Answer to Question #124087 in Trigonometry for Arafeen

Question

4. (a) 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 x^2/a^2+y^2/ b^2 = 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

This is a question of parameters so we will arrange the variables to get the desired equation of ellipse.

We will attempt all the questions sectionwise:

(i) x = 4cos2t and y = 7sin2t

Rearranging the terms

$\frac{x}{4}$ = cos2t ....(i)

$\frac{y}{7}$ = sin2t ....(ii)

Squaring and adding equations (i) and (ii),we get

$\frac{x^2}{16}$ + $\frac{y^2}{49}$ = cos²2t + sin²2t = 1 ; $\frac{x^2}{16}$ + $\frac{y^2}{49}$ = 1

Therefore comparing to standard equation of ellipse $\frac{x^2}{a^2}$ + $\frac{y^2}{b^2}$ = 1, we can say

a = 4 ; b =7 and equation $\frac{x^2}{16}$ + $\frac{y^2}{49}$ = 1 resembles the equation of ellipse. ...(answer)

(ii) L(t) is distansce between origin(0,0) and P(x,y) which is given by:

L(t) = $\sqrt{(x-0)^2 + (y-0)^2}$ = $\sqrt{x^2+y^2}$

Now putting x = 4cos2t and y = 7sin2t, we get

L(t) = $\sqrt{16cos^22t+49sin^22t}$ ..... (iii) .....(answer)

Therefore equation (iii) above is required equation of L(t)

(iii) To indicate how fast is distance between P and origin is changing we need to differentiate L(t) which represents the distance between them.

So differentiating equation(iii) both sides we get,

$\frac{dL(t)}{dt}$ = $\frac{1}{2}$ $\frac{1}{\sqrt{16cos^22t + 49sin^22t}}$ ($16*2cos2t*(-2sin2t)$+ $49*2sin2t*2cos2t)$

Now rearranging and putting t=$\frac{\Pi}{8}$ we get

$​$ $\frac{dL(t)}{dt}$ =$\frac{1}{2}$ $\frac{1}{\sqrt{16cos^22t + 49sin^22t}}$ ($98sin4t-32sin4t)$

$\frac{dL(t)}{dt}$ =$\frac{1}{2}$ $\frac{1}{\sqrt{16cos^2\frac{\Pi}{4} + 49sin^2\frac{\Pi}{4}}}$ ($98sin\frac{\Pi}{2}-32sin\frac{\Pi}{2})$

$\frac{dL(t)}{dt}$ =$\frac{1}{2}$ $\frac{1}{\sqrt{16*\frac{1}{2} + 49*\frac{1}{2}}}$ (98-32)

​$\frac{dL(t)}{dt}$ =$\frac{1}{2}$ $\frac{1}{\sqrt{8+ 24.5}}$ *66 = 33*$\frac{1}{\sqrt{32.5}}$ = 5.788

Hence,how fast is the distance between P and the origin changing when t = $\frac{\Pi}{8}$ is given by:

$\frac{dL(t)}{dt}$ = 5.788 .....(answer)

Thanks and hope all doubts related are solved.