5.6 - applications of trig functions (1.) lamaj rides his bike over a piece of gum and continues riding his…

5.6 - applications of trig functions (1.) lamaj rides his bike over a piece of gum and continues riding his bike at a constant rate. at time = 1.25 seconds,the gum is at a maximum height above the ground and 1 second later the gum is on the ground again. a.) if the diameter of the wheel is 68 cm, write an equation that models the height of the gum in centimeters above the ground at any time, t, in seconds. b.) what is the height of the gum when lamaj gets to the end of the block at t = 15.6 seconds? (use a calculator for this question)

5.6 - applications of trig functions (1.) lamaj rides his bike over a piece of gum and continues riding his bike at a constant rate. at time = 1.25 seconds,the gum is at a maximum height above the ground and 1 second later the gum is on the ground again. a.) if the diameter of the wheel is 68 cm, write an equation that models the height of the gum in centimeters above the ground at any time, t, in seconds. b.) what is the height of the gum when lamaj gets to the end of the block at t = 15.6 seconds? (use a calculator for this question)

Answer

Explanation:

Step1: Encontrar la frecuencia y la fase inicial

El período $T$ del movimiento del chicle se obtiene del hecho de que desde el máximo hasta tocar el suelo es 1 segundo. El período completo es $T = 2$ segundos. La frecuencia angular $\omega=\frac{2\pi}{T}=\pi$ rad/s. Sabemos que en $t = 1.25$ s hay un máximo, para una función senoidal $y = A\sin(\omega t+\varphi)+k$, en un máximo $\sin(\omega t+\varphi)=1$. Si $\omega=\pi$ y $t = 1.25$, entonces $\pi\times1.25+\varphi=\frac{\pi}{2}+2n\pi$, $n\in\mathbb{Z}$. Resolviendo para $\varphi$: $\varphi=\frac{\pi}{2}- 1.25\pi=-\frac{3\pi}{4}$.

Step2: Encontrar la amplitud y la altura media

El diámetro de la rueda es $d = 68$ cm, entonces la amplitud $A=\frac{d}{2}=34$ cm y la altura media $k = 34$ cm. La ecuación que modela la altura $h(t)$ del chicle es $h(t)=34\sin(\pi t-\frac{3\pi}{4}) + 34$.

Step3: Calcular la altura en $t = 15.6$ s

Sustituir $t = 15.6$ en la ecuación $h(t)=34\sin(\pi\times15.6-\frac{3\pi}{4})+34$. $\pi\times15.6-\frac{3\pi}{4}=15.6\pi - 0.75\pi=14.85\pi$. $\sin(14.85\pi)=\sin(14\pi + 0.85\pi)=\sin(0.85\pi)$. $h(15.6)=34\sin(0.85\pi)+34$. Calculando con una calculadora: $\sin(0.85\pi)\approx0.9877$, entonces $h(15.6)=34\times0.9877 + 34=34\times(0.9877 + 1)=34\times1.9877\approx67.58$ cm.

Answer:

a. $h(t)=34\sin(\pi t-\frac{3\pi}{4})+34$ b. Aproximadamente $67.58$ cm