More Arc Length

[kneaiak]

This looks like it should be fairly easy but I'm getting stuck. Here's my work

y=(3/5)x^(5/3) - (3/4)x^(1/3) + 5, [1,8]

y'=x^(2/3) - (1/4)x^-(2/3)

y'^2=x^(4/3) + (1/16)x^-(4/3) - 1/2

int from 1 to 8 of sqrt(1 + x^(4/3) + (1/16)x^-(4/3) - 1/2) dx

int from 1 to 8 of sqrt(x^(4/3) + (1/16)x^-(4/3) + 1/2) dx

and I'm stuck........

 

[MarkFL]

We are asked to find the arc-length along \(\displaystyle \displaystyle y=\frac{3}{5}x^{\frac{5}{3}}-\frac{3}{4}x^{\frac{1}{3}}+5\) on the interval \(\displaystyle [1,8]\).

Step 1: compute the derivative:

\(\displaystyle \displaystyle \frac{dy}{dx}=x^{\frac{2}{3}}-\frac{1}{4}x^{-\frac{2}{3}}=\frac{4x^{\frac{4}{3}}-1}{4x^{\frac{2}{3}}}\)

Step 2: square the derivative and add 1:

\(\displaystyle \displaystyle 1+\left(\frac{dy}{dx} \right)^2=1+\left(\frac{4x^{\frac{4}{3}}-1}{4x^{\frac{2}{3}}} \right)^2=\left(\frac{4x^{\frac{4}{3}}+1}{4x^{ \frac{2}{3}}} \right)^2\)

Step 3: Take the square root of this as the integrand, and the end-points of the interval as the limits of integration:

\(\displaystyle \displaystyle s=\int_1^8\frac{4x^{\frac{4}{3}}+1}{4x^{ \frac{2}{3}}}\,dx=\int_1^8x^{\frac{2}{3}}+\frac{1}{4}x^{-\frac{2}{3}}\,dx\)

Step 4: Evaluate the definite integral, which I'll leave for you.

 

[Deleted member 4993]

This looks like it should be fairly easy but I'm getting stuck. Here's my work

y=(3/5)x^(5/3) - (3/4)x^(1/3) + 5, [1,8]

y'=x^(2/3) - (1/4)x^-(2/3)

y'^2=x^(4/3) + (1/16)x^-(4/3) - 1/2

int from 1 to 8 of sqrt(1 + x^(4/3) + (1/16)x^-(4/3) - 1/2) dx

int from 1 to 8 of sqrt(x^(4/3) + (1/16)x^-(4/3) + 1/2) dx

and I'm stuck........

This is exactly like one of your previous problems!!

(x^(4/3) + (1/16)x^-(4/3) + 1/2 = {x^(2/3)}^2 + {(1/4)x^-(2/3)}^2 + 2*{x^(2/3)}*{(1/4)x^-(2/3)} = [{x^(2/3)} + {(1/4)x^-(2/3)}]^2

You need to learn from your previous work!!

 

[kneaiak]

Thanks Mark. Very helpful. Looks like I need to slow down with the Algebra.

Kahn-I'm not trying to waste anyone's time. I'm juggling a highly demanding client driven career, a 4 yo on my own and school after 8 years of being out of it. I'm trying! Sometimes I get tunnel vision and focused on the problem at hand. From now on I'll reference previous questions. Thanks for the tip.

 

[MarkFL]

I wanted to add a post to call attention to the fact that I edited my post above to include the correct boundaries, as the boundaries I initially mistakenly used would lead to an improper integral. I apologize if this has caused the OP any additional problems.

 

[HallsofIvy]

This looks like it should be fairly easy but I'm getting stuck. Here's my work

y=(3/5)x^(5/3) - (3/4)x^(1/3) + 5, [1,8]

y'=x^(2/3) - (1/4)x^-(2/3)

y'^2=x^(4/3) + (1/16)x^-(4/3) - 1/2

int from 1 to 8 of sqrt(1 + x^(4/3) + (1/16)x^-(4/3) - 1/2) dx

int from 1 to 8 of sqrt(x^(4/3) + (1/16)x^-(4/3) + 1/2) dx

Recall that y'^2 was x^(4/3)- 1/2+ (1/16)x^-(4/3)- that is, basically (a- b)^2= a^2- 2ab+ b^2.
Now, compare that to x^(4/3)+ 1/2+ (1/16)x^-(4/3). That is, a^2+ 2ab+ b^2= (a+ b)^2.

and I'm stuck........