$A$ and $B$ of a rod of length of $\sqrt{5}$ are sliding along the curve $y=2x^2$ . Let $x_A$ and $x_B$ be the $x$ -coordinates of the ends. At the moment when $A$ is at $(0,0)$ and $B$ is at $(1,2)$ . Find the value of the derivative $\dfrac{dx_B}{dx_A}$ .

The ends
$\frac{1}{3}$
$\frac{1}{\sqrt{5}}$
$9$
$\frac{1}{9}$

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Let a = x

a, b = xb, and d(xb)/d(xa) = db/da = b'. The length of the rod is expressible as:5 = (b - a)^2 + (2

b^2 - 2a^2)^2 = (b - a)^2 + 4*(b^2 - a^2)^2 (i)and implicitly differentiating both sides of (i) with respect to a yields:

0 = 2

(b-a)(b' - 1) + 8(b^2 - a^2)(2bb' - 2a);or 2

(b-a) + 16a(b^2 - a^2) = [2(b-a) + 16b(b^2 - a^2)]*b';or b' = [2

(b-a) + 16a(b^2 - a^2)] / [2(b-a) + 16b(b^2 - a^2)] (ii)A final substitution of a = x

a = 0 and b = xb = 1 into (ii) gives b' = 2/(2 +16) = 1/9, or choice D as a correct answer.