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A particle moves along the curve 6y = x^3 +2. Find the points

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A particle moves along the curve $6 y = x^{3} + 2$ . Find the points on the curve at which the $y$ -coordinate is changing $8$ times as fast as the $x$ -coordinate.

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A particle moves along the curve

6 y = x^{3} + 2

. Find the points on the curve at which the

y

-coordinate is changing

8

times as fast as the

x

-coordinate.

9 mins ago

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9 mins ago

Text solutionVerified

The equation of the curve is given as,

6 y = x^{3} + 2

Differentiating both sides w.r.t.

t

6 \frac{d y}{d t} = 3 x^{2} \frac{d x}{d t} + 0

\Rightarrow 2 \frac{d y}{d t} = x^{2} \frac{d x}{d t}

It is given that,

(\frac{d y}{d t} = 8 \frac{d x}{d t})

Thus we have,

2 (8 \frac{d x}{d t}) = x^{2} \frac{d x}{d t}

\Rightarrow 16 \frac{d x}{d t} = x^{2} \frac{d x}{d t}

\Rightarrow (x^{2} - 16) \frac{d x}{d t} = 0

\Rightarrow x^{2} - 16 = 0 \Rightarrow x = \pm 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}{6} = - \frac{31}{3}

Hence, the points required on the curve are

(4, 11)

and

(- 4, \frac{- 31}{3})

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Question Text	A particle moves along the curve $6 y = x^{3} + 2$ . Find the points on the curve at which the $y$ -coordinate is changing $8$ times as fast as the $x$ -coordinate.
Answer Type	Text solution:1
Upvotes	150