Using Maclaurin's theorem to estimate second derivative

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Homework Help Overview

The discussion revolves around using Maclaurin's theorem to estimate the second derivative of the deflection of a beam at a specific point. The context involves understanding the relationship between the beam's deflection in millimeters and the distance along the beam in meters.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of Maclaurin's theorem and the relationship between coefficients in the polynomial expansion and the second derivative. There are attempts to clarify the importance of units in the final answer and questions about whether to convert measurements from millimeters to meters.

Discussion Status

Participants are actively engaging with the problem, discussing the implications of unit conversions on the numerical value of the second derivative. Some guidance has been offered regarding the need to include units in physical measurements, and there is an ongoing exploration of how to properly apply the theorem to derive the second derivative.

Contextual Notes

There is a noted ambiguity regarding the units of measurement for the deflection and distance, which affects the interpretation of the second derivative's value. Participants are considering the implications of these units on their calculations.

Roomie
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Homework Statement



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Use Maclaurin's theorem to estimate d^{2}y/dx^{2} at x=0

It's the deflection of a 2m beam, where x is the distance along the beam, y is the deflection in mm.

Homework Equations



maclaurin1.gif


The Attempt at a Solution



I didn't know where to start so I tried to solve it a different way, by figuring out the formula for the sequence and then just differentiating it twice.. I got the answer of -200 or -0.2 depending upon if I needed to convert the mm to m?
 

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You have y=f(x), so y''(0)=f''(0). If you can find the coefficient of x2, you can solve for the second derivative.

The expansion f(x) = f(0) + f'(0)x + f''(0)/2 x2 is the equation of a parabola. Fit your data points to a parabola. The coefficients you get will correspond to f(0), f'(0), and f''(0)/2.
 
vela said:
You have y=f(x), so y''(0)=f''(0). If you can find the coefficient of x2, you can solve for the second derivative.

The expansion f(x) = f(0) + f'(0)x + f''(0)/2 x2 is the equation of a parabola. Fit your data points to a parabola. The coefficients you get will correspond to f(0), f'(0), and f''(0)/2.

So the answer is -200 then? Since y = -100x^2 + 100

So f''(0)/2 = -100

so f''(0) = -200

?? Or should I convert the mm to m?
 
The answer isn't a pure number. It has to have units. The numerical value will depend on what units you decide to use.
 
vela said:
The answer isn't a pure number. It has to have units. The numerical value will depend on what units you decide to use.

Does that mean the -200 was right :P?

Well if I converted it to metres then it would be a pure number wouldn't it? As it'd be m/m otherwise it's mm/m?

Thanks a lot for your help!
 
No, it's not right because you left off the units! You always have to include units on physical measurements. For example, the length of the bar in this problem is 2 meters or 200 cm. The answers are equivalent, but you only know that because the units were included. A pure number is typically meaningless for physical measurements.

Regarding the units of y'', if y is measured in millimeters and x is in meters, you'd have the following units:

y => mm
y' => mm/m
y'' => mm/m2

Every time you differentiate, you're essentially dividing by x, so you get another power of meters on the bottom.
 
vela said:
No, it's not right because you left off the units! You always have to include units on physical measurements. For example, the length of the bar in this problem is 2 meters or 200 cm. The answers are equivalent, but you only know that because the units were included. A pure number is typically meaningless for physical measurements.

Regarding the units of y'', if y is measured in millimeters and x is in meters, you'd have the following units:

y => mm
y' => mm/m
y'' => mm/m2

Every time you differentiate, you're essentially dividing by x, so you get another power of meters on the bottom.
Thanks for all your help!
 

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