the acceleration y, in meters per second squared, of an object after x seconds is given by ( y =…

the acceleration y, in meters per second squared, of an object after x seconds is given by ( y = 7sinleft(\frac{pi}{4}x\right) ).\nduring the first 10 seconds, over which intervals is the acceleration increasing?\n○ (2,6)\n○ (4,8)\n○ (0,2) and (6,10)\n○ (0,4) and (8,10)

the acceleration y, in meters per second squared, of an object after x seconds is given by ( y = 7sinleft(\frac{pi}{4}x\right) ).\nduring the first 10 seconds, over which intervals is the acceleration increasing?\n○ (2,6)\n○ (4,8)\n○ (0,2) and (6,10)\n○ (0,4) and (8,10)

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))