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Question

Let x=f′′(t)cost+f(t)sint and y=f′′(t)sint+f(t)cost. Then [(dxdt)2+(dydt)2]12dt equals
(Note : f(x),f(x),f′′(x),f′′′(x)>0 )

A
f(t)+f′′(t)+c
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B
f′′(t)+f′′′(t)+c
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C
f(t)+f′′(t)+c
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D
f(t)f′′(t)+c
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Solution

The correct option is C f(t)+f′′(t)+c

x=f′′(t)cost+f(t)sintdxdt=f′′′(t)costf′′(t)sint+f′′(t)sint+f(t)costdxdt=f′′′(t)cost+f(t)costdxdt=(f′′′(t)+f(t))cost..........(1)y=f′′(t)sint+f(t)costdydt=f′′′(t)sintf′′(t)cost+f′′(t)costf(t)sintdydt=f′′′(t)sintf(t)sintdydt=sint(f′′′(t)+f(t))........(2)addsquaresofeq.1and2(dxdt)2+(dydt)2=(f′′′(t)+f(t))2((dxdt)2+(dydt)2)12=f′′′(t)+f(t)integratebothsides((dxdt)2+(dydt)2)12dt=f′′(t)+f(t)+c


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