to estimate the error of v and t = 6, if g = 9.8 and m = 50 but c = 12.5 + or - 1.5.

I'm having a tough time starting this one. Can someone nudge me in the right direction?

EDIT: I just realized I posted this is the wrong section, it's clearly not an ODE can someone please move it?
• Jan 14th 2011, 05:02 PM
dwsmith
Quote:

Originally Posted by jegues
Given that the velocity of the falling parachutist can be computed by,

$\displaystyle v(t) = \frac{gm}{c}(1-e^{(\frac{c}{m})t})$

Use a first-order error analysis to estimate the error of v and t = 6, if g = 9.8 and m = 50 but c = 12.5 + or - 1.5.

I'm having a tough time starting this one. Can someone nudge me in the right direction?

EDIT: I just realized I posted this is the wrong section, it's clearly not an ODE can someone please move it?

Click the traingle with ! in the middle and type in the forum it should be in and hit submit.
• Jan 14th 2011, 11:44 PM
CaptainBlack
Quote:

Originally Posted by jegues
Given that the velocity of the falling parachutist can be computed by,

$\displaystyle v(t) = \frac{gm}{c}(1-e^{(\frac{c}{m})t})$

Use a first-order error analysis to estimate the error of v and t = 6, if g = 9.8 and m = 50 but c = 12.5 + or - 1.5.

I'm having a tough time starting this one. Can someone nudge me in the right direction?

EDIT: I just realized I posted this is the wrong section, it's clearly not an ODE can someone please move it?

Start by expanding as a Taylor series (in c) about c=12.5 and truncate after the first term.

CB
• Jan 15th 2011, 07:12 AM
jegues
Quote:

Originally Posted by CaptainBlack
Start by expanding as a Taylor series (in c) about c=12.5 and truncate after the first term.

CB

So,

$\displaystyle v(t) = f(t) + R_{n}$

$\displaystyle v(6) = g(50)(1-e^{\frac{12.5 \times 6}{50}}) + R_{n} = -1706.03 + R_{n}$

How do I get the error from this?
• Jan 15th 2011, 09:37 AM
CaptainBlack
Quote:

Originally Posted by jegues
So,

$\displaystyle v(t) = f(t) + R_{n}$

$\displaystyle v(6) = g(50)(1-e^{\frac{12.5 \times 6}{50}}) + R_{n} = -1706.03 + R_{n}$

How do I get the error from this?

After the first non-constant term.

CB
• Jan 15th 2011, 12:54 PM
jegues
Quote:

Originally Posted by CaptainBlack
After the first non-constant term.

CB

Here's what I came up with.

I'm pretty sure I must be misunderstanding something because I dont see any nonconstant terms.

We know the values for every variable so we should just end up with a number, no?

Where did I go wrong?

EDIT: Don't worry about the -1706.03 in the first line, it's not supposed to be there.
• Jan 15th 2011, 02:17 PM
CaptainBlack
Quote:

Originally Posted by jegues
Given that the velocity of the falling parachutist can be computed by,

$\displaystyle v(t) = \frac{gm}{c}(1-e^{(\frac{c}{m})t})$

Use a first-order error analysis to estimate the error of v and t = 6, if g = 9.8 and m = 50 but c = 12.5 + or - 1.5.

I'm having a tough time starting this one. Can someone nudge me in the right direction?

EDIT: I just realized I posted this is the wrong section, it's clearly not an ODE can someone please move it?

$\displaystyle v(t,c=12.5+\varepsilon)=\dfrac{gm}{12.5+\varepsilo n}(1+e^{(12.5+\varepsilon)t/m})$

Now expand the right hand side as a series in $\displaystyle \varepsilon$

CB
• Jan 15th 2011, 03:19 PM
jegues
Quote:

Originally Posted by CaptainBlack
$\displaystyle v(t,c=12.5+\varepsilon)=\dfrac{gm}{12.5+\varepsilo n}(1+e^{(12.5+\varepsilon)t/m})$

Now expand the right hand side as a series in $\displaystyle \varepsilon$

CB

Okay so,

$\displaystyle v(t) = \dfrac{gm}{12.5+\varepsilon}(1+e^{(12.5+\varepsilo n)t/m}) + \dfrac{gm}{12.5+\varepsilon}(1+\frac{(12.5+\vareps ilon)}{m}e^{(12.5+\varepsilon)t/m}) + ... + R_{n}$

What's next?
• Jan 16th 2011, 02:16 AM
CaptainBlack
Quote:

Originally Posted by jegues
Okay so,

$\displaystyle v(t) = \dfrac{gm}{12.5+\varepsilon}(1+e^{(12.5+\varepsilo n)t/m}) + \dfrac{gm}{12.5+\varepsilon}(1+\frac{(12.5+\vareps ilon)}{m}e^{(12.5+\varepsilon)t/m}) + ... + R_{n}$

What's next?

Is that a power series in $\displaystyle \varepsilon$?

CB
• Jan 16th 2011, 07:13 AM
jegues
Quote:

Originally Posted by CaptainBlack
Is that a power series in $\displaystyle \varepsilon$?

CB

I'm confused as to what you want me to do.

Do you mean this,

$\displaystyle e^{12.5 + \varepsilon } = \sum \frac{(12.5 + \varepsilon)^{n}}{n!}$

The sum starts from n=0 and goes to infinity, I can't seem to put it in.