the acceleration y, in meters per second squared, of an object after x seconds is given by ( y = 7sinleft(\frac{pi}{4}x
ight) ).
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)

Question

Accepted Answer

# Answer:
C. (0, 2) and (6, 10)

# Explanation:
## Step1: Recall the derivative of sine function
The derivative of $y = A\sin(Bx)$ is $y'=AB\cos(Bx)$. For $y = 7\sin(\frac{\pi}{4}x)$, $A = 7$ and $B=\frac{\pi}{4}$, so $y'=\frac{7\pi}{4}\cos(\frac{\pi}{4}x)$.
## Step2: Find where $y'\gt0$ (function is increasing)
We know that $\cos t> 0$ when $2k\pi-\frac{\pi}{2}<t<2k\pi+\frac{\pi}{2},k\in\mathbb{Z}$. Let $t = \frac{\pi}{4}x$, then $2k\pi-\frac{\pi}{2}<\frac{\pi}{4}x<2k\pi+\frac{\pi}{2}$.
Multiply through by $\frac{4}{\pi}$: $8k - 2<x<8k + 2$.
## Step3: Consider the interval $0\leq x\leq10$
When $k = 0$: $0 - 2<x<0 + 2$, i.e., $0<x<2$ (since $x\geq0$).
When $k = 1$: $8\times1- 2<x<8\times1 + 2$, i.e., $6<x<10$ (since $x\leq10$).

So the acceleration $y = 7\sin(\frac{\pi}{4}x)$ is increasing on the intervals $(0,2)$ and $(6,10)$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the derivative of sine function

Step2: Find where \(y'\gt0\) (function is increasing)

Step3: Consider the interval \(0\leq x\leq10\)

Answer: