How to Find the Length of a Graph Using Calculus?

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Homework Help Overview

The discussion revolves around determining the length of a graph defined by the function f(x) = (x^5)/10 + (1/(6x^3)) using calculus, specifically through the application of the arc length formula.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the derivative f'(x) and its square, questioning the correctness of the initial computations. There is mention of a potential trick to simplify the expression under the radical in the arc length formula.

Discussion Status

Some participants are exploring the correctness of the derivative and its implications for the arc length calculation. Suggestions have been made to express the term under the radical as a perfect square, indicating a potential direction for further exploration.

Contextual Notes

There is an indication of confusion regarding the manipulation of the function and its derivative, as well as the need for clarification on the steps involved in the arc length calculation.

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Homework Statement


Determine the length of the following graph:

[tex]f(x) \ = \ \frac{x^5}{10} + \frac{1}{6x^3}[/tex]


Homework Equations



length of a graph: [tex]\int \sqrt{1 + f'(x)^2}dx[/tex]

The Attempt at a Solution



so [tex]f'(x) = \frac{x^4}{2} -\frac{1}{18x^4}[/tex]

[tex]f'(x)^2 = \frac{x^8}{4} + \frac{1}{18} + \frac{1}{324x^8}[/tex]

Is f'(x)^2 correct?

Did I even need to expand, or is there some trick to this?
 
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There is almost always a trick to length of arc problems and the trick is almost always make the expression under the radical into a perfect square. Add 1 to f'^2 and you will see that you can. Except fix f'(x) first. I get x^4/2-1/(2*x^4). Why don't you?
 
Last edited:
I see what I did wrong. I "brought up" the 6 as well, so I wrote it as 6x^-3
 
Ok, so can you square it and then express the expression under the radical as a perfect square? I'm betting you can.
 

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