anonymous
  • anonymous
need help with a line integral ( i think). The problem is as follows: find the mass of the wire AB ( picture below ) using the density function f(x,y)=x
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1378661580329:dw|
anonymous
  • anonymous
Do you know how to parameterize a line segment?
anonymous
  • anonymous
that would be the issue, mainly.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Well there are two ways to do it. One would be to find the equation of the line, another would be to set up a vector equation. The equation of the line connecting (1,0) and (3,1) is \(y=\dfrac{1}{3}x-\frac{1}{3}\). So you could use the following parameterization: \[C:=x=t,~y=\frac{1}{3}t-3,~1\le t\le3\\ \text{which gives you}\\ \frac{dx}{dt}=1,~\frac{dy}{dt}=\frac{1}{3}\] So the integral would be \[M=\int_Cf(x,y)~ds=\int_1^3t\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt\]
anonymous
  • anonymous
Disregard that vector setup I mentioned. I don't think you have the right conditions to do it.
anonymous
  • anonymous
the line equation i got there was \[y=\frac{ 1 }{2 } x - \frac{ 1 }{2 } \] ( used the formulae ( x2-x1)(y-y1)=(x-x1)(y2-y1). regardless of that, i still don't understand why after x=t y=1/3t-3 and goes between 1 and 3.
anonymous
  • anonymous
That's what I get for trying to do simple arithmetic in my head -_- \(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles The equation of the line connecting (1,0) and (3,1) is \(\color{red}{y=\dfrac{1}{2}x-\dfrac{1}{2}}\). So you could use the following parameterization: \[C:=x=t,~\color{red}{y=\frac{1}{2}t-\frac{1}{2}},~1\le t\le3\\ \text{which gives you}\\ \frac{dx}{dt}=1,~\color{red}{\frac{dy}{dt}=\frac{1}{2}}\] So the integral would be \[M=\int_Cf(x,y)~ds=\int_1^3t\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt\] \(\color{blue}{\text{End of Quote}}\) \(t\) is a dummy variable. You could continue to use \(x\); I use \(t\) out of habit, as these line integral questions usually provide more complicated functions. Suppose we just use \(x\). What would the lower/upper limits of this variable be? The first point is (1,0), whose x-coordinate is 1. The second is (3,1), with x=3. Does that make sense? Next, you differentiate \(x\) and \(y\) with respect to the dummy variable. In my work, it's t, so you find \(dx/dt\) and \(dy/dt\), then plug them into the \(ds\) formula.
anonymous
  • anonymous
And finally, in the integral, since \(f(x,y)=x\), you would simply substitute and say \(f(x,y)=t\).
anonymous
  • anonymous
so, if instead of x we would have done the parametrization by y, with y=t and x=2y+1, t would have ranged from 0 to 1? Of course, the solution would have been different, that much I understand, it's just the parameter that I want to make sure of.
anonymous
  • anonymous
Actually, the solution would be the same. Just a different but equivalent integral.
anonymous
  • anonymous
That's what I wanted to say, sorry. same result, different way to get to it.
anonymous
  • anonymous
Yes, that's it.
anonymous
  • anonymous
And here's a check with Wolfram: http://www.wolframalpha.com/input/?i=Integrate%5Bt*Sqrt%5B5%2F4%5D%2C%7Bt%2C1%2C3%7D%5D%2C+Integrate%5B%282t%2B1%29*Sqrt%5B5%5D%2C%7Bt%2C0%2C1%7D%5D
anonymous
  • anonymous
Thank you very much for the explanation. I don't know exactly why I got scared of that t there and kept looking for a complicated solution for it, but now I got it. Thank you again.
anonymous
  • anonymous
You're welcome!

Looking for something else?

Not the answer you are looking for? Search for more explanations.