The Length of a Curve: Calculating Using a Non-Traditional Approach

[ProBasket]

Find the length of the curve

$$x = 3y^{4/3} - 3/32y^{2/3}$$

from $$-8 \leq y \leq 64$$

dy/dx = $$4*y^{1/3}-\frac{1}{16y^{1/3}}$$ <-- don't need to check this, i used a math tool to check it.

if i plug all of this into a integral, its going to be crazy. so i was wondering if there is a different approach to this problem

$$\int_{-8}^{64} \sqrt{1+{(4y^{1/3}-\frac{1}{16y^{1/3}})}^2}$$

 

[cepheid]

DO it. Why would it be "crazy?" You've just got some power functions.

 

[ProBasket]

is my integral setup correctly?

 

[HallsofIvy]

So far, yes.

Actually MOST functions would give "crazy" arclength integrals- so teachers choose functions that simplify. Go ahead, square the formula and add the "1"- and see what happens (hint: the result is a perfect square).

 

[Nenad]

your integral is correct, you just forgot the dy at the end.

$$\int_{-8}^{64} \sqrt{1+{(4y^{1/3}-\frac{1}{16y^{1/3}})}^2} dy$$

 

[ZapperZ]

Find the length of the curve

$$x = 3y^{4/3} - 3/32y^{2/3}$$

from $$-8 \leq y \leq 64$$

dy/dx = $$4*y^{1/3}-\frac{1}{16y^{1/3}}$$ <-- don't need to check this, i used a math tool to check it.

if i plug all of this into a integral, its going to be crazy. so i was wondering if there is a different approach to this problem

$$\int_{-8}^{64} \sqrt{1+{(4y^{1/3}-\frac{1}{16y^{1/3}})}^2}$$

I don't want to be picky, but what you differentiated was dx/dy, not dy/dx. That is how you could dx = {garbage}*dy.

Zz.