Let ${\nabla f} = -4xe^{-x^{2}}\sin\!\left(3y\right)\,\vec i + 6e^{-x^{2}}\cos\!\left(3y\right)\, \vec j$. Find the change in $f$ between $(0,0)$ and $(1,\pi/2)$ in two ways.

(a) First, find the change by computing the line integral $\int_C \nabla f\cdot d\vec r$, where $C$ is a curve connecting $(0,0)$ and $(1,\pi/2)$.

The simplest curve is the line segment joining these points. Parameterize it:
with $0\le t\le 1$, $\vec r(t) =$ $\vec i +$ $\vec j$
So that $\int_C\nabla f\cdot d\vec r = \int_0^1$ $dt$
Note that this isn't a very pleasant integral to evaluate by hand (though we could easily find a numerical estimate for it). It's easier to find $\int_C\nabla f\cdot d\vec r$ as the sum $\int_{C_1}\nabla f\cdot d\vec r + \int_{C_2}\nabla f\cdot d\vec r$, where $C_1$ is the line segment from $(0,0)$ to $(1,0)$ and $C_2$ is the line segment from $(1,0)$ to $(1,\pi/2)$. Calculate these integrals to find the change in $f$.
$\int_{C_1}\nabla f\cdot d\vec r =$
$\int_{C_2}\nabla f\cdot d\vec r =$
So that the change in $f = \int_C\nabla f\cdot\vec r = \int_{C_1}\nabla f\cdot d\vec r + \int_{C_2}\nabla f\cdot d\vec r =$

(b) By computing values of $f$. To do this,
First find $f(x,y) =$
Thus
$f(0,0) =$ and $f(1,\pi/2) =$ ,
and the change in $f$ is .