Answer
The work done along the parabolic path will be minimum for $c=\frac{1}{4}$ and the minimum value of work done is 119.5 units.
If the object is moved along a straight line path connecting the two points then the work required is 120 units. so the work requirement is less along the parabolic path with a suitable choice of $c$.
Work Step by Step
The parabolic path is given by $y = c(1-x^2)$
$\therefore \textbf r(x,y) = x\textbf i + y\textbf j = x\textbf i + c(1-x^2)\textbf j\\
\therefore d\textbf r = (\textbf i - 2cx\textbf j) dx$
Force field $\textbf F = 15[(4-x^2y)\textbf i - xy \textbf j]$
Workdone $W_1 = \int_C \textbf F \cdot d\textbf r\\
= 15\int_{-1}^1 [(4-x^2y)\textbf i - xy \textbf j] \cdot (\textbf i - 2cx\textbf j) dx\\
= 15 \int_{-1}^1 [4+(c-2c^2)(x^4 - x^2)]dx\\
= 15[4x+(c-2c^2)(\frac{x^5}{5} - \frac{x^3}{3})]_{-1}^1\\
= 60[2+\frac{1}{15}(c-2c^2)$
$W_1 $ will be minimum when $\frac{dW_1}{dc} = 0\\
\Rightarrow 4c-1 =0 \Rightarrow c = \frac{1}{4}$
$\therefore$ minimum work done $= 60[2+\frac{1}{15}(\frac{2}{16}- \frac{1}{4})] = 119.5$ units
the straight line conecting (-1,0) and (1,0) can be written as $\textbf r = x\textbf i$ for $-1\le x \le 1$
$therefore d\textbf r = dx \textbf i$, along this path $y = 0$
Work done in that case, $W_2 = \int \textbf F \cdot d\textbf r = \int 15[(4-x^2y)\textbf i - xy \textbf j] \cdot (dx \textbf i)\\
= \int _{-1}^1 60 dx = 120$ units.
So, in the parabolic path , the work done is less than the work required to move the object along the straight line path.
