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Quick math question - Just look over my work.

R

RESmonkey

Diamond Member
The function v(x) is the velocity in m/sec of a particle moving along the x-axis. Find the total distance traveled by the particle.

Given: v(x) = 5(cos(x)^2)sin(x) , 0 <= x <= 2pi

A. 20/3 <--the answer
B. 0
C. 20
D. 10/3

My work:
----------
Since velocity is given, I just integrated it from 0 to 2pi. But I got 1.7 something, which fits NONE of the answers.

So then, I decided to integrate using the distance formula, since we're on that part of the chapter: (1 + v(x)^2)^.5 integrated from 0 to 2pi. I got 6.58037, which is close to 20/3, which is the answer.

***Question: WHY didn't my first approach work? Can't you just integrate velocity to get distance traveled? Why did it fail?


 
Well I can tell you that integrating that equation will NOT give you position... what you have now is velocity as a funtion of x, NOT time. Not sure how you do it exactly, but that's why you were wrong at first.
 
You want the total distance moved. You found the final resting spot of the object. Integrating velocity gives you position at time t.

To get the total distance, you need to integrate the absolute value of the velocity.

Hint: Find where the velocity is negative and break up your integral accordingly.
 
1. Velocity isn't negative in that graph.

2. I replaced X in for t (it's written like v(t) instead of v(x) on the paper, but that shouldn't matter).

3. There is a similar problem at the bottom of the page, same instructions, and I got the right answer by integrating.

This leads me to believe my original problem may be flawed; the instructions are asking for something else. Correct?
 
The velocity should be negative. You're squaring that cosine, so thats always going to be positive. But the sine goes negative in 0 to 2pi.
 
Yes it does. Is your calculator in degrees by chance? 😀

Plug in 4 and you'll get a negative number
 
In fact, if you integrate that, you get 0.
If you do absolute value, you get 20/3.

Edit: If you accidentally leave your calculator in degrees and integrate that, you get 1.71 😀
 
That's why! Argghhh! Thank you, I don't know how it went to degree mode. I haven't used any programs that do that to the calc. Hmm...
 
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