Correct Answer - Option 3 : x(t) = ½ + cos t, y(t) = 2 sin t
Explanation:
Translation of the follower is, d = \(√{{x^2}+{y^2}}\)
Case 1:
x(t) = cos t, y(t) = sin t
\(d=√{{cos^2t}+{sin^2t}} = 1\)
dmax = 1
Case 2:
x(t) = cos t, y(t) = 2 sin t
\(d=√{{cos^2t}+{\left(2\;sin\;t\right)^2}} = √{{cos^2t}+{4\;sin^2t}} = √{1+3\;sin^2t} \)
'd' will be maximum when sin2t will be maximum.
sin t will be maximum at t = π /2 , 3π /2 in the interval of (0, 2π)
\(d_{max}=√{1+3} = 2\)
Case 3:
x(t) = ½ + cos t, y(t) = 2 sin t
\(d=√{{(\frac{1}{2}+cos\;t)^2}+{\left(2\;sin\;t\right)^2}} = √{\frac{1}{4}+cos\;t+{cos^2t}+{4\;sin^2t}} = √{\frac{5}{4}+cos\;t+3\;sin^2t} \)
'd' will be maximum when leading term 'sin2t' will be maximum.
sin2t will be maximum at t = π /2 , 3π /2 in the interval of (0, 2π)
\(d_{max}=√{\frac{5}{4}+0+3} = 2.06\)
Case 4:
x(t) = ½ + cos t, y(t) = sin t
\(d=√{{(\frac{1}{2}+cos\;t)^2}+{\left(sin\;t\right)^2}} = √{\frac{1}{4}+cos\;t+{cos^2t}+{sin^2t}} = √{\frac{5}{4}+cos\;t} \)
'd' will be maximum when cos t will be maximum.
cos t will be maximum at t = 0 , 2π in the interval of (0, 2π)
\(d_{max}=√{\frac{5}{4}+1} = 1.5\)
Option ‘C’ has maximum net amplitude.