Question

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

Answer 1

Let P(x₁, y₁) be the point on the curve 6y = x³ + 2 whose y-coordinate is changing 8 times as fast as the coordinate.

Then `(dy/dx)_("at"(x_1, y_1)) = 8(dx/dt)_("at"(x_1, y_1)`       ...(1)

Differentiating 6y = x³ + 2 w.r.t. t, we get

`6dy/dt = d/dt(x^3 + 2) = 3x^2dx/dt + 0`

∴ `2dy/dt = x^2dx/dt`

∴ `2(dy/dt)_("at"(x_1, y_2)) = x_1^2.(dx/dt)_("at"(x_1, y_1)`

∴ `2 xx 8(dx/dt)_("at"(x_1, y_2)) = x_1^2.(dx/dt)_("at"(x_1, y_1)`  ...[By (1)]

∴ x₁² = 16

∴ x₁ = ± 4

Now, (x₁, y₂) lies on the curve 6y = x³ + 2.

∴ 6y₁ = x₁³ + 2

When x₁ = 4, 6y₁ = (4)³ + 2 = 66

∴ y₁ = 11

When x₁ = - 4, 6y₁ = (– 4)³ + 2 = – 62

∴ y₁ = `-(31)/(3)`

Hence, the required points on the curve are (4, 11) and `(-4, (-31)/3)`.