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Sagot :
A parametrization of path C is x= 4t and y = 8t
Line segments on nearly trivial to parametrize. It is often it easiest to do in interval t∈ [0,1]
Then we just need to think about what we need to add to each first coordinate to get the second
Since, 0+ 4 = 4 and 0+ 8 = 8
So, parametrization is
=> x = 4t
y = 6t
(b) To evalute the line integral, we have to derivate parametrize
dx/dt = 4
dy/dt = 8
So, the differential line element is
[tex]ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}[/tex]
=> √(4)²+(8)²
=> 4√5
Integrating ,
[tex]\int\limits_{c} {x^2 + y^2} \, ds\\ = \int\limits^1_0 {(4t)^2 + (8t)^2 } 4\sqrt{5} \,dt\\=4\sqrt{5} [\frac{80}{3} t^3]^{1}_{0}\\[/tex]
Putting the limits
=> 285.51
To know more about Parametrization here
https://brainly.com/question/14666291
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