Vector Calc-Line Integral

In summary, the mass of the wire lying along the piecewise linear curve extending from the point (2,4) to the point (8,6) to the point (8,11) can be found by parametrizing the curve and using a line integral with the integrand \rho ds=\rho \frac{ds}{dt}dt, where ds is a piece of length from the curve, to get a result of 266.
  • #1
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A wire lies along the piecewise linear curve extending from the point (2,4) to the point (8,6) to the point (8,11). If the density of the wire is given by p(x,y)=2xy+6x, use a line integral to find the mass of the wire.

Here is what I have tried:

C1: (1-t)<2,4>+t<8,6>
C1: x=2+6t y=4+2t

[tex]\int_{0}^{1} 2xy+6y dt [/tex]
[tex]\int_{0}^{1} 2(2+6t)(4+2t)+6(2+6t) dt = 82[/tex]

C2: (1-t)<8,6> + t<8,11>
C2: x=8 y=6+5t

[tex]\int_{0}^{1} 2xy+6y dt [/tex]
[tex]\int_{0}^{1} 2(8)(6+5t)+6(8) dt = 184[/tex]

82+184 = 266

However this is incorrect :(
 
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  • #2
Tom McCurdy said:
C1: (1-t)<2,4>+t<8,6>
C1: x=2+6t y=4+2t

[tex]\int_{0}^{1} 2xy+6y dt [/tex]
I think the integrand should be [tex]\rho ds=\rho \frac{ds}{dt}dt[/tex]
where ds is a piece of length from your curve. So if you parametrized the line by [tex]\vec r(t)[/tex], then [tex]ds=|\frac{d\vec r}{dt}|dt[/tex]
 
  • #3
Are you saying I need to do

[tex] \int_{0}^{1} (2xy+6x)* \sqrt{(2y+6)^2+(2x)^2} [/tex]

C1: (1-t)<2,4>+t<8,6>
C1: x=(2+6t) y=(4+2t)


[tex]\int_{0}^{1} (2(2+6t)(4+2t)+6(2+6t)) * \sqrt{(2(4+2t)+6)^2+(2(2+6t))^2} [/tex]

C2: (1-t)<8,6> + t<8,11>
C2: x=(8) y=(6+5t)

[tex]\int_{0}^{1} 2xy+6y dt [/tex]
[tex]\int_{0}^{1} (2(8)(6+5t)+6(8)) * \sqrt{(2(6+5t)+6)^2+(2(8))^2} = [/tex]

Then add the results of the two integrals?

However this is incorrect :(
 
  • #4
Galileo said:
I think the integrand should be [tex]\rho ds=\rho \frac{ds}{dt}dt[/tex]
where ds is a piece of length from your curve. So if you parametrized the line by [tex]\vec r(t)[/tex], then [tex]ds=|\frac{d\vec r}{dt}|dt[/tex]

Any ideas... I was not quite sure what you meanat by this post...
 
  • #5
Ahhh I figured it out... I was doing the wrong dervatives...
 

1. What is a vector in vector calculus?

A vector in vector calculus is a quantity that has both magnitude and direction. It is represented by an arrow with a specific length and direction in a coordinate system.

2. What is a line integral in vector calculus?

A line integral in vector calculus is a mathematical tool used to calculate the total value of a vector field along a given curve. It takes into account both the direction and the length of the curve.

3. How is a line integral calculated?

A line integral is calculated by breaking up the curve into small segments and approximating the value of the vector field at each point. The values are then summed up to get the total value of the line integral.

4. What is the significance of line integrals in physics?

Line integrals have many applications in physics, particularly in the fields of electromagnetism and fluid mechanics. They are used to calculate work, flux, and circulation, which are important concepts in these areas.

5. Can line integrals be calculated in three-dimensional space?

Yes, line integrals can be calculated in three-dimensional space. In this case, the curve is represented by a parametrization using three variables, and the line integral is calculated by taking into account the three-dimensional vector field at each point on the curve.

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