Distance of Particle from Origin

[syeh]

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.

 

[Mark44]

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.

I don't see how that can be right. The graph of the cosine function is always between 2 and 4, so the farthest a particle on that path could be from the origin would be 4, which is quite a bit less than the number you got.

I'll take a closer look at your calculations and report back.

 

[Mark44]

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

There are an infinite number of values for which the derivative is zero: {..., 7 - 2$\pi$, 7 - $\pi$, 7, 7 + $\pi$, 7 + 2$\pi$, ...}.

Also, there is a much easier way to find the distance from (7, 2) to the origin. Using Pythagoras, the distance is $\sqrt{7^2 + 2^2}$. In any case, this point isn't the closest to the origin.

 

[syeh]

There are an infinite number of values for which the derivative is zero: {..., 7 - 2$\pi$, 7 - $\pi$, 7, 7 + $\pi$, 7 + 2$\pi$, ...}.

Also, there is a much easier way to find the distance from (7, 2) to the origin. Using Pythagoras, the distance is $\sqrt{7^2 + 2^2}$. In any case, this point isn't the closest to the origin.

thanks. i see how Pythagorean's theorem works. But i am still confused how to find when the function is closest to the origin.??

 

[Dick]

thanks. i see how Pythagorean's theorem works. But i am still confused how to find when the function is closest to the origin.??

If you mean that the particle A is sitting at (0,0) and particle B is somewhere along the curve (x,-cos(x-7)+3) then you can graph it and try to estimate closest distance. It's a bit over 2. You can also use Pythagoras to get an expression for the distance and try to minimize it with calculus. But it will be too complicated to solve with algebra. You'd have to use numerical methods to get an estimate.

 

[syeh]

If you mean that the particle A is sitting at (0,0) and particle B is somewhere along the curve (x,-cos(x-7)+3) then you can graph it and try to estimate closest distance. It's a bit over 2. You can also use Pythagoras to get an expression for the distance and try to minimize it with calculus. But it will be too complicated to solve with algebra. You'd have to use numerical methods to get an estimate.

thanks. i used the Pythagorean theorem and found the length to be 2.084!

 

[Dick]

thanks. i used the Pythagorean theorem and found the length to be 2.084!

Well, I get something a lot closer to 2.1. I'm not sure what you did. But if you realize you have to estimate the answer that's a start.

 

[HallsofIvy]

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.

WHY?? You are asked to minimize the distance to the origin, d= \sqrt{x^2+ y^2}= \sqrt{x^2+ f^2(x)}, not y= f(x).

Of course, minimizing the distance will also minimize the square of the distance so you can drop the square root and look at d^2= x^2+ f^2(x)= x^2+ (3- cos(x- 7))^2= x^2+ 9- 6 cos(x- 7)+ cos^2(x- 7).

Differentiate that.

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.