Line integral and continuous gradient

[nlsherrill]

Homework Statement

A table of values of a function f with continuous gradient is given. Find the line integral over C of "gradient F dr" where C has parametric equations x = t2 + 1, y = t3 + t, 0<=t<= 1.


Sorry, don't know latex.

But here's a picture of the table and values

Homework Equations

The Attempt at a Solution

I'm not even sure what the problem wants me to do with the table. I "assume" since the problem says the table represents values of a function, that I should look at the table and construct a function from it. Well I have tried a few different ways and none of them can stay consistent for more than a line. I feel like this problem is easier than it seems.

 

[gabbagabbahey]

HINT: What does the fundamental theorem of gradients tell you?

 

[nlsherrill]

HINT: What does the fundamental theorem of gradients tell you?

So basically with that definition and the table provided, they were looking for you to just evaluate the function at its endpoints correct? I got the right answer, which is 6, but I don't know if what I did was the right way to do it. The parameter t ranges from 0 to 1, so I just plugged in 0 to the x and y components to get (1,0), which on the table=4. Then plugging in 1 for t in the components I got (2,2), which on the table =10. Subtracting the endpoint from the starting point gave me 6. Is this the correct way to do it?

And thank you for your help so far.

 

[HallsofIvy]

Yes, the as long as f(x,y)dx+ g(x,yy)dy is an "exact differential", that is, there exist F(x,y) such that dF= f(x)dx+ g(y)dy, then
\int_{t_0}^{t_1} f(x)dx+ g(y)dy= F(x(t_1),y(t_1))- F(x(t_0), y(t_0))

Here, you are given that the F whose values are tabulated gives the required gradient. Since the path is apparently given by x = t^2 + 1, y = t^3 + t, 0\le t\le 1, the endpoints are at (1, 0) and (2, 2).

Your integral is just F(2, 2)- F(1, 0) both of which can be read off the table.