Question

The ends A and B of a rod of length $\sqrt{5}$ are sliding along the curve y = 2x². 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 $\frac{dx_{B}}{dx_{A}}$ has the value equal to –

• A. 1/3
• B. 1/5
• C. 1/8
• D. 1/9

Answer 1

Answer: (D)

1/9

Answer 2

Given, y = 2x²

Now, (AB)² = (X_B – X_A) + (2x_B² – 2x_A² ) = 5

or (x_B – x_A)² + 4 (x_B² – X_A²)² = 5

On differentiating wrt X_A and denoting$\frac{dx_{B}}{dx_{A}}$=D

2(x_B – x_A) (D – 1) + 8(x_B² – x_A²) (2x_BD – 2x_A) = 0

On putting x_A = 0; x_B = 1, then

⇒ 2(1 – 0) (D – 1) + 8(1 – 0) (2D – 0) = 0

⇒ 2D – 2 + 16D = 0

⇒ D = $\frac{1}{9}$.