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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. |
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Answer» Let the required point be (x, y) Given: `(dy)/(dt) = 8 (dx)/(dt)`...(i) Given curve is `6y = x^(3) + 2` ...(ii) On differentiating both sides of (ii) w.r.t. t, we get `6(dy)/(dt) = 3x^(2) (dx)/(dt)` `rArr 6(8 (dx)/(dt)) = 3x^(2) (dx)/(dt)` [using (i)] `rArr 3x^(2) = 48 rArr x^(2) = 16 rArr x = +-4` Putting x = 4 in (ii), we get y = 11 Putting `x = -4` in(ii), we get `y = (-62)/(6) = (-31)/(3)` Hence, the required points are (4,11) and `(-4, (-31)/(3))` |
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