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A particle travels on the curve y = x³ + 2. Find the point(s) P on the curve where the y-coordinate changes at a rate 8 times the rate of change of the x-coordinate. The points are (4, 11) and (-4, -31/3).
- The x-values of point P are ±4
- The y-values of point P are 11 and -31/3
- Both the x-values in (a) and the y-values in (b)
- None of these
Correct answer: None of these
Solution
dy/dt = 8 dx/dt means dy/dx = 8, and dy/dx = 3x^2, so 3x^2 = 8 -> x = +-sqrt(8/3) ≈ +-1.63. The quoted points (4,11) and (-4,-31/3) do not satisfy this (at x=4, dy/dx=48), so neither (a) nor (b) is correct: None of these.
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