Question

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