这是一份manchester曼切斯特大学 MATH20411作业代写的成功案例
$$
\text { Force } \times \text { Distance } \approx f\left(x_{i_{*}}, y_{i *}\right) \sqrt{\Delta x_{i}^{2}+\Delta y_{i}^{2}} \text {, }
$$
where $\left(x_{i *}, y_{i *}\right)=\left(x\left(t_{i} *\right), y\left(t_{i} *\right)\right)$ for some $t_{i} *$ in $\left[t_{i}, t_{i+1}\right]$, and so
$$
W \approx \sum_{i=0}^{n-1} f\left(x_{i *}, y_{i *}\right) \sqrt{\Delta x_{i}^{2}+\Delta y_{i}^{2}}
$$
is approximately the total amount of work done over the entire curve. But since
$$
\sqrt{\Delta x_{i}^{2}+\Delta y_{i}^{2}}=\sqrt{\left(\frac{\Delta x_{i}}{\Delta t_{i}}\right)^{2}+\left(\frac{\Delta y_{i}}{\Delta t_{i}}\right)^{2}} \Delta t_{i}
$$
where $\Delta t_{i}=t_{i+1}-t_{i}$, then
$$
W \approx \sum_{i=0}^{n-1} f\left(x_{i *}, y_{i *}\right) \sqrt{\left(\frac{\Delta x_{i}}{\Delta t_{i}}\right)^{2}+\left(\frac{\Delta y_{i}}{\Delta t_{i}}\right)^{2}} \Delta t_{i} .
$$
MATH20411 COURSE NOTES ：
Solution: We need to find a real-valued function $F(x, y)$ such that
$$
\frac{\partial F}{\partial x}=x^{2}+y^{2} \quad \text { and } \quad \frac{\partial F}{\partial y}=2 x y .
$$
Suppose that $\frac{\partial F}{\partial x}=x^{2}+y^{2}$, Then we must have $F(x, y)=\frac{1}{3} x^{3}+x y^{2}+g(y)$ for some function $g(y)$. So $\frac{\partial F}{\partial y}=2 x y+g^{\prime}(y)$ satisfies the condition $\frac{\partial F}{\partial y}=2 x y$ if $g^{\prime}(y)=0$, i.e. $g(y)=K$, where $K$ is a constant. Since any choice for $K$ will do (why?), we pick $K=0$. Thus, a potential $F(x, y)$ for $\mathbf{f}(x, y)=\left(x^{2}+y^{2}\right) \mathbf{i}+2 x y \mathbf{j}$ exists, namely
$$
F(x, y)=\frac{1}{3} x^{3}+x y^{2} .
$$