Question

Suppose a particle P is moving in the plane so that its coordinates are given by P(x; y),

where x = 4 cos 2t, y = 7 sin 2t.

(i) By finding a; b ϵ 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

Solution a. We know that the parametric equation of an ellipse is

 x= acosθ and y =bsinθ

where a =semi major axis and

b= semi minor axis

therefore, comparing with

 x= 4 cost 2t, and y = 7 sint 2t

we get a = 4, b=7 and θ = 2t

therefore, using L.H.S from equation of an ellipse :

$(\frac{x^2}{a^2}) + (\frac{y^2}{b^2})$

=$\frac{(16 cos^22t)}{16 }+\frac{(49sint^22t)}{49}$

= 1= R.H.S

Hence, the particle is moving in an elliptical path.

ii) $L(t) =( (x- 0)^2 + (y - 0)^2)$ ^0.5

= $((16 cos^22t + 49 sin^2t))$ ^0.5

=$(16cost^22t + 49 (1- cost ^22t))$ ^0.5

$=(49 - 33 cost^22t)$ ^0.5

(Answer)

iii) Rate of change of L(t) = $\frac{dL(t)}{dt}$

$=\frac{d[\frac{(49 - 33(1+cost4t)^\frac{1}{2}}{2}]} {dt}$

$=\frac{[\frac{d (98 - 33- 33cos 4t)}{2}]^\frac{1}{2} }{ dt}$

$= \frac{[\frac{d(65-33cos 4t)}{2}]^\frac{1}{2}}{dt}$

$=(\frac{1}{2}^\frac{1}{2}) * d[(65 - 33cos4t)]^\frac{1}{2}$

$$=(\frac{1}{2 }^\frac{1}{2} * 2 * (65 - 33cos4t)^\frac{1}{2}) * [33*4*sin4t]$$

$=\frac{66 sin4t}{(130-66cos4t)^\frac{1}{2}}$

Putting 

t=π / 8 or (π /8)*(180/π )= 22.5 degree

we get ,

$\frac{d L(t)}{dt }= \frac{(66 sin(4*22.5))}{(130 - 66 cos (4*22.5))^\frac{1}{2}}$

= $\frac{66} {1300.5}$

= 5.788 $\frac{units}{rad}$ (Answer)