a particle moves along the x-axis so that its velocity at any time t is equal to or > 0 is given by v(t)= (2pie-5)t-sin(piet)
a.find the acceleration at any time t.
b.find the minimum acceleration of the particle over the interval [0,3]
c.find the maximum velocity of the particle over the interval [0,2]
I will read that as
v(t) = (2π - 5)t - sin(πt)
= 2πt - 5t - sin(πt)
a(t) = 2π - 5 - πcos(πt)
for a min of a , a ' (t) = 0
π^2 sin(πt) = 0
sin(πt) = 0
πt = 0 or πt = π or πt = 2π
t = 0 , t = 1 , t = 2
a(0) = 2π - 5 - πcos0 = π - 5
a(1) = 2π - 5 - πcosπ = 2π-5+π = 3π-5
a(2) = 2π - 5 - πcos2π = 2π - 5 - π = π - 5
a(3) = ... = 3π - 5
looks like the min is π-5
c) for max velocity, a(t) = 0
2πt - 5t - sin(πt) = 0
sin (πt) = 2πt - 5t
very messy equation to solve,
I ran it through Wolfram's "magic" webpage and got
t = 0 and t = .670349
so
v(.670349) = 0
v(0) = sin0 = 0
v(2) = (2π-5)(2) - sin 2π
- 2.566.. - 0 = 2.566
check my arithmetic
To find the acceleration at any time t, we need to find the derivative of the velocity function v(t) with respect to time.
a. acceleration a(t) = v'(t) = d/dt[(2πe-5)t - sin(πet)]
To find the derivative of (2πe-5)t, apply the power rule: the derivative of t^n is n*t^(n-1). Since n = 1 and the coefficient is (2πe-5), the derivative is (2πe-5)*1 = 2πe-5.
Next, to find the derivative of -sin(πet), we use the chain rule. The derivative of the outer function -sin(u) is -cos(u). The derivative of the inner function πet is πe^(πet) by applying the chain rule. Therefore, applying the chain rule, the derivative of -sin(πet) is -πe^(πet)*cos(πet).
Combining these derivatives, we get:
a(t) = (2πe-5) - πe^(πet)*cos(πet)
b. To find the minimum acceleration of the particle over the interval [0,3], we need to find the critical points of the acceleration function a(t) and evaluate them within the given interval.
To find the critical points, we set the derivative of a(t) equal to zero and solve for t:
(2πe-5) - πe^(πet)*cos(πet) = 0
Unfortunately, this equation does not have a simple algebraic solution. Instead, we can use numerical methods or a graphing calculator to find the approximate values of t where a(t) = 0. Once we have these critical points, we can then evaluate the acceleration at these points and determine the minimum value within the interval [0,3].
c. To find the maximum velocity of the particle over the interval [0,2], we need to find the maximum values of the velocity function v(t) within the given interval.
We can start by finding the derivative of the velocity function v(t) with respect to time to find critical points. Once we have the critical points, we can evaluate the velocity at these points and determine the maximum value within the interval [0,2].
To find the critical points, we set the derivative of v(t) equal to zero and solve for t:
(2πe-5) - πe^(πet)*cos(πet) = 0
Again, this equation does not have a simple algebraic solution, so we need to use numerical methods or a graphing calculator to find the approximate values of t where v'(t) = 0. Once we have these critical points, we can then evaluate the velocity at these points and determine the maximum value within the interval [0,2].