Respuesta :
Answer:
The time [tex]t = \frac{3}{2}[/tex] seconds elapses between two consecutive times that the swing is at its maximum height 'h' = 2
Step-by-step explanation:
Explanation:-
Step(i):-
Given function [tex]h(t) = 3 cos (\frac{3\pi }{2 t} ) +5[/tex] ....(i)
By using derivative formulas
[tex]\frac{d cosx }{d x} = -sinx[/tex]
[tex]\frac{d x^{n} }{d x} = n x^{n-1}[/tex]
[tex]\frac{d t^{-1} }{d x} = -1 t^{-1-1} = - t^{-2} = \frac{-1}{t^{2} }[/tex]
Step(ii):-
Differentiating equation(i) with respective to 't'
[tex]h^{l} (t) = 3(-sin(\frac{3\pi }{2t})\frac{d}{dt} (\frac{3\pi }{2t } )+0[/tex] ...(ii)
[tex]h^{l} (t) = 3(-sin(\frac{3\pi }{2t})(\frac{-3\pi }{2t^{2} } )+0[/tex]
Equating zero
[tex]h^{l} (t) = 3(-sin(\frac{3\pi }{2t})(\frac{-3\pi }{2t^{2} } )=0[/tex]
[tex]3(-sin(\frac{3\pi }{2t})(\frac{-3\pi }{2t^{2} } ) = 0[/tex]
on simplification , we get
[tex](sin(\frac{3\pi }{2t}) = 0[/tex]
now we use formulas
sin 0 = 0 and sinπ = 0
General solution
[tex](sin(\frac{3\pi }{2t}) = sin\pi[/tex]
[tex](\frac{3\pi }{2t}) = \pi[/tex]
Cancellation 'π' on both sides, we get
[tex]3 = 2 t[/tex]
Dividing '2' on both sides , we get
[tex]t = \frac{3}{2}[/tex]
Again differentiating with respective to 't' , we get
[tex]h^{ll} (t) = 3(-cos(\frac{3\pi }{2t})(\frac{-3\pi }{2t^{2} } )+ (-3)(-sin(\frac{3\pi }{2t} )(\frac{6\pi }{2t^{3} }[/tex]
Put t= 3/2 and simplification
[tex]h^{ll} (t) < 0[/tex]
The maximum height
[tex]h(t) = 3 cos (\frac{3\pi }{2 t} ) +5[/tex]
[tex]h(\frac{3}{2} ) = 3 cos (\frac{3\pi }{2(\frac{3}{2} )} )+5[/tex]
[tex]h(\frac{3}{2} ) = 3 cos (\pi )+5 = -3+5 =2[/tex]
[tex]t = \frac{3}{2}[/tex] seconds elapses between two consecutive times that the swing is at its maximum height 'h' = 2
Conclusion:-
The time [tex]t = \frac{3}{2}[/tex] seconds elapses between two consecutive times that the swing is at its maximum height 'h' = 2
