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

In this problem, we are given the equation of the path line of the particle. To form a relationship between the velocities along the two axes, we differentiate both sides of the equation with respect to t (time). Then, we put the condition that $\dfrac{dy}{dt}=8\dfrac{dx}{dt}$ and then find the required points.

Complete step by step solution:
In this problem, we have been given the path line along which the particle moves. From the given equation, it is clearly visible that the path is nothing but a cubic equation. The equation is of such a form that we are given a relation between the y coordinate and the x coordinate. The particle moves on a two-dimensional plane with both the x and the y coordinates.
Velocity is nothing but the rate of change of displacement with time. Since the particle has two sets of displacements – one along the x-axis and the other along the y-axis, it experiences two sets of velocities – one along the x-axis and the other along the y-axis. So, we differentiate both sides of the equation with respect to t (time) and get,

& 6\dfrac{dy}{dt}=3{{x}^{2}}\dfrac{dx}{dt} \\\ & \Rightarrow 2\dfrac{dy}{dt}={{x}^{2}}\dfrac{dx}{dt}....\left( i \right) \\\ \end{aligned}$$ Also, we are given the condition that the rate of change of the y-coordinate is $8$ times of that of the rate of change of the x coordinate. This means that, $\dfrac{dy}{dt}=8\dfrac{dx}{dt}....\left( ii \right)$ Inserting equation ii in equation I, we get, $$\begin{aligned} & \Rightarrow 2\left( 8\dfrac{dx}{dt} \right)={{x}^{2}}\dfrac{dx}{dt} \\\ & \Rightarrow 16={{x}^{2}} \\\ & \Rightarrow x=\pm 4 \\\ \end{aligned}$$ Putting these two values of x in the given equation, we get, $$\begin{aligned} & 6y={{4}^{3}}+2,6y={{\left( -4 \right)}^{3}}+2 \\\ & \Rightarrow y=\dfrac{66}{6},y=\dfrac{-62}{6} \\\ & \Rightarrow y=11,-10.33 \\\ \end{aligned}$$ **Therefore, we conclude that the required points are $\left( 4,11 \right)$ and $\left( -4,-10.33 \right)$.** **Note:** These problems are prone to a lot of mistakes. Firstly, most of the students differentiate with respect to x rather than t which is a grave mistake. Secondly, we often do not consider both the values, positive and negative, which leaves us with only half the answer. We should be careful not to commit these mistakes.