Finding Arc Length?

[justinram22]

[SOLVED] Finding Arc Length?

Hello, I usually don't like my first post to be a question but I can't seem to figure this problem out.

Ok so the arc-length formula is: ∫ sqrt(1 + f'(x)²)

So I have to find the arclength between 1 and 5 for the function f(x)=(1/4)x⁴+(1/8)x^-2

f'(x) = x³ - (1/4)x^-3

∫ sqrt(1 + (x³-(1/4)x^-3)²)
= ∫ sqrt(1 + (x⁶-(1/2)+(1/16)x^-6))
= ∫ sqrt(x⁶ + (1/2) + (1/16)x^-6)
= ∫ sqrt(( (1/4)x^-3 + x³)² )
= ∫ (1/4)x^-3+ x³

= (-1/8)x^-2 + (1/4)x⁴ | 1 to 5

So I plug in the numbers and get 156.245 - .125 = 156.12

The answer though is 624.12, so I'm not sure where I've went wrong. The decimal seems to be right but the rest not even close...

Thanks guys for any help you can give!

 

[tkhunny]

You must learn to be a little more sturdy. When the "answer" is wrong, why not prove it so?

f(1) = 3/8 = 0.375
f(5) = 31251/200 = 156.255

Length of a straight line:

\(\displaystyle \sqrt{(5-1)^{2}+(156.255-0.375)^{2}} = \sqrt{4^{2}+155.88^{2}} = \sqrt{24314.5744} = 155.931\)

That should be a lower bound.

Now, let's take a path we KNOW to be longer than the given curve. How about horizontal displacement followed by vertical displacement?

5 - 1 = 4 -- There's the horizontal.
156.255 - 0.375 = 155.88 -- That's the vertical
4 + 155.88 = 159.88

That should be an upper bound.

Now, tell me again that the answer is 624.12.

 

[willmoore21]

I've check this by hand, MAPLE and R, and they all give me the answer you got, 156.12, so unless you wrote down the question wrong, this is right.

 

[justinram22]

Thank you both, and I never thought of checking it that way so thank you for that too!

 

[tkhunny]

ALWAYS keep "reasonableness" in mind. ALWAYS!!!