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Final review potential function

mearing

Is this the correct way to solve question 2 from the “a bit more” section of the review sheet?

\vec{F}(x,y)=\langle2xy^2+1,3x^2y^2+1\rangle

\int 2x^2y^3+1 \space dx=x^2y^3+x+c(y)

\int 3x^2y^2+1\space dy=x^2y^3+y

c(y)=y

f(x,y)=x^2y^3+x+y

f(b)-f(a) :[(1)^2(1)^3+1+1]-[(0)^2(0)^3+0+0]=3

mark

Yes, that’s correct!! Of course, you can always check by computing \nabla f and checking that you do, indeed get \vec{F}.

myost

I am still confused on the whole +c(y) part and how you get f(x,y)

mark

Let’s see what happens if we don’t use the +c(y) part. Starting with the vector field

\vec{F}(x,y) = \left\langle 2 x y^3+1,3 x^2 y^2+1\right\rangle,

we’re looking for a function f whose gradient is \vec{F}. That is

\left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \left\langle 2 x y^3+1,3 x^2 y^2+1\right\rangle.

Well, if we take the anti-derivative of 2xy^3+1 with respect to x but ignore the constant of intgration, we get

f(x,y) = x^2 y^3 + x.

So that’s our candidate for f If we compute the gradient, though, we get

\nabla f = \left\langle 2 x y^3+1,3 x^2 y^2\right\rangle,

which is just not right since it’s missing the +1 on the second component.

Had we included the +c, we would’ve had

f(x,y) = x^2 y^3 + x + c(y).

Again, the constant can potentially include a y, since we’ve computed an antiderivative with respect to x and y is constant, as far as x is concerned. When we differentiate that with respect to y, we then see that c'(y) is exactly the 1 that we were missing.