Arc length of parametric curve x = t³, y = t² from t = 0 to 4

[Stratosphere]

Homework Statement

Find the length of
$$\ x =t^{3}$$
$$\ y =t^{2}$$
0 $$\leq$$t$$\leq$$ 4

Homework Equations

I would write the formula for the arc length but I don't know how to make a definite integral.

The Attempt at a Solution

I have the whole thing set up and I'm ready to integrate but I seem to have forgotten how to integrate $$\int\ \sqrt{9t^{4}+2t^{2}}dt$$

 

[darkmagic]

From the integral you set up, first factor t^2, then it will become t by square root. After that, use trigonometric substitution. 3t=sqrt(2)tan(theta), 3dt=sqrt(2)sec^2(theta), then substitute them to the integral

 

[Stratosphere]

Unfortunately I don't follow you, after I factor out the t^2 I get
$$\int\sqrt{t^{2}(9t^{2}+2)}$$ after that I don't know what your trying to say.

 

[Mark44]

Bring the t² factor out of the radical so that the integrand becomes
$$t\sqrt{9t^2 + 2}$$.

Also, it is much simpler to evaluate this integral using an ordinary substitution instead of the more complicated trig substitution that darkmagic suggested.

 

[Mark44]

Also, here is the definite integral. Click on it to see the LaTeX code that I used.
$$\int_{0}^{4} t\sqrt{9t^2 + 2}~dt$$

 

[Stratosphere]

I hate to ask again but I think I rushed through U-substitution to fast. I don't see anything that's the derivative of the otherthing so I'm not sure what to pick for U.

 

[slider142]

I hate to ask again but I think I rushed through U-substitution to fast. I don't see anything that's the derivative of the otherthing so I'm not sure what to pick for U.

Recall the chain rule: If f is a function of g and g is a function of t, and we define h(t) = f(g(t)), then h'(t) = f'(g(t))g'(t).
What is the derivative of the inside of that radicand in your integral? Can you modify the outside of the integral so that the integrand resembles a chain rule?

 

[Mark44]

I hate to ask again but I think I rushed through U-substitution to fast. I don't see anything that's the derivative of the otherthing so I'm not sure what to pick for U.

Then you should review this technique. For your problem, let u = 9t² + 2. Then du = ______ dt ?

 

[Stratosphere]

I got the answer to be

$$\frac{(9t^{3}+2)^{3/2}}{18}$$

is this right?

 

[Mark44]

I got the answer to be

$$\frac{(9t^{3}+2)^{3/2}}{18}$$

is this right?

It's easy enough to check for yourself. Just differentiate what you have and if it's correct, you should get the integrand you started with.

I will say, though, that it appears you have made an error in your integration. Also, when you do get the correct antiderivative, you will still need to evaluate it at 4 and at 0 -- you're working with a definite integral, so you should get a number for your arc length.

Show us how you did the substitution and the work after that, and we'll help you out.

 

[darkmagic]

ok. I do not see that. Just a u-substitution.

 

[Stratosphere]

Here are my steps

$$\ \int_{0}^{4} t\sqrt{9t^2 + 2}~dt$$

$$\ \frac{1}{18}\int 18t\sqrt{9t^2 + 2}~dt$$

$$\ u=9t^2 + 2$$
$$\ du=18t$$

$$\ \frac{1}{18} \int (u)^{1/2} du$$
$$\ \frac{1}{18} \frac{(9t^{2}+2)^{3/2}}{3/2}$$

I can't see where I went wrong.

 

[Mark44]

Here are my steps

$$\ \int_{0}^{4} t\sqrt{9t^2 + 2}~dt$$

$$\ \frac{1}{18}\int 18t\sqrt{9t^2 + 2}~dt$$

$$\ u=9t^2 + 2$$
$$\ du=18t$$

The last line should be du = 18t dt. If you get in the habit of omitting this differential, it can come back around and bite you in more complicated integrals.

$$\ \frac{1}{18} \int (u)^{1/2} du$$
$$\ \frac{1}{18} \frac{(9t^{2}+2)^{3/2}}{3/2}$$
Now all you need to do to finish this is to evaluate your antiderivative at the two endpoints, 0 and 4.

I can't see where I went wrong.

That's slightly different from what you had in post #9. If you look closely, you should see the difference.

To continue where you left off, and doing a little simplification

$$\ \int_{0}^{4} t\sqrt{9t^2 + 2}~dt$$
(using your substitution and then undoing the substitution)
$$\ = \ \frac{1}{27} ~(9t^{2}+2)^{3/2}\vert_0^4$$
Now, just evaluate this antiderivative at 4 and at 0 and subtract.