Question

A particle moves along the curve 6y=x^3 +2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer 1

A particle moves along the curve 6y = x³ + 2
differentiate with respect to time,
e.g., $\bf{6\frac{dy}{dt}=\frac{dx^3}{dt}+\frac{d2}{dt}}$
$\bf{6\frac{dy}{dt}=3x^2\frac{dx}{dt}}----(1)$

A/C to question,
we have to find out the point on the curve at which the y coordinate is changing 8 times as fast as the x - coordinate.
e.g., $\bf{\frac{dy}{dt}=8\frac{dx}{dt}}$
now put it in equation (1),
$\bf{6.8\frac{dx}{dt}=3x^2\frac{dx}{dt}}\\\\\bf{(16-x^2)\frac{dx}{dt}=0}\\\\\implies(x^2-16)=0\\\\\bf{x=\pm 4}$

now, put x = 4 in 6y = x³ + 2
6y = 4³ + 2 = 66
6y = 66 => y = 11
hence, a point on the curve is (4,11)

put x = -4 in 6y = x³ + 2
6y = (-4)³ + 2 = -64 + 2 = -62
6y = -62 => y = -31/3
hence, another point on the curve is (-4,-31/3)

hence, there are two points (4,11) and (-4,-31/3)