Question #124087
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?
1
Expert's answer
2020-06-28T18:49:33-0400

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

x4\frac{x}{4} = cos2t ....(i)

y7\frac{y}{7} = sin2t ....(ii)

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

x216\frac{x^2}{16} + y249\frac{y^2}{49} = cos22t + sin22t = 1 ; x216\frac{x^2}{16} + y249\frac{y^2}{49} = 1

Therefore comparing to standard equation of ellipse x2a2\frac{x^2}{a^2} + y2b2\frac{y^2}{b^2} = 1, we can say

a = 4 ; b =7 and equation x216\frac{x^2}{16} + y249\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) = (x0)2+(y0)2\sqrt{(x-0)^2 + (y-0)^2} = x2+y2\sqrt{x^2+y^2}

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

L(t) = 16cos22t+49sin22t\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,

dL(t)dt\frac{dL(t)}{dt} = 12\frac{1}{2} 116cos22t+49sin22t\frac{1}{\sqrt{16cos^22t + 49sin^22t}} (162cos2t(2sin2t)16*2cos2t*(-2sin2t)+ 492sin2t2cos2t)49*2sin2t*2cos2t)


Now rearranging and putting t=Π8\frac{\Pi}{8} we get

dL(t)dt\frac{dL(t)}{dt} =12\frac{1}{2} 116cos22t+49sin22t\frac{1}{\sqrt{16cos^22t + 49sin^22t}} (98sin4t32sin4t)98sin4t-32sin4t)


dL(t)dt\frac{dL(t)}{dt} =12\frac{1}{2} 116cos2Π4+49sin2Π4\frac{1}{\sqrt{16cos^2\frac{\Pi}{4} + 49sin^2\frac{\Pi}{4}}} (98sinΠ232sinΠ2)98sin\frac{\Pi}{2}-32sin\frac{\Pi}{2})


dL(t)dt\frac{dL(t)}{dt} =12\frac{1}{2} 11612+4912\frac{1}{\sqrt{16*\frac{1}{2} + 49*\frac{1}{2}}} (98-32)


dL(t)dt\frac{dL(t)}{dt} =12\frac{1}{2} 18+24.5\frac{1}{\sqrt{8+ 24.5}} *66 = 33*132.5\frac{1}{\sqrt{32.5}} = 5.788


Hence,how fast is the distance between P and the origin changing when t = Π8\frac{\Pi}{8} is given by:

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


Thanks and hope all doubts related are solved.

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