name: \n2. the temperature in sams home varies sinusoidal over time. given the equation t = 4cos(30(t…

Explanation:

Step1: Solve for part b of temperature - problem

Set $T<14$. So, $4\cos(30(t - 8))+16<14$. Subtract 16 from both sides: $4\cos(30(t - 8))<-2$. Divide both sides by 4: $\cos(30(t - 8))<-\frac{1}{2}$. We know that $\cos\theta<-\frac{1}{2}$ when $120^{\circ}+360^{\circ}n<\theta < 240^{\circ}+360^{\circ}n,n\in\mathbb{Z}$. So, $120 < 30(t - 8)<240$ (ignoring the $360^{\circ}n$ for the first - cycle). Divide through by 30: $4 < t - 8<8$. Add 8 to all parts: $12 < t<16$. The length of time is $16 - 12=4$ hours.

Step2: Solve for part c of temperature - problem

4:15 is 4.25 hours after 12 midnight. Substitute $t = 4.25$ into $T = 4\cos(30(t - 8))+16$. First, calculate $30(4.25 - 8)=30\times(-3.75)=-112.5$. $T = 4\cos(-112.5)+16$. Since $\cos(-\alpha)=\cos\alpha$, $T = 4\cos(112.5)+16$. $\cos(112.5)=\cos(90 + 22.5)=-\sin(22.5)=-\frac{\sqrt{2-\sqrt{2}}}{2}$. $T = 4\times(-\frac{\sqrt{2 - \sqrt{2}}}{2})+16=16 - 2\sqrt{2-\sqrt{2}}\approx16-2\times0.76=16 - 1.52 = 14.48^{\circ}C$.

Step3: Solve for part c of water - wheel problem

The general form of a sinusoidal function for the height of the boot $h(t)=A\sin(\omega t+\varphi)+k$. The radius of the water - wheel is $r = 3$ m, the bottom is 2 m under water, so the center of the wheel is $3 + 2=5$ m above the water level. So $k = 5$, $A = 3$. At $t = 0$, the boot is at the bottom, and at $t = 5$ it is at the high - point. The period $T=2\times5 = 10$ s, and $\omega=\frac{2\pi}{T}=\frac{\pi}{5}$. When $t = 0$, $h(0)=2$, so $2=3\sin(\varphi)+5$, $\sin(\varphi)= - 1$, $\varphi=-\frac{\pi}{2}$. Then $h(t)=3\sin(\frac{\pi}{5}t-\frac{\pi}{2})+5$. Set $h(t)=2 + 2=4$. So, $3\sin(\frac{\pi}{5}t-\frac{\pi}{2})+5 = 4$. $3\sin(\frac{\pi}{5}t-\frac{\pi}{2})=-1$. $\sin(\frac{\pi}{5}t-\frac{\pi}{2})=-\frac{1}{3}$. $\frac{\pi}{5}t-\frac{\pi}{2}=\arcsin(-\frac{1}{3})+2\pi n$ or $\frac{\pi}{5}t-\frac{\pi}{2}=\pi-\arcsin(-\frac{1}{3})+2\pi n$. $\frac{\pi}{5}t=\frac{\pi}{2}+\arcsin(-\frac{1}{3})+2\pi n$ or $\frac{\pi}{5}t=\frac{\pi}{2}+\pi-\arcsin(-\frac{1}{3})+2\pi n$. $t=\frac{5}{2}+\frac{5}{\pi}\arcsin(-\frac{1}{3}) + 10n$ or $t=\frac{5}{2}+\frac{5}{\pi}(\pi-\arcsin(-\frac{1}{3}))+10n$.

Step4: Solve for part d of water - wheel problem

Set $h(t)=5$. $3\sin(\frac{\pi}{5}t-\frac{\pi}{2})+5 = 5$. $\sin(\frac{\pi}{5}t-\frac{\pi}{2})=0$. $\frac{\pi}{5}t-\frac{\pi}{2}=n\pi$. $\frac{\pi}{5}t=n\pi+\frac{\pi}{2}$. $t = 5n+\frac{5}{2}$. The boot enters the water when $h(t)=2$ and leaves when $h(t)=8$. Set $h(t)=2$: $3\sin(\frac{\pi}{5}t-\frac{\pi}{2})+5 = 2$, $\sin(\frac{\pi}{5}t-\frac{\pi}{2})=-1$, $\frac{\pi}{5}t-\frac{\pi}{2}=-\frac{\pi}{2}+2\pi n$, $t = 10n$. Set $h(t)=8$: $3\sin(\frac{\pi}{5}t-\frac{\pi}{2})+5 = 8$, $\sin(\frac{\pi}{5}t-\frac{\pi}{2}) = 1$, $\frac{\pi}{5}t-\frac{\pi}{2}=\frac{\pi}{2}+2\pi n$, $t = 5 + 10n$. The length of time above water is 5 seconds.

Step5: Solve for part e of water - wheel problem

Substitute $t = 12$ into $h(t)=3\sin(\frac{\pi}{5}t-\frac{\pi}{2})+5$. $\frac{\pi}{5}\times12-\frac{\pi}{2}=\frac{24\pi}{10}-\frac{5\pi}{10}=\frac{19\pi}{10}$. $\sin(\frac{19\pi}{10})=\sin(2\pi-\frac{\pi}{10})=-\sin(\frac{\pi}{10})$. $h(12)=3\times(-\sin(\frac{\pi}{10}))+5\approx3\times(-0.309)+5=5 - 0.927 = 4.073$ m. The boot is above the water.

Answer:

b) 4 hours c) Approximately $14.48^{\circ}C$ c) $t=\frac{5}{2}+\frac{5}{\pi}\arcsin(-\frac{1}{3}) + 10n$ or $t=\frac{5}{2}+\frac{5}{\pi}(\pi-\arcsin(-\frac{1}{3}))+10n$ d) 5 seconds e) Above the water