Line integral of a conservative vector field

stevecallaway
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Homework Statement



This is an example in my book, and this is the information in the question.

Find the work done by thr force field F(x,y) = (1/2)xy i + (1/4)x^2 j (with i and j vectors) on a particle that moves from (0,0) to (1,1) along each path (graph shows a x=y^2 curve from (0,0) to (1,1)). This is the information in the question.

Homework Equations


This is the answer given in this example. r(t)=t i + sqrt(t) j for 0<=t<=1, so that dr = (i + 1/(2sqrt(t))dt and F(x,y)=(1/2)t^(3/2) i + (1/4)t^2 j. Then the work done is integral from 0 to 1, (5/8)t^(3/2)dt=(1/4)t^(5/2) = 1/4


The Attempt at a Solution


My question is how do you attain the r(t)=t i + sqrt(t) j equation from the information given?
 
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so
r(t) = (x(t), y(t)) = (t,\sqrt{t})
is a parametric representation of x = y^2, try substituting in and see if it satisfies the equation

there are actually infinite ways to paremeterise the curve, this one was chosen by starting with x(t) = t, and finding y(t)
 
As landance said, there are an infinite number of ways to parameterize a curve. Another perfectly valid one would be x= t^2, y= t.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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