the acceleration y, in meters per second squared, of an object after x seconds is given by y = 7 sin(π/4 x)…

the acceleration y, in meters per second squared, of an object after x seconds is given by y = 7 sin(π/4 x). during the first 10 seconds, over which intervals is the acceleration increasing? (2, 6) (4, 8) (0, 2) and (6, 10) (0, 4) and (8, 10)

the acceleration y, in meters per second squared, of an object after x seconds is given by y = 7 sin(π/4 x). during the first 10 seconds, over which intervals is the acceleration increasing? (2, 6) (4, 8) (0, 2) and (6, 10) (0, 4) and (8, 10)

Answer

Explanation:

Step1: Find the derivative of $y$

We know that if $y = 7\sin(\frac{\pi}{4}x)$, using the chain - rule. If $y = A\sin(u)$ and $u = Bx$ ($A = 7$, $B=\frac{\pi}{4}$), then $y^\prime=A\cdot B\cos(u)$. So $y^\prime=\frac{7\pi}{4}\cos(\frac{\pi}{4}x)$.

Step2: Determine when $y^\prime>0$

We want to find when $\frac{7\pi}{4}\cos(\frac{\pi}{4}x)>0$. Since $\frac{7\pi}{4}>0$, we just need to find when $\cos(\frac{\pi}{4}x)>0$. The general solution for $\cos(t)>0$ is $2k\pi-\frac{\pi}{2}<t<2k\pi + \frac{\pi}{2},k\in\mathbb{Z}$. Substituting $t = \frac{\pi}{4}x$, we get $2k\pi-\frac{\pi}{2}<\frac{\pi}{4}x<2k\pi+\frac{\pi}{2}$.

Step3: Solve the inequality for $x$

First, multiply each part of the inequality $2k\pi-\frac{\pi}{2}<\frac{\pi}{4}x<2k\pi+\frac{\pi}{2}$ by $\frac{4}{\pi}$. We have $8k - 2<x<8k + 2$.

Step4: Find the intervals for $0\leq x\leq10$

When $k = 0$, $- 2<x<2$. Since we are considering $0\leq x\leq10$, the relevant part is $0\leq x<2$. When $k = 1$, $8 - 2<x<8 + 2$, i.e., $6<x<10$.

Answer:

C. $(0, 2)$ and $(6, 10)$