Answer :
Answer:
0.41 (correct to 2 decimal place)
Step-by-step explanation:
If [tex]B(t)=4.0+0.35sin\frac{2\pi t }{5.4}[/tex]
Let [tex]u=\frac{2\pi t }{5.4}[/tex], then [tex]B(u)=4.0+0.35sin u[/tex]
We want to determine the rate of increase [tex]\frac{dB}{dt}[/tex] after one day
[tex]du=\frac{2\pi dt }{5.4}[/tex] and [tex]\frac{dB}{du} =0.35cos u[/tex]
[tex]\frac{dB}{dt}=\frac{2\pi }{5.4}0.35cos (\frac{2\pi t }{5.4})=0.407cos (\frac{2\pi t }{5.4})[/tex]
[tex]B^{'} (t)=0.407cos (\frac{2\pi t }{5.4})[/tex]
Rate of increase after one day, i.e. t=1
[tex]B^{'} (1)=0.407cos (\frac{2\pi X 1 }{5.4})[/tex]= 0.407 =0.41 (correct to 2 decimal place)
