The position of two particles on the s-axis are s1= cos(t) and s2= cos(t+(symbol for pie)/4)
a. what is the farthest apart the particles ever get?
b. when do the particles collide?
I think it would be of more benefit if you posted your thinking or work, and we critiqued it.
Well I am in AP Calculus and I don't understand the class to well.
difference = y = cos ( t + pi/4) - cos t
for part a
take dy/dt and set to 0
dy/dt = -sin (t+pi/4) + sin t
0 = sin t - sin (t+pi/4)
sin t = sin (t+pi/4)
sin (a+b) =sin a cos b + cos a sin b
so
sin t = sin t cos pi/4 + cos t sin pi/4
sin t = sin t /sqrt 2 + cos t/sqrt 2
sqrt 2 sin t = sin t + cos t
(sqrt 2-1) = cos t/sin t = cot t
cot t = .4142
t = 67.5 degrees = 1.178 radians
also in other quadrants where cot t =.4142
NOW, go back and find y
for part b,just find when cos (t+pi/4) = cos t
Thank You
a. The particles are really spinning around on the s-axis like a wild disco dance party! To find out the farthest distance between them, we can use some trigonometry magic. The distance between the particles at any given time t is given by d(t) = |s1 - s2|.
Using the cosine difference formula, we have d(t) = |cos(t) - cos(t + π/4)|. Now, let's flex our math muscles:
d(t) = |cos(t) - cos(t)cos(π/4) + sin(t)sin(π/4)|
= |cos(t) - (cos(t) + sin(t))/√2|
Applying the absolute value function on both sides helps us eliminate the pesky negative sign. Now we have:
d(t) = |2cos(t)/√2 - sin(t)/√2|
= |(2cos(t) - sin(t))/√2|
To find the farthest distance, we need to find the maximum value of d(t). Hmm, this looks like a good opportunity for our buddy, Calculus. By taking the derivative of d(t) and setting it equal to zero, we can find the critical points.
After some grinding with calculus, the critical point occurs when sin(t) = 2cos(t), which is t = π/6. Plugging t = π/6 into our original equation, we find that the farthest distance is:
d(π/6) = |2cos(π/6) - sin(π/6)|/√2
= |√3 - 1|/√2
So, the particles get as far apart as |√3 - 1|/√2.
b. Ah, the inevitable collision moment! We are looking for the time when the particles meet each other with open arms. To find out when they collide, we need to solve s1 = s2.
cos(t) = cos(t + π/4)
If we apply some geometric thinking, we know that this happens when the angle t and t + π/4 differ by some multiple of 2π. So, we can write:
t + π/4 = t + 2πn (where n is an integer)
Simplifying, we have:
π/4 = 2πn
Dividing through by π, we find:
1/4 = 2n
Since n has to be an integer, the only solution is n = 0. Substituting back into the equation, we get:
t + π/4 = t
This simplifies to:
π/4 = 0
However, this is not a valid solution as π/4 ≠ 0. Therefore, the particles never collide on the s-axis.
Science fact-turned-humor: These particles seem to be playing a perpetual "catch me if you can" game. Always dodging each other, those tricky little particles!
To find the farthest distance between the two particles, we need to find the maximum difference between their positions at any given time. Let's start by expressing their positions as functions of time.
Given:
s1 = cos(t)
s2 = cos(t + π/4)
a. To find the farthest apart the particles ever get, we need to find the maximum value of (s2 - s1) over a certain time interval. This can be done by finding the maximum value of the function (cos(t + π/4) - cos(t)).
To find this maximum value, we can take the derivative of (cos(t + π/4) - cos(t)) with respect to t, set it equal to zero, and solve for t. This will give us the time when the particles are farthest apart.
Let's calculate the derivative:
f(t) = cos(t + π/4) - cos(t)
f'(t) = -sin(t + π/4) - (-sin(t))
Setting f'(t) = 0:
-sin(t + π/4) + sin(t) = 0
Now, we can use trigonometric identities to simplify this equation:
-sin(t)cos(π/4) + cos(t)sin(π/4) + sin(t) = 0
(-sin(t) + cos(t)) / √2 + sin(t) = 0
Simplifying further:
√2(sin(t) + cos(t)) + √2sin(t) = 0
√2sin(t) + √2sin(t) + √2cos(t) = 0
2√2sin(t) + √2cos(t) = 0
Now, we can express the equation in terms of tan(t/2):
2√2tan(t/2) + √2 = 0
Dividing both sides by √2:
2tan(t/2) + 1 = 0
tan(t/2) = -1/2
Using the properties of the tangent function, we know that the angle t/2 must be in the second or fourth quadrant, where tan(t/2) = -1/2.
In the second quadrant:
t/2 = π + arctan(1/2)
In the fourth quadrant:
t/2 = -π + arctan(1/2)
Now, we can find the values of t:
t = 2(π + arctan(1/2))
t = 2(-π + arctan(1/2))
Now that we have the values of t, we can calculate the positions s1 and s2 at those moments and find the maximum distance between them.
b. To determine when the particles collide, we need to find the time when their positions are equal. Mathematically, this means finding the values of t for which the equation s1 = s2 is satisfied.
Setting s1 = s2:
cos(t) = cos(t + π/4)
Using the trigonometric identity cos(A) = cos(B) implies A = ±B + 2πn, where n is an integer:
t = ±(t + π/4) + 2πn
Simplifying:
0 = ±π/4 + 2πn
±π/4 = 2πn
Dividing both sides by π:
±1/4 = 2n
±1/4 = n
This means that the particles collide at t = ±π/4 + 2πn, where n is an integer.
Note: In both parts (a) and (b), it's important to consider the values of t that fall within a specific interval, if provided.