computing work using line integral of vector fields- check?

[pockystix]

Hi,
I've been working on this one problem for 2 days now and I still can't seem to get the right answer. It's driving me crazy! And everytime I try to re-do it, I seem to be getting a different answer....Here is the work from my latest attempt. Can someone please check it and tell me what I am doing wrong...


Original Problem: Compute the work performed in a particle moving along the path c(t) = (cos(2t), sin(2t), t) 0?t? ?/4 by the force F=<2yz,xz, 4xy>

I'm pretty sure I did everything in the beginning correctly, i.e. finding c'(t) and F(c(t)) and the dot product between F(c(t).c'(t)...It's just my integration is a little off... I shouldn't have an extra ?. The answer in the book is 1/32(20-?^2)
Thanks in advance!

WORK-please look and check
http://img40.imageshack.us/img40/1866/hw10.jpg
http://img4.imageshack.us/img4/9555/hw11.jpg
the answer-http://img3.imageshack.us/img3/4928/hw12.jpg

 

[galactus]

We have \(\displaystyle (2yzi+xzj+4xyk)\cdot (dxi+dyj+dzk)\)

Substitute, \(\displaystyle x=cos(2t), \;\ y=sin(2t), \;\ z=t, \;\ dx=-2sin(2t), \;\ dy=2cos(2t), \;\ dz=1\)

Then, take the dot product:

\(\displaystyle DotP\left[2tsin(2t), \;\ tcos(2t), \;\ 4cos(2t)sin(2t)\right]\cdot \left[-2sin(2t), \;\ 2cos(2t), \;\ 1\right]=4sin(2t)cos(2t)-6tsin^{2}(2t)+2t\)

\(\displaystyle \int_{0}^{\frac{\pi}{4}}\left[4sin(2t)cos(2t)-6tsin^{2}(2t)+2t\right]dt\)

\(\displaystyle =\frac{5}{8}-\frac{{\pi}^{2}}{32}\)

Factor:

\(\displaystyle \boxed{\frac{20-{\pi}^{2}}{32}}\)

Run the integral through a calculator or computer program once you have it set up. Laboriously doing these things by hand anymore is anachronistic.

That's what technology is for. Setting up the problem is the trick, then solve the integral the easy way. Unless your instructor is one of those who lives in the past and wants you to do it by hand.

 

[pockystix]

thank you! thank you! thank you galactus! You have just saved my life.

I calculated my dot product incorrectly....Your answer is right, but I don't understand how you got 4sin(2t)cos(2t)-6tsin^2(2t)+2t for your dot product.
When I calculated it, I got -4tsin^2(2t)+2tcos^2(2t)+4cos(2t)sin(2t)...

Isn't it true that when you calculate the dot product, it's a1a2+b1b2+c1c2?

 

[galactus]

Yes, that is the dot product. What I have is just simplified a bit using the identity \(\displaystyle cos^{2}(2t)=1-sin^{2}(2t)\)

You can just do it without that. Just dot it and then do the integration without simplifying: \(\displaystyle \int_{0}^{\frac{\pi}{4}}\left[-4tsin^{2}(2t)+2tcos^{2}(2t)+4cos(2t)sin(2t)\right]dt\)

Try and see what I done to get it down to \(\displaystyle 4sin(2t)cos(2t)-6tsin^{2}(2t)+2t\), if you wish.