Find the exact arc length of the parametric curve

warwick

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Find the exact arc length of the parametric curve without eliminating the parameter.

11. x = cos (2t), y = sin (2t) (0 less than or equal to t less than or equal to (pi/2))

L = definite integral from 0 to pi/2; square root of [-2 sin (2t))^2 + (2 cos (2t))^2] with respect to y.

= square root of [- 4 sin^2 (2t) + 4 cos^2 (2t)]

= square root of [4 (cos^2 (2t) - sin^2 (2t)]

= square root of [4 (cos (4t)]

= 2 square root of [cos (4t)]

I'm stuck. Maybe I did it incorrectly from the start. Learning process FTW!!

5. y = x^(2/3) from x = 1 to x = 8.

L = definite integral from 1 to 8 with respect to x;

square root of [1 + ((2/3x^(1/3))^2]

= square root of [1 + ((4/9x^(2/3)]

= square root of [1 + ((4/9x^(-2/3)]

If I use u-substitution, I don't see how I can get it into the u du form.

13. x = e^t cos t, y = e^t sin t (0 less than or equal to t less than or equal to pi/2)

L = definite integral from 0 to pi/ with respect to t;

square root of [(e^t(cos t - sin t))^2 + ((e^t(cos t + sin t))^2]

= square root of [2((e^t)^2)]

?????????????????
 
\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt\)

\(\displaystyle \L\\\frac{dx}{dt}=-2sin(2t)\)

\(\displaystyle \L\\\frac{dy}{dt}=2cos(2t)\)

\(\displaystyle \L\\(-2sin(2t))^{2}+(2cos(2t))^{2}=4\)

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}\sqrt{4}dt\)
 
galactus said:
\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt\)

\(\displaystyle \L\\\frac{dx}{dt}=-2sin(2t)\)

\(\displaystyle \L\\\frac{dy}{dt}=2cos(2t)\)

\(\displaystyle \L\\(-2sin(2t))^{2}+(2cos(2t))^{2}=4\)

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}\sqrt{4}dt\)

How does \(\displaystyle \L\\(-2sin(2t))^{2}+(2cos(2t))^{2}=4\) ?

Wait. I think I squared those two terms incorrectly initially.

Yes, I did. The sine and cosine squares simplify to one with the 4 factored out. Thanks, G.
 
do a tan substitution such that you get sec^2 under the radical
 
Here's a hint:

Your arc length integral simplifies to:

\(\displaystyle \L\\\frac{1}{3}\int\frac{\sqrt{9x^{\frac{2}{3}}+4}}{x^{\frac{1}{3}}}dx\)

Let \(\displaystyle \L\\u=\sqrt{9x^{\frac{2}{3}}+4}\)

Make the substitutions:

\(\displaystyle \L\\\frac{1}{9}\int{u^{2}}du\)

\(\displaystyle \L\\\frac{1}{27}u^{3}\)

Now resub.
 
galactus said:
Here's a hint:

Your arc length integral simplifies to:

\(\displaystyle \L\\\frac{1}{3}\int\frac{\sqrt{9x^{\frac{2}{3}}+4}}{x^{\frac{1}{3}}}dx\)

Let \(\displaystyle \L\\u=\sqrt{9x^{\frac{2}{3}}+4}\)

Make the substitutions:

\(\displaystyle \L\\\frac{1}{9}\int{u^{2}}du\)

\(\displaystyle \L\\\frac{1}{27}u^{3}\)

Now resub.

Well, I was able to work my way up to this, but I wasn't able to appropriately make the substitutions.

\(\displaystyle \L\\\frac{1}{3}\int\frac{\sqrt{9x^{\frac{2}{3}}+4}}{x^{\frac{1}{3}}}dx\)
 
Hey Warwick:

The trick is to get something to look like the 1/x^3 term.

It may be a little tricky. I was seeing if you could spot it.

Let \(\displaystyle \L\\u=\sqrt{9x^{\frac{2}{3}}+4}\)

\(\displaystyle \L\\u^{2}=9x^{\frac{2}{3}}+4\)

\(\displaystyle \L\\2udu=\frac{6}{x^{\frac{1}{3}}}\)

\(\displaystyle \L\\\frac{1}{3}udu=\frac{1}{x^{\frac{1}{3}}}\)

Now make the subs and you have it.
 
galactus said:
Hey Warwick:

The trick is to get something to look like the 1/x^3 term.

It may be a little tricky. I was seeing if you could spot it.

Let \(\displaystyle \L\\u=\sqrt{9x^{\frac{2}{3}}+4}\)

\(\displaystyle \L\\u^{2}=9x^{\frac{2}{3}}+4\)

\(\displaystyle \L\\2udu=\frac{6}{x^{\frac{1}{3}}}\)

\(\displaystyle \L\\\frac{1}{3}udu=\frac{1}{x^{\frac{1}{3}}}\)

Now make the subs and you have it.

Haha. Yeah, I actually thought about squaring it and going that route, but obviously I didn't or I MIGHT have seen it. At any rate, I'm stuck in 13, too. Should I use some kind of u-substitution there as well?
 
It simplifies down to a nice little easy integral also.

Find the derivative of x and square it. Find the derivative of y and square it. Add them and take the square root. You should get something small and easy to integrate.
 
galactus said:
It simplifies down to a nice little easy integral also.

Find the derivative of x and square it. Find the derivative of y and square it. Add them and take the square root. You should get something small and easy to integrate.

Well, I did that. Maybe I did it wrongly and came up with something extraneous. This is what I have

square root of [(e^t(cos t - sin t))^2 + ((e^t(cos t + sin t))^2]

= square root of [2((e^t)^2)]
 
\(\displaystyle \L \sqrt {2\left( {e^t } \right)^2 } = \left( {\sqrt 2 } \right)e^t\)
 
pka said:
\(\displaystyle \L \sqrt {2\left( {e^t } \right)^2 } = \left( {\sqrt 2 } \right)e^t\)

Wow. Hello! I can separate them now that there is a product under the radical. Now, I just have to bring out the constant and integrate e^t, correct?
 
Ok. I got #13 worked out.

Galactus, I basically copied/followed your hints but had a discrepancy with substituting (1/3)u du into the integral. Where does the dx go?
 
warwick said:
Ok. I got #13 worked out.

Galactus, I basically copied/followed your hints but had a discrepancy with substituting (1/3)u du into the integral. Where does the dx (you are substituting for x and dx) go?
 
Subhotosh Khan said:
warwick said:
Ok. I got #13 worked out.

Galactus, I basically copied/followed your hints but had a discrepancy with substituting (1/3)u du into the integral. Where does the dx (you are substituting for x and dx) go?

(1/3)u du = 1/(x^1/3) dx

Correct?
 
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