Question

Suppose a particle P is moving in the plane so that its coordinates are given by P(x,y), where x = 4cos2t, y = 7sin2t. (i) By finding a,b ∈ R such that x2 /a2 + y2 /b2 = 1, show that P is travelling on an elliptical path. (ii) Let L(t) be the distance from P to the origin. Obtain an expression for L(t). (iii) How fast is the distance between P and the origin changing when t = π/8?

Answer 1

(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 $L(t) \;\; \text{in} \;\; \dfrac{\pi}{8}:$

$L'(t) = \dfrac{66\sin2t\cos2t}{\sqrt{16 + 33\sin^22t}} = \dfrac{33\sin4t}{\sqrt{16 + 33\sin^22t}} = \dfrac{33\sin\frac{\pi}{2}}{\sqrt{16 + 33\sin^2\frac{\pi}{4}}} = \dfrac{33}{\sqrt{32.5}} \approx 5.79.$