The point on the curve $y^2=16 x$ 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)
$y^2 = 16x$
$2y\frac{dy}{dt} = 16\frac{dx}{dt}$
$\frac{dy}{dt} = \frac{8}{y}\frac{dx}{dt}$
$\text{Given } \frac{dy}{dt} = 2\frac{dx}{dt}$
$\frac{8}{y}\frac{dx}{dt} = 2\frac{dx}{dt}$
$\frac{8}{y} = 2$
$y = 4$
$y^2 = 16x \Rightarrow 16 = 16x \Rightarrow x = 1$
The required point is $(1, 4)$.