Question

The ends A and B of a rod of length $\sqrt{5}$ are sliding along the curve $y=2x^2$. Let $x_A$ and $x_B$ be the x-coordinate of the ends. At the moment when A is at ( 0, 0) and B is at (1, 2) the derivative $\frac{d{x}_B}{d{x}_A}$ has the value equal to

• A. 1/3
• B. 1/5
• C. 1/8
• D. 1/9

Answer 1

Answer: (D)

1/9

Let $y_B=2{x_B}^2,y_A=2{x_A}^2$

Given length of rod is $\sqrt{5}$

$⟹(x_A-x_B{)}^2+(y_A-y_B{)}^2=5$

$⟹(x_A-x_B{)}^2+4(x_A^2-x_B^2{)}^2=5$

$⟹2(x_A-x_b)(1-\frac{d{x}_B}{d{x}_A})+8(x_A^2-x_B^2)(2x_A-2x_B\frac{d{x}_B}{d{x}_A})=0$

$x_a=0,x_B=1$

$⟹2(-1)(1-\frac{d{x}_B}{d{x}_A})-8(-1{)}^2(0-2\frac{d{x}_B}{d{x}_A})=0⟹\frac{d{x}_B}{d{x}_A}=\frac{2}{18}=\frac{1}{9}$