The variable x is increasing at the rate of 2 units per second. Match each function y in List-I with the rate of change of y (in units per second) at the instant when x = 3 in List-II
| List-I | List-II | ||
| (A) | \(y = x^2\) | (I) | \(-\displaystyle\frac{2}{9}\) |
| (B) | \(y = x^3\) | (II) | \(\displaystyle\frac{1}{2}\) |
| (C) | \(y = \displaystyle\frac{1}{x}\) | (III) | 54 |
| (D) | \(y = \sqrt{x + 1}\) | (IV) | 12 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
By the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\), with \(\frac{dx}{dt} = 2\) and x = 3.
(A) \(y = x^2\)
\(\frac{dy}{dt} = 2x \times 2 = 2(3)(2) = 12\)
This matches (IV) 12.
(B) \(y = x^3\)
\(\frac{dy}{dt} = 3x^2 \times 2 = 3(9)(2) = 54\)
This matches (III) 54.
(C) \(y = \frac{1}{x}\)
\( \displaystyle \frac{dy}{dt} = -\frac{1}{x^2} \times 2 = -\frac{2}{9} \)
This matches (I) \(-\frac{2}{9}\); y decreases as x increases.
(D) \(y = \sqrt{x + 1}\)
\( \displaystyle \frac{dy}{dt} = \frac{1}{2\sqrt{x + 1}} \times 2 = \frac{1}{\sqrt{4}} = \frac{1}{2} \)
This matches (II) \(\frac{1}{2}\).