Answer
a. See explanations.
b. $0.682327804$
Work Step by Step
a. Using the figure given in the Exercise, for a point $P(x,y)$ on the parabola, we have $y=x^2$. The distance D between point P and the location of the buoy $(2, -\frac{1}{2})$ satisfies $D^2=(x-2)^2+(y+\frac{1}{2})^2=(x-2)^2+(x^2+\frac{1}{2})^2=x^2-4x+4+x^4+x^2+\frac{1}{4}=x^4+2x^2-4x+\frac{17}{4}$. Let $s=D^2$; to minimize $D^2$, we need to set $s'=0$, which gives $4x^3+4x-4=0$ or $x^3+x-1=0$ (which is the same as $x=\frac{1}{x^2+1}$).
b. To solve the above equation is the same as finding zeros of the function $f(x)=x-\frac{1}{x^2+1}$. We have $f'(x)=1+\frac{2x}{(x^2+1)^2}$. Test the signs of $f(x)$ to get $f(0)=-1\lt0, f(1)=1-\frac{1}{2}=\frac{1}{2}\gt0$; thus there is a zero in the interval of $(0,1)$. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a starting point $x_0=1$, we can obtain the zero shown in the table as $0.682327804$.
