I need to find the work required to move a certain block from point A to point B. $\vec{F}(x,y)=2y^{3/2}\ \textbf{i}+3x\sqrt{y} \ \textbf{j}$, where point A is $(1,1)$ and point B is $(2,9)$. I do know how to solve this problem, but I thought of another way to solve it.
The first method I used to solve this problem: $P=2y^{3/2}$ and $Q=3x\sqrt{y}$ Because the force field $\vec{F}$ has to be conservative, $\frac{\partial P}{\partial x}=2y^{3/2}$ and $\frac{\partial Q}{\partial y}=3x\sqrt{y}$.
I integrated $\frac{\partial f}{\partial x}$ and got $f(x,y)=2xy^{3/2}+h(y)$. I differentiated both sides with respect to $y$ and got $\frac{\partial f}{\partial y}=3y\sqrt{x}+\frac{\partial h(y)}{\partial y}$.
I get that $\frac{\partial h(y)}{\partial y}=0$, which means that $h(y)=c$. If we assume that $c=0$, $f(x,y)=2xy^{3/2}$.
Using the FTC of Line Integrals, I get that $$\int\limits_C \nabla f\cdot d\vec{r}=f\left(\vec{r}\left(b\right)\right)-f\left(\vec{r}\left(a\right)\right)$$. And found work as $$W=\int\limits_C \nabla f\cdot d\vec{r}=f\left(2,9\right)-f\left(1,1\right) = 2\cdot2\cdot9^{3/2}-2\cdot1\cdot1^{3/2} = 106$$
The second method that I thought of was using parametric equations. Finding a parametric equation that passes through point A and B, through time $t = 0$ to $t = 1$, I get $x=1+t$ and $y=8t+1$. I use a different way to evaluate the FTC Line Integral: $$\int\limits_C \vec{F}\cdot d\vec{r} = \int_0^1 \begin{bmatrix} 2y^{3/2} \\ 3x\sqrt{y} \\ \end{bmatrix} \cdot \frac{d}{dt}\left(\begin{bmatrix} 1+t \\ 1+8t \\ \end{bmatrix}\right) dt = \int_0^1 \begin{bmatrix} 2y^{3/2} \\ 3x\sqrt{y} \\ \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 8 \\ \end{bmatrix} dt = \int_0^1 2y^{3/2}+24x\sqrt{y} \ dt$$.
I substitute $x=1+t$ and $y=8t+1$ into the integral and evaluate the following integral. $$\int_0^1 {2\left(8t+1\right)^{3/2}+24\cdot\left(t+1\right)\cdot\sqrt{8t+1}} \ dt = 106$$, thus giving the same answer.
I want to know why the ways of solving the problem results in the same solution, but the first way of solving the problem requires me to assume that $c=0$? Am I missing something that allows me to safely assume that the constant of integration is 0?
In fact it doesn’t matter what constant of integration $c$ you have chosen, if you realize in your last step of method 1, the constant cancel out of each other when you subtract the potential.
Physically you may just treat the constant as some sort of reference value, and it is immaterial of which reference value you have chosen.