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 $± 4$
• 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
$6 \frac{dy}{dt}=3x^{2} \frac{dx}{dt} \Rightarrow 6\times8 \frac{dx}{dt}=3x^{2} \frac{dx}{dt}$
$\Rightarrow 3x^{2}=48 \Rightarrow x^{2}=16$
$\Rightarrow x=\pm4$
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 $\left(-4, \frac{-31}{3}\right)$