An object travels the path given by the parametric equations x=ln2t + t^3 and y=(e^t) - 2 sint. At t=2.1 sec, what is the acceleration of the particle?

Question

An object travels the path given by the parametric equations x=ln2t + t^3 and y=(e^t) - 2 sint. At t=2.1 sec, what is the acceleration of the particle?

Answer 1

x=ln2t + t^3

dx/dt = 1/t + 3t^2

y=(e^t) - 2 sint
dy/dt = e^t - 2cost
v = dy/dx = (dy/dt) / (dx/dt) = (e^t - 2cost)/((1/t + 3t^2)
a = dv/dt = [(1/t + 3t^2)(e^t + 2sint) - (e^t - 2cost)(-1/t^2 + 6t)]/(1/t + 3t^2)^2
phewww
now let t = 2.1, I am sure you have a calculator, make sure you are set to radians

Answer 2

v = dy/dx = dy/dt / dx/dt = (e^t-2cost)/(1/t + 3t^2)

a = dv/dx = dv/dt / dx/dt
= [(e^t+2sint)(1/t + 3t^2) - (6t - 1/2)(e^t-2cost)]/(1/t + 3t^2)^3

This is just a stupid exercise -- needlessly complicated.

Answer 3

To find the acceleration of the particle at t = 2.1 sec, we need to differentiate the parametric equations twice with respect to time (t).

Given the parametric equations:
x = ln(2t) + t^3
y = e^t - 2sin(t)

Let's start by finding the first derivatives of x and y with respect to t:

dx/dt = d/dt (ln(2t) + t^3)
dy/dt = d/dt (e^t - 2sin(t))

To find the second derivatives, we differentiate the first derivatives:

d^2x/dt^2 = d/dt (dx/dt)
d^2y/dt^2 = d/dt (dy/dt)

First, let's find d^2x/dt^2:

d^2x/dt^2 = d/dt (dx/dt)
= d/dt (d/dt (ln(2t) + t^3))
= d/dt (1/t + 3t^2)
= -1/t^2 + 6t

Now, let's find d^2y/dt^2:

d^2y/dt^2 = d/dt (dy/dt)
= d/dt (d/dt (e^t - 2sin(t)))
= d/dt (e^t - 2cos(t))
= e^t + 2sin(t)

Now, we can substitute t = 2.1 sec into the expressions for d^2x/dt^2 and d^2y/dt^2:

d^2x/dt^2 = -1/(2.1)^2 + 6(2.1) ≈ 12.828 sec^-2
d^2y/dt^2 = e^(2.1) + 2sin(2.1) ≈ 15.616

So, the acceleration of the particle at t = 2.1 sec is approximately 12.828 sec^-2 in the x-direction and 15.616 in the y-direction.

Answer 4

To find the acceleration of the particle at t = 2.1 sec, we need to find its second derivative with respect to time.

Given the parametric equations:
x = ln(2t) + t^3
y = e^t - 2sin(t)

First, let's find the first derivatives of x and y with respect to t to find the velocity:

dx/dt = (1/t) + 3t^2
dy/dt = e^t - 2cos(t)

Next, let's find the second derivatives of x and y with respect to t to find the acceleration:

d^2x/dt^2 = d/dt[(1/t) + 3t^2] = -1/t^2 + 6t
d^2y/dt^2 = d/dt[e^t - 2cos(t)] = e^t + 2sin(t)

Now, substitute t = 2.1 sec into the equations to find the acceleration at t = 2.1 sec:

d^2x/dt^2 = -1/(2.1)^2 + 6(2.1)
d^2y/dt^2 = e^2.1 + 2sin(2.1)

Using a calculator or a mathematical software, evaluate the values to find the acceleration at t = 2.1 sec.