Question

A particle moves along the curve $6y=x^3+2$. The point $‘P’$ on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate, are $(4,11)$ and $\left(-4,-\frac{31}{3}\right).$

• A. x-coordinates at the point P are $\pm 4$
• B. y-coordinates at the point P are 11 and $\frac{-31}{3}$
• C. Both (a) and (b)
• D. None of the above

Answer 1

Answer: (C)

Given, $6y=x^3+2$
On differentiating w.r.t. t, we get
$6\frac{dy}{dt}=3x^2\frac{dx}{dt}\Rightarrow 6\times 8\frac{dx}{dt}=3x^2\frac{dx}{dt}$
$\Rightarrow 3x^2=48\Rightarrow x^2=16$
$\Rightarrow x=\pm 4$
When $x=4$, then $6y={\left(4\right)}^3+2$
$\Rightarrow 6y=64+2\Rightarrow y=\frac{66}{6}=11$
When $x=-4$, then $6y={\left(-4\right)}^3+2$
$\Rightarrow 6y=-64+2$
$\Rightarrow y=\frac{-62}{6}=\frac{-31}{3}$
Hence, the required points on the curve are $\left(4,11\right)$ and $\left(-4,\frac{-31}{3}\right)$