How do I get the parametrization?

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SUMMARY

The discussion focuses on finding the correct parametrization for the line integral of the scalar function f(x,y,z) = xe^{z^2} along a specified path. The initial attempt at parametrization was c(t) = <0,t,1-\frac{y}{2>}, which was identified as incorrect. The correct parametrization is c(t) = <0, 2t, 1 - 2t>, as it accurately represents the line segment from (0,0,1) to (0,2,0) and maintains the correct endpoints when t is evaluated from 0 to 1.

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Homework Statement


Compute the line integral of the scalar function.
f(x,y,z) = x[itex]e^{z^2}[/itex], piecewise linear path from (0,0,1) to (0,2,0) to (1,1,1)

Homework Equations





The Attempt at a Solution


In this problem, all I need is a parametrization. First I drew the line from (0,0,1) to (0,2,0) xyz-plane. I got the slope as z = 1 - [itex]\frac{y}{2}[/itex]. So I set y = t then z will be 1 - [itex]\frac{t}{2}[/itex]. I got parametrization as c(t) = <0,t,1-[itex]\frac{y}{2}[/itex]>. But it's wrong. It's c(t) = <0,2t,1-2t>. Would anyone help me how to get parametrization ??
 
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Yes, the line from (0, 0, 1) to (0, 2, 0) can be written as z= 1- y/2. Since you are using t as parameter, that would be <0, t, 1- t/2>. Why do you say that is wrong?
When t= 0, your parameterization gives (0, 0, 1) and when t= 2, it gives (0, 2, 1). A line is determined by two points so it gives the correct line.

It is c(t)= <0, 2t, 1- 2t> that is wrong. In order that y= 2t= 2, t must be 1. But then z= 1- 2= -1, not 0.
 

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