Question: Particle A is resting on the origin. Particle B travels along the path of f(x)=-cos(x-7)+3. What is the minimum distance that the particles can be apart? Attempt: I tried to find the minimum of f(x)=-cos(x-7)+3. So I found when the derivative is 0: dy/dx= sin(x-7) 0=sin(x-7) x=7 So I plugged that into the original equation: f(7)= -cos0+3= -1+3= 2. So the point (7,2) is a possible minimum/maximum. Using the First Derivative Number Line Test, I found that the point (7,2) is a minimum. Does that mean that is the point where Particle B is closest to the origin?? Assuming it does, I attempted to find the line connecting (7,2) and the origin, and then to find the length of that curve on the interval x=[0,7]: y=mx+b to find m using the 2 points (0,0) and (7,2): m=(2-0)/(7-0)= 2/7 b is 0 because it does through (0,0) y=(2/7)x Length=∫[from 0 to 7] sqrt(1+(dy/dx)^2)dx dy/dx=2/7 so, Length=∫[from 0 to 7] sqrt(1+(2/7)^2)dx =∫[from 0 to 7] 1.04 dx =[1.04x]from 0 to 7 =(using f(b)-f(a)): (1.04*7)-(1.04*0) =7.28 That was my final answer: when Particle B is closest to the origin/Particle A, it is 7.28 units away. Is this right? Or on the right path? I don't know.