Finding Arc Length of x = (y^4)/8 + 1/(4y^2) from 1 to 2

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SUMMARY

The arc length of the function defined by x = (y^4)/8 + 1/(4y^2) over the interval from y = 1 to y = 2 can be determined using the arc length formula. The derivative of the function, when simplified using negative exponents, reveals a perfect square, allowing for straightforward integration. The expression simplifies to (y^3)/2 + 1/2y^3, which can then be integrated to find the arc length. This method emphasizes the importance of algebraic manipulation in solving arc length problems.

PREREQUISITES
  • Understanding of arc length formulas in calculus
  • Familiarity with derivatives and their applications
  • Knowledge of algebraic manipulation, including factoring and simplifying expressions
  • Experience with integration techniques in calculus
NEXT STEPS
  • Study the arc length formula in detail, focusing on its derivation and applications
  • Practice integrating functions that involve perfect squares and negative exponents
  • Explore advanced integration techniques, such as substitution and integration by parts
  • Review calculus concepts related to derivatives and their geometric interpretations
USEFUL FOR

Students studying calculus, particularly those focusing on arc length problems, as well as educators seeking to enhance their understanding of integration techniques and algebraic manipulation.

MHrtz
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Homework Statement



Determine the arc length of the function on the given interval

x = (y^4)/8 + 1/(4y^2) from 1 to 2

The arc length formula

\int (f'(x)2 + 1).5 dx

The Attempt at a Solution



I used the arc length formula but don't know where to go from here. Usually these problems can't be done unless a form of cancellation takes place and I can't seem to find it. Below is what I input in the formula but I can not figure out how to integrate this.

\int((y6 - 1)2/(2y3)) + 1))).5 dy
 
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MHrtz said:

Homework Statement



Determine the arc length of the function on the given interval

x = (y^4)/8 + 1/(4y^2) from 1 to 2

The arc length formula

\int (f'(x)2 + 1).5 dx

The Attempt at a Solution



I used the arc length formula but don't know where to go from here. Usually these problems can't be done unless a form of cancellation takes place and I can't seem to find it. Below is what I input in the formula but I can not figure out how to integrate this.

\int((y6 - 1)2/(2y3)) + 1))).5 dy

It's helpful in this problem to use negative exponents instead of fractions.

(x')2 + 1 = (1/4)y6 - 1/2 + (1/4)y-6 + 1
= (1/4)y6 + 1/2 + (1/4)y-6

This turns out to be a perfect square, so you can readily take its square root.
 
I factored and then took the square root and came up with this:

(y^3)/2 + 1/2y^3
 
Yes, that is correct. Now integrate.
 

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