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. [10 marks] (ii) Let L(t) be the distance from P to the origin. Obtain an expression for L(t).[8 marks] (iii) How fast is the distance between P and the origin changing when t = π/8?[7 marks]

Answer 1

ANSWER

(i)a=4, b=7

Explanation: $cos^22t+sin^22t=1$ . Therefore $\frac { { x }^{ 2 } }{ { 4 }^{ 2 } } +\frac { { y }^{ 2 } }{ { 7 }^{ 2 } } =\cos ^{ 2 }{ 2t } +\sin ^{ 2 }{ 2t=1 }$ ,or $\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 49 } =1$

ANSWER

(ii )The distance $L$ from the point $P(x,y)$ to the origin is calculated by the formula $L=\sqrt { { x }^{ 2 }{ +y }^{ 2 } }$ . Therefore $L(t)=\sqrt { { x }^{ 2 }(t){ +y }^{ 2 }(t) } =\sqrt { 16\cos ^{ 2 }{ 2t+49\sin ^{ 2 }{ 2 } t } }$ or $L(t)=\ \sqrt { 16+33\sin ^{ 2 }{ 2t } }$

ANSWER

(iii)$\frac { 33\sqrt { 2 } \ }{ \sqrt { 65 } }$

Explanation:

Let $V(t)$ denote the rate of change of distance between points $P$ and the origin $V(t)=\ \frac { dL(t) }{ dt } =\frac { 33\cdot 2\cdot \left( \sin { 2t } \right) \cdot \left( 2\cos { 2t } \right) }{ 2\sqrt { 16+33\sin ^{ 2 }{ 2t } } } =\frac { 33\sin { 4t } }{ \sqrt { 16+33\sin ^{ 2 }{ 2t } } }$ .

$\sin { \frac { \pi }{ 2 } } =1,\quad \sin { \frac { \pi }{ 4 } } =\frac { 1 }{ \sqrt { 2 } }$ , so $V\left( \frac { \pi }{ 8 } \right) =\frac { 33\sin { \frac { 4\pi }{ 8 } } }{ \sqrt { 16+33\sin ^{ 2 }{ \frac { 2\pi }{ 8 } } } } =\frac { 33 }{ \sqrt { 16+\frac { 33 }{ 2 } } } =\frac { 33\sqrt { 2 } \quad }{ \sqrt { 65 } }$