The point on the curve \(y^2=16x\) for which the y-coordinate is changing 2 times as fast as the x-coordinate is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (1, 4)
Differentiating \(y^2 = 16x\) with respect to time t (rates of change):
\( \displaystyle 2y\,\frac{dy}{dt} = 16\,\frac{dx}{dt} \)
The y-coordinate changes 2 times as fast as the x-coordinate:
\( \displaystyle \frac{dy}{dt} = 2\,\frac{dx}{dt} \)
Substituting:
\( \displaystyle 2y \times 2\,\frac{dx}{dt} = 16\,\frac{dx}{dt} \)
\(4y = 16,\quad y = 4\)
From the curve:
\(16 = 16x,\quad x = 1\)
So the point is (1, 4).