Distance Traveled by a Particle & Curve Length

[jenn9580]

Given:
x=5(sin t)^2
y=5(cos t)^2
0<t<pi

I am looking for the distance traveled by a particle, but seem to be stuck. I have the formula, but I think the problem is in my derivative. (dx/dt)=(10sin(t)cos(t))^2 (by the chain rule). I would like to then convert using the double-angle formula of 2sinxcosx=sin2x. I got 5sin^2(2t), but I think I am wrong. I can do the rest of the problem, I am just stuck with this part. Thanks!!

 

[daon]

The derivative of x is 10sin(t)cos(t), no squares should be present.
The derivative of y is -10sin(t)cos(t).

You then do:

\(\displaystyle \int_0^{\pi} \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt = \int_0^{\pi}\sqrt{200(\sin t \cos t)^2}dt\)

oh, also since you will get an absolute value, break it up:

\(\displaystyle \int_0^{\pi}|\sin t \cos t|dt = \int_0^{\pi/2}\sin t \cos t dt - \int_{\pi/2}^{\pi}\sin t \cos tdt\)

 

[jenn9580]

thank you very much!

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ x \ = \ 5sin^{2}(t), \ y \ = \ 5cos^{2}(t), \ 0<t<\pi, \ find \ Arc \ Length.\)

\(\displaystyle Arc \ Length \ = \ \int_{a}^{b}\bigg[\bigg(\frac{dx}{dt}\bigg)^{2}+\bigg(\frac{dy}{dt}\bigg)^{2}\bigg]^{1/2}dt\)

\(\displaystyle = \ \int_{0}^{\pi}[(10sin(t)cos(t))^{2}+(-10sin(t)cos(t))^{2}]^{1/2}dt\)

\(\displaystyle = \ \int_{0}^{\pi}[200sin^{2}(t)cos^{2}(t)]^{1/2}dt \ = \ 2\int_{0}^{\pi/2}[200sin^{2}(t)cos^{2}(t)]^{1/2}dt^{*}\)

\(\displaystyle = \ 20\sqrt2\int_{0}^{\pi/2}[sin^{2}(t)cos^{2}(t)]^{1/2}dt\)

\(\displaystyle = \ 20\sqrt2 \int_{0}^{\pi/2}\bigg[\bigg(\frac{1-cos(2t)}{2}\bigg)\bigg(\frac{1+cos(2t)}{2}\bigg)\bigg]^{1/2}dt\)

\(\displaystyle = \ 20\sqrt2 \int_{0}^{\pi/2}\bigg[\frac{1-cos^{2}(2t)}{4}\bigg]^{1/2}dt\)

\(\displaystyle = \ 10\sqrt2 \int_{0}^{\pi/2}sin(2t)dt \ = \ 10\sqrt2\bigg[\frac{-cos(2t)}{2}\bigg]_{0}^{\pi/2} \ = \ 5\sqrt2(1+1) \ = \ 10\sqrt2\)

\(\displaystyle ^{*}Note: \ Function \ is \ smooth \ from \ 0 \ to \ \pi/2 \ and \ then \ from \ \pi/2 \ to \ \pi.\)

[attachment=0:f0f4vdjj]abc.jpg[/attachment:f0f4vdjj]

 

[BigGlenntheHeavy]

Another and easier way to find arc length:

\(\displaystyle Given: \ x = 5sin^{2}(t), \ y \ = \ 5cos^{2}(t), \ and \ 0<t<\pi.\)

\(\displaystyle \frac{x}{5} \ = \ sin^{2}(t) \ and \ \frac{y}{5} \ = \ cos^{2}(t), \ ergo \ \frac{x+y}{5} \ = \ 1, \ y \ = \ 5-x.\)

\(\displaystyle When \ t \ goes \ from \ 0 \ to \ \pi/2, \ x \ goes \ from \ 0 \ to \ 5 \ and \ when \ t \ goes \ from \ \pi/2\)

\(\displaystyle to \ \pi, \ x \ goes \ from \ 5 \ to \ 10.\)

\(\displaystyle Hence, \ Arc \ Length \ = \ \int_{0}^{10}[(-1)^{2}+1]^{1/2}dx \ = \ \int_{0}^{10}2^{1/2}dx\)

\(\displaystyle = \ [2^{1/2}x]_{0}^{10} \ = \ 10\sqrt2. \ Removing \ the \ parameters \ is \ quicker.\)

[attachment=0:10akzjwd]def.jpg[/attachment:10akzjwd]

\(\displaystyle Note: \ The \ easiest, \ by \ inspection \ of \ the \ graph \ (y \ = \ 5-x), \ we \ have \ two \ isosceles \ right \ triangles, \ each\)

\(\displaystyle each \ with \ hypotenuse \ of \ 5\sqrt(2).\)

 

[jenn9580]

Thanks! I knew there must be an easier way!