x = 2t^2 + 1, y = (1/3)t^3 - t; t = 3

preview_player
Показать описание
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
x = 2t^2 + 1, y = (1/3)t^3 - t; t = 3
Рекомендации по теме