Find the length of the curve. Check my work.

[shivers20]

I think the answer is wrong, I made a mistake somewhere, not sure where.


x= (y^3/3)+ 1/(4y), from y=1 to y=4.

x' = x^2 + 1/4

1+ (x^2+1/4)^2

= 1+ (x^4 + 1/16)

1+ x^4 -1/2 +1/16

x^4 + 1/2 + 1/16

(x^2 + 1/4) ^2

L = 1 and 4 (x^2 + 1/4)^2

(x^2 + 1/4)dx

[x^3 + x/4 ] 1 and 4

8/3 + 1/4 - 64/3 + 1 = -209/12

 

[tkhunny]

x= (y^3/3)+ 1/(4y), from y=1 to y=4.

x' = x^2 + 1/4

Oops.

1) How did 'y' change to 'x'?
2) That '4y' appears to be in the denominator. Your derivative of that piece is not good.

Take another shot at it.

 

[shivers20]

Sorry x= (x^2) + (-1/4y^2) is this correct?

 

[tkhunny]

You seem to have tackled objection #2. How about #1?

 

[soroban]

Hello, shivers20!

Let's start over . . .

\(\displaystyle x\;= \;\frac{y^3}{3}\,+\,\frac{1}{4y}\;\;\) from \(\displaystyle y=1\) to \(\displaystyle y=4\)

We have: \(\displaystyle \L\,x\:=\:\frac{1}{3}y^3\,+\,\frac{1}{4}y^{-1}\)

Then: \(\displaystyle \L\,\frac{dx}{dy}\:=\:y^2\,-\,\frac{1}{4}y^{-2}\)

And: \(\displaystyle \L\,\left(\frac{dx}{dy}\right)^2\:=\:\left(y^2\,-\,\frac{1}{4}y^{-2}\right)^2 \;= \;y^4 \,-\,\frac{1}{2}\,+\,\frac{1}{16}y^{-4}\)

So: \(\displaystyle \L\,1\,+\,\left(\frac{dx}{dy}\right)^2\;= \;1\,+\,\left(y^4 \,-\,\frac{1}{2}\,+\,\frac{1}{16}y^{-4}\right)\)

\(\displaystyle \L\;\;\;\;= \;y^4\,+\,\frac{1}{2}\,+\,\frac{1}{16}y^{-4}\;=\;\left(y^2\,+\,\frac{1}{4}y^{-2}\right)^2\)


Therefore, the integral is: \(\displaystyle \L\:L \;= \;\int^{\;\;\;4}_1\left(y^2\,+\,\frac{1}{4}y^{-2}\right)\,dy\)