A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by \(y = y(x) = x + \sqrt{2x - 3x^2}\) where ‘x’ (in meters) is the horizontal distance moved and ‘y’ (in meters) is the corresponding height. Assuming that his initial point is at origin.
Based on the above information answer the Questions
The value of \(\displaystyle\frac{dy}{dx}\) when \(x = \displaystyle\frac{1}{6}\), is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
By the chain rule,
\( \displaystyle \frac{dy}{dx} = 1 + \frac{2 - 6x}{2\sqrt{2x - 3x^2}} = 1 + \frac{1 - 3x}{\sqrt{2x - 3x^2}} \)
At \(x = \frac{1}{6}\):
\( \displaystyle 1 - 3x = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \displaystyle 2x - 3x^2 = \frac{1}{3} - \frac{1}{12} = \frac{1}{4},\quad \sqrt{\frac{1}{4}} = \frac{1}{2} \)
So
\( \displaystyle \frac{dy}{dx} = 1 + \frac{1/2}{1/2} = 2 \)