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

The correct answer is \((4, 11)\) and \((-4,\frac{-31}{3}).\)
The equation of the curve is given as:
\(6y=x^3+2\)
The rate of change of the position of the particle with respect to time \((t)\) is given by,
\(6\frac{dy}{dt}=3x^2 \frac{dx}{dt}+0\)
\(=2 \frac{dy}{dt}=x^2 \frac{dx}{dt}\)
When the y-coordinate of the particle changes 8 times as fast as the
\(x\)-coordinate i.e.,\((\frac{dy}{dt}=8\frac{dx}{dt})\),we have:
\(2(8\frac{dx}{dt})=x^2 \frac{dx}{dt}\)
\(\implies 16\frac{dx}{dt}=x^2 \frac{dx}{dt}\)
\((x^2-16)\frac{dx}{dt}=0\)
\(\implies x^2=16\)
\(x=±4\)
When \(x=4,y=\frac{4^3+2}{6}=\frac{66}{6}=11\)
When \(x=-4,y=\frac{(-4)^3+2}{6}=\frac{-62}{2}=\frac{-31}{3}\)
Hence, the points required on the curve are \((4, 11)\) and \((-4,\frac{-31}{3}).\)