please answer this?

Question

x=3cost+4sint is the position of an oscillating particle at time t.Find the acceleration of the particle at two extreme points of oscillation.

Answer 1

Here is one more example to hammer on that point: Let F be thevector field F(x,y) = (x + y)i + (xy)j. Let C be the arc of the ellipsex = 3cost, y = 4sint, t = 0...π/2. Then the vector version of thisparametrization isr(t) = (3cost)i + (4sint)j ,the velocity isv(t) = (−3sint)i + (4cost)j ,and we haveCF · T ds =π/2t=0F · v dt =π/2t=0(3cost + 4sint)(−3sint) + ((3cos t)(4sint))(4cost) dt .(I’ll leave it to you and your computer, or table of integrals, to workthis out.)

Answer 2

x= 3 cos t+ 4 sin t
dx/dt= -3 sint + 4 cos t
acceleration= d2x/dt2 = -3 cost - 4sin t= -(3 cost +4sin t)= -x

at the two extreme positions of the particle,
x = x(max) = a = amplitude of the oscillation
at the two extreme positions of the particle,
acceleration= - x(max)= -a
and the amplitude of the oscillation = (3^2 + 4^2)^(1/2)= 5
hence at the two extreme positions, the acceleration of the particle is -5 units, the negative sign indicates that it is opposite directional with displacement.

Answer 3

a===5

Answer 4

easy

the velocity, v, is the derivative of the position, with respect to time. so we have
v = dx/dt=-3sint + 4cost

the acceleration, a, is the derivative of the velocity w respect to time, so we have
a = dv/dt =-3cost -4sint

now, at extreme points of oscillation, by definition the speed is zero

so from above you have
-3sint + 4cost = 0

divide by cost
3tant = 4

tant = 4/3

so t = .927 radians

now plug this into the above function for a, so you have
a (extreme point) = -3cos(.927)-4sin(.927)
a (extreme point) = -5.0


hope this helps

Answer 5

Hint:
It is a Maxima - Minima problem at extreme Velocity is Zero and acceleration changes sign. If it is linear, then length of the pendulum is infinite.

Find the equation for pendulum and apply this logical parameter to solve for requested parameters.

Answer 6

x = 3cos(t) + 4sin(t)
dx/dt = -3sin(t) + 4cos(t) = velocity (v)
d2x/dt2 = -3cos(t) - 4sin(t) = acceleration (a)

set dx/dt = 0 and solve for t:
t ~ 0.9273 radians

plug that t value into the equation for acceleration:
a = 5

Answer 7

You should state your question instead of "please answer this?"

I feel like I have wasted valuable time just to educate you in this matter.

Answer 8

yes
Ask me a question
you don't know nothing