Question

A particle moves along the curve 6y = x³ + 2, Find the points on the curve at which y-co-ordinate is changing 8 times as fast as the x-co-ordinate.

Answer 1

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Answer 2

Answer:

Points are (-4, -10.33) and (4, 11).

Step-by-step explanation:

In the question,

The equation of th movement of the particle is given along the curve by,

$6y = x^{3} + 2$

So,

$y=\frac{x^{3}+2}{6}$

Now, on differentiating 'y' w.r.t 'x' we get,

$y=\frac{x^{3}+2}{6}\\\frac{dy}{dx}=\frac{1}{6}\frac{d(x^{3}+2)}{dx}\\So,\\\frac{dy}{dx}=\frac{1}{6}(3x^{2})=\frac{x^{2}}{2}$

Now,

We have already been provided that rate of change of 'y' w.r.t 'x' is = 8.

So,

$8=\frac{x^{2}}{2}\\x^{2}=16\\x=-4,4$

Therefore, the points on the curve are given at,

x = -4 and x = 4

So,

At, x = -4, y = -10.33

And,

At, x = 4, y = 11

So,

Points are (-4, -10.33) and (4, 11).