- #1
syeh
- 15
- 0
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.
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.