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
x 4 \frac{x}{4} 4 x = cos2t ....(i)
y 7 \frac{y}{7} 7 y = sin2t ....(ii)
Squaring and adding equations (i) and (ii),we get
x 2 16 \frac{x^2}{16} 16 x 2 + y 2 49 \frac{y^2}{49} 49 y 2 = cos2 2t + sin2 2t = 1 ; x 2 16 \frac{x^2}{16} 16 x 2 + y 2 49 \frac{y^2}{49} 49 y 2 = 1
Therefore comparing to standard equation of ellipse x 2 a 2 \frac{x^2}{a^2} a 2 x 2 + y 2 b 2 \frac{y^2}{b^2} b 2 y 2 = 1, we can say
a = 4 ; b =7 and equation x 2 16 \frac{x^2}{16} 16 x 2 + y 2 49 \frac{y^2}{49} 49 y 2 = 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) = ( x − 0 ) 2 + ( y − 0 ) 2 \sqrt{(x-0)^2 + (y-0)^2} ( x − 0 ) 2 + ( y − 0 ) 2 = x 2 + y 2 \sqrt{x^2+y^2} x 2 + y 2
Now putting x = 4cos2t and y = 7sin2t, we get
L(t) = 16 c o s 2 2 t + 49 s i n 2 2 t \sqrt{16cos^22t+49sin^22t} 16 co s 2 2 t + 49 s i n 2 2 t ..... (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,
d L ( t ) d t \frac{dL(t)}{dt} d t d L ( t ) = 1 2 \frac{1}{2} 2 1 1 16 c o s 2 2 t + 49 s i n 2 2 t \frac{1}{\sqrt{16cos^22t + 49sin^22t}} 16 co s 2 2 t + 49 s i n 2 2 t 1 (16 ∗ 2 c o s 2 t ∗ ( − 2 s i n 2 t ) 16*2cos2t*(-2sin2t) 16 ∗ 2 cos 2 t ∗ ( − 2 s in 2 t ) + 49 ∗ 2 s i n 2 t ∗ 2 c o s 2 t ) 49*2sin2t*2cos2t) 49 ∗ 2 s in 2 t ∗ 2 cos 2 t )
Now rearranging and putting t=Π 8 \frac{\Pi}{8} 8 Π we get
d L ( t ) d t \frac{dL(t)}{dt} d t d L ( t ) =1 2 \frac{1}{2} 2 1 1 16 c o s 2 2 t + 49 s i n 2 2 t \frac{1}{\sqrt{16cos^22t + 49sin^22t}} 16 co s 2 2 t + 49 s i n 2 2 t 1 (98 s i n 4 t − 32 s i n 4 t ) 98sin4t-32sin4t) 98 s in 4 t − 32 s in 4 t )
d L ( t ) d t \frac{dL(t)}{dt} d t d L ( t ) =1 2 \frac{1}{2} 2 1 1 16 c o s 2 Π 4 + 49 s i n 2 Π 4 \frac{1}{\sqrt{16cos^2\frac{\Pi}{4} + 49sin^2\frac{\Pi}{4}}} 16 co s 2 4 Π + 49 s i n 2 4 Π 1 (98 s i n Π 2 − 32 s i n Π 2 ) 98sin\frac{\Pi}{2}-32sin\frac{\Pi}{2}) 98 s in 2 Π − 32 s in 2 Π )
d L ( t ) d t \frac{dL(t)}{dt} d t d L ( t ) =1 2 \frac{1}{2} 2 1 1 16 ∗ 1 2 + 49 ∗ 1 2 \frac{1}{\sqrt{16*\frac{1}{2} + 49*\frac{1}{2}}} 16 ∗ 2 1 + 49 ∗ 2 1 1 (98-32)
d L ( t ) d t \frac{dL(t)}{dt} d t d L ( t ) =1 2 \frac{1}{2} 2 1 1 8 + 24.5 \frac{1}{\sqrt{8+ 24.5}} 8 + 24.5 1 *66 = 33*1 32.5 \frac{1}{\sqrt{32.5}} 32.5 1 = 5.788
Hence,how fast is the distance between P and the origin changing when t = Π 8 \frac{\Pi}{8} 8 Π is given by:
d L ( t ) d t \frac{dL(t)}{dt} d t d L ( t ) = 5.788 .....(answer)
Thanks and hope all doubts related are solved.
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