Using integrals to calculate arc length

[Pascal's Pal]

Homework Statement

Just started Calc II last month, it's been smooth so far but I've run into a bit of snag involving the application of integrals in the calculation of arc length.

The formula you use is the definite integral of (1+(d/dx)^2)^.5.

Often once you derive the d/dx and square it, you're left with a somewhat nasty looking equation under the radical. Deriving this square root is what's giving me the trouble. Is there any particular technique?

Here's one example:

Y= ((x^4)/8) + 1/4x^2, In the interval [1,2]

Y'= (x^3)/2 - 1/2x

The Attempt at a Solution

(1+(x^(3/2) - 1/2x)^2)^.5

I need to integrate this from 1 to 2, but how does can one easily transform ut into an integrable form using algebra!

EDIT: I should mention that the textbook indicated that integrals involving arc lentgh are "often very difficult to evaluate" yet proceeded to present examples where such integration was smoothly carried out.

 

[ziad1985]

y' is wrong , the derivative of 1/4x^2 is -1/2x^3

 

[ranger]

y' is wrong , the derivative of 1/4x^2 is -1/2x^3

Er, you need to subtract 1 from the exponent. What you did is wrong. You'll get x/2.

 

[ziad1985]

if it was (1/4)x^2 i would be wrong
but if you notice he wrote:
Y= ((x^4)/8) + 1/4x^2
Y'= (x^3)/2 - 1/2x
the nagative sign for 1/2x seem to mean he have 1/(4*x) and derived it wrong
that's how i saw it..
anyway however you look at it he did a mistake , whether it was the derivation it self , or putting a minus from no where..

 

[Daverz]

if

[itex]y = \frac{x^4}{8} + \frac{x^2}{4}[/itex]

then

[itex]y' = \frac{x^3}{2} + \frac{x}{2}[/itex].

 

[ziad1985]

BTW taking it as (1/(4*x^2)) makes the integral very easy to solve.
But we need his input on this..

 

[Pascal's Pal]

Sometimes I guess it helps to reassess your assumptions...

I did indeed have the derivative wrong, which is why the integral was giving me such a hard time . But I solved it, so thanks for the help!

 

[Pascal's Pal]

These were my steps:

d/dx = (x^3)/2 -1/2x^3

You take out the one half and you get: (1/2)(X^3 - 1/x^3)

Now you square it, and you get: (1/4)(X^6 - 2 + 1/x^6)

It is this plus 1 that we take the square root of, so we incorpate the 1 by adding four to the product (balanced out by (1/4)) so new your radical looks like: ((1/4)(X^6 + 2 + 1/x^6))^.5

which equals: (1/2)((X^3)+(1/X^3))



integral is (X^4)/8 - 1/4x^2) from one to two, final answer 33/16...