(i) We know that "\\sin^2x+\\cos^2x=1" . Therefore, "\\dfrac{(4\\cos 2t)^2}{4^2} + \\dfrac{(7\\sin 2t)^2}{7^2} = 1." So a = 4, b = 7. Therefore, the path of the point is elliptical with semiaxes 4 and 7.
(ii) The distance between two points can be calculates as "L(t)=\\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} = \\sqrt{(x-0)^2+(y-0)^2} = \\sqrt{16\\cos^22t + 49\\sin^22t} = \\sqrt{16 + 33\\sin^22t}."
(iii) Let us calculate the derivative of "L(t) \\;\\; \\text{in} \\;\\; \\dfrac{\\pi}{8}:"
"L'(t) = \\dfrac{66\\sin2t\\cos2t}{\\sqrt{16 + 33\\sin^22t}} = \\dfrac{33\\sin4t}{\\sqrt{16 + 33\\sin^22t}} = \\dfrac{33\\sin\\frac{\\pi}{2}}{\\sqrt{16 + 33\\sin^2\\frac{\\pi}{4}}} = \\dfrac{33}{\\sqrt{32.5}} \\approx 5.79."
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