We have the following function:
[tex]H(t)=10\sin (2\pi(t-\frac{1}{4}))+15[/tex]a) The initial height occurs at t=0. By substituting this value into the function, we get
[tex]H(0)=10\sin (2\pi(0-\frac{1}{4}))+15[/tex]which gives
[tex]\begin{gathered} H(0)=10\sin (-2\pi\frac{1}{4})+15 \\ H(0)=10\sin (-\frac{2\pi}{4})+15 \\ H(0)=10\sin (-\frac{\pi}{2})+15 \end{gathered}[/tex]since sin(-Pi/2) is equal to -1, we have
[tex]H(0)=-10+15[/tex]Then, H(0)= 5, so the height is 5 feets.
b) The car will make a full rotation when the argument is shifted 2Pi:
[tex]H(t)=10\sin (2\pi(t-\frac{1}{4})+2\pi)+15[/tex]at this point H(t) must be equal to 5 feets. Then, we have
[tex]5=10\sin (2\pi(t-\frac{1}{4})-2\pi)+15[/tex]and we must find t. If we move +15 to the left hand side, we obtain
[tex]\begin{gathered} 5-15=10\sin (2\pi(t-\frac{1}{4})-2\pi) \\ \text{then} \\ 10\sin (2\pi(t-\frac{1}{4})-2\pi)=-10 \end{gathered}[/tex]By moving the coefficient 10 the the right hand side, we get
[tex]\begin{gathered} \sin (2\pi(t-\frac{1}{4})-2\pi)=-\frac{10}{10} \\ \sin (2\pi(t-\frac{1}{4})-2\pi)=-1 \end{gathered}[/tex]we can note that when
[tex]\sin \theta=-1\Rightarrow\theta=-90=-\frac{\pi}{2}[/tex]this implies that the argument of our last result is
[tex]2\pi(t-\frac{1}{4})-2\pi=-\frac{\pi}{2}[/tex]By moving 2Pi to the right hand side, we have
[tex]\begin{gathered} 2\pi(t-\frac{1}{4})=-\frac{\pi}{2}+2\pi \\ 2\pi(t-\frac{1}{4})=\frac{3\pi}{2} \end{gathered}[/tex]Now, by moving 2Pi to the right hand side, we have
[tex]t-\frac{1}{4}=-\frac{\frac{3\pi}{2}}{2\pi}[/tex]which gives
[tex]t-\frac{1}{4}=\frac{3}{4}[/tex]so, t is given by
[tex]\begin{gathered} t=\frac{3}{4}+\frac{1}{4} \\ t=1 \end{gathered}[/tex]That is, n 1 minute, the car will take one full rotation.
c) The maximum height of the car ocurrs at t=0.5 min because in one minute it take one full rotation. Then, at t=1/2 we get
[tex]H(\frac{1}{2})=10\sin (2\pi(\frac{1}{2}-\frac{1}{4}))+15[/tex]which gives
[tex]\begin{gathered} H(\frac{1}{2})=10\sin (2\pi(\frac{1}{4}))+15 \\ H(\frac{1}{2})=10\sin (\frac{\pi}{2})+15 \\ H(\frac{1}{2})=10(1)+15 \\ H(\frac{1}{2})=25 \end{gathered}[/tex]that is, the maximum height is 25 feets
