A particle moves along the x-axis with the velocity given by v(t)=3t/(1+t^2) for t >or equal to 0. When t=0, the particle is at the point (4,0). 1. Determine the maximum velocity for the particle. Justify your answer. 2. Determine the position of the particle at any time t. 3. Find the limit of the velocity as t->infinity. 4. Find the limit of the position as t->infinity.
Use Steve's answer for integral !
v(t) = 3t/(1+t^2)
1.
a(t) = 3(1-t^2)/(1+t^2)^2
max v at a=0, or t=1
v(1) = 3/2
2.
x(t) = 3/2 log(1+t^2)+C
at t=0, x=4, so
4 = 3/2 log(1+0) + C
C = 4
x(t) = 3/2 log(1+t^2) + 4
3.
v(t) → 0 as t→∞
4.
x(t) → ∞
dv/dt = [(1+t^2)(3) - 3t(2t)] / (t^2+1)^2
=[3 t^2 + 3 - 6 t^2 ] /(t^2+1)^2
zero when t= 1 (since t always >/=0)
so the max v is at t = 1
then at t = 1
v(1) = 3/2
x = integral v dt
= (3/2)(t^2+1)^1 from t = 4 to t = t
= (3/2)(t^2+1) - 3*17/2
as t --> oo, v--> 3t/t^2 = 3/t = 0
as t -->oo, x --> (3/2)t^2 = oo
To determine the maximum velocity of the particle, we need to find the maximum value of the velocity function v(t) = 3t/(1+t^2).
Step 1: Find the derivative of v(t) with respect to t.
The derivative of v(t) can be found using the quotient rule:
v'(t) = [3(1+t^2) - 3t(2t)] / (1+t^2)^2
Simplifying, we get v'(t) = 3(1-t^2) / (1+t^2)^2
Step 2: Find the critical points.
To find the critical points, we set v'(t) equal to zero and solve for t:
3(1-t^2) / (1+t^2)^2 = 0
Since the numerator is always nonzero, the critical points occur when the denominator is zero:
1+t^2 = 0
t^2 = -1
This equation has no real solutions, so there are no critical points.
Step 3: Analyze the endpoints.
To analyze the endpoint t = 0, we substitute it back into the original velocity function:
v(0) = 3(0) / (1 + 0^2)
v(0) = 0
Therefore, the initial velocity of the particle is 0.
Step 4: Determine the limits of the velocity.
As t approaches positive infinity, the velocity function v(t) approaches:
lim(t->∞) (3t/(1+t^2)) = 0/(1+∞^2) = 0
Therefore, the limit of the velocity as t approaches infinity is 0.
Step 5: Determine the maximum velocity.
Since there are no critical points and the limit as t approaches infinity is 0, the maximum velocity must occur at one of the endpoints. In this case, the maximum velocity is 0.
2. To determine the position of the particle at any time t, we can integrate the velocity function. Since the initial position is given as (4,0), we can apply the initial condition.
∫(0 to t) (3t/(1+t^2)) dt = ∫(4 to x) 1 dx
[3/2 * ln|1+t^2|] [0 to t] = [x] [4 to x]
[3/2 * ln(1+t^2)] - [0] = x - 4
x = 3/2 * ln(1+t^2) + 4
Therefore, the position of the particle at any time t is x = 3/2 * ln(1+t^2) + 4.
3. As we found in Step 4, the limit of the velocity as t approaches infinity is 0.
4. To find the limit of the position as t approaches infinity, we substitute t = ∞ into the position function:
x = 3/2 * ln(1+∞^2) + 4
The natural logarithm of a large number approaches infinity, so the limit of the position as t approaches infinity is infinity.