Arc Length Q's: y = x^(5/2) on [1, 3], etc

[hingis]

I noticed that other posters have included the integral symbol. How would I do that on here?

Arc Length Questions:

1) Set the integral up:

. . .y = x^(5/2) on [1,3]

I got integral (1+25/4x^3)^.5 dx

Is that right?

2) This one wanted me to solve it, my answer came out sort of odd, but maybe its right, im not 100% sure.

. . .x = 2/3(y - 1)^(3/2) [1,5]

The derivative is (y - 1)^0.5, so it's:

. . .integral y^0.5 dy

. . .= (2/3)(5^3)^0.5 - 2/3 ...?

Last one!

3) x = 1/8y^4 + 1/4y^(-2) [1,2]

My derivative is 1/4y^(3) - 0.5y^(-3) but then I got:

. . .integral (1/16y^6 + 1/4y^(-6) + 3/4)^0.5 dy

But I don't think I can do anything with this. Am I setting it up wrong or doing the derivative incorrectly? Thanks for the help in advance guys

 

[soroban]

Hello, hingis!

You did quite good!

Arc Length Questions:

1) Set up the integral: \(\displaystyle \,y \:= \:x^{\frac{5}{2}}\) on [1,3]

I got: \(\displaystyle \L\,\int^{\;\;\;3}_1 (1\,+\,\frac{25}{4}x^3)^{\frac{1}{2}}\,dx\)

Is that right? . . . Yes!

2) \(\displaystyle x \:=\:\frac{2}{3}(y\,-\,1)^{\frac{3}{2}}\;\;\;\;[1,\,5]\)

The derivative is: \(\displaystyle \,\frac{dy}{dx} \,= \, (y - 1)^{\frac{1}{2}}\)

so it's: \(\displaystyle \L\,\int^{\;\;\;5}_1y^{\frac{1}{2}}\,dy \;=\;\frac{2}{3}(5^3)^{\frac{1}{2}}\,-\,\frac{2}{3}\)

Correct! I'd write it: \(\displaystyle \,\frac{2}{3}\left(5^{\frac{3}{2}}\,-\,1\right)\)

\(\displaystyle 3)\;x\:=\:\frac{1}{8}y^4\,+\,\frac{1}{4}y^{-2}\;\;\;\;[1,\,2]\)

We must be very careful with this one . . .

The derivative is: \(\displaystyle y'\;=\;\frac{1}{2}y^3\,-\,\frac{1}{2}y^{-3}\)

Then: \(\displaystyle \,1\,+\,(y')^2\;=\; 1\,+\,\left(\frac{1}{2}y^3\,-\,\frac{1}{2}y^{-3}\right)^2 \;= \;1 \,+\,\frac{1}{4}y^6\,-\,\frac{1}{2}\,+\,\frac{1}{4}y^{-6}\) . . . did you square correctly?

\(\displaystyle \;\;\)and we have: \(\displaystyle \,\frac{1}{4}y^6\,+\,\frac{1}{2} \,+\,\frac{1}{4}y^{-6}\) . . . which factors: \(\displaystyle \,\left(\frac{1}{2}y^3 \,+\,\frac{1}{2}y^{-3}\right)^2\) . . . did you follow that?


Take the square root: \(\displaystyle \,\sqrt{1 + (y')^2} \;= \;\sqrt{\left(\frac{1}{2}y^3 \,+\,\frac{1}{2}y^{-3}\right)^2} \;= \;\frac{1}{2}y^3 \,+\,\frac{1}{2}y^{-3}\) . . . neat, huh?


So the integral is: \(\displaystyle \L\,\int^{\;\;\;2}_1\left(\frac{1}{2}y^3\,+\,\frac{1}{2}y^{-3}\right)\,dy\)