The Question: A particle moves along the X-axis so that at time t > or equal to 0 its position is given by x(t) = cos(√t). What is the velocity of the particle at the first instance the particle is at the origin?

So far I was able to determine that the velocity of the particle would be undefined but, I don't understand what it means in the context of this problem for velocity to be undefined. What is the particle doing if its velocity is undefined?

I believe your equation for velocity is incorrect.

In this case
V=-sin(sqrt(t))*1/2(sqrt(t))^-1/2

Your equation for velocity is missing raising to the -1/2 power at the end. applying that piece will make velocity undefined at t=0.

Nope.

x= cos(t^1/2)
v=-sin(t^1/2) *1/2*1/t^1/2
= -sin(sqrtt)*1/(2sqrtt) which is what I have. Now, how does having
sqrt(PI/2) in the denominator make it undefined. I must not be seeing your point.

Oh lol, I think I see what we're doing differently.

I thought that the first instance that the particle would be at the origin was when t=0 but it really is when t=pi/2.

Basically, I substituted the wrong number.

Thank you, and I'm sorry for not seeing your point.

Ah, the good old undefined velocity question. Well, think of it this way: when the velocity is undefined, the particle is going through a bit of an existential crisis. It's like when a clown loses their sense of humor - it just doesn't know which way to go!

In this specific case, if the velocity is undefined at the first instance the particle is at the origin, it means that the particle is momentarily stuck, unsure of which direction to move in. It's having a brief pause, contemplating the mysteries of the universe... or at least the mysteries of its own motion.

But fear not! This existential crisis is temporary. After a split second of confusion, the particle will eventually regain its velocity and continue on its merry way, exploring the wonders of the X-axis. It's all just a part of life's ups and downs, you know?

So, to sum it up, if the velocity is undefined at the first instance the particle is at the origin, it simply means the particle is experiencing a momentary pause or indecisiveness. It's like it's saying, "Hold on, folks! Let me think for a moment before I figure out where to go next!"

To determine the velocity of the particle, we need to find the derivative of the position function, x(t), with respect to time, t.

The position function is given as x(t) = cos(√t).

To find the derivative of x(t), we can apply the chain rule:

Let's rewrite the position function as x(t) = cos((t)^(1/2)).

Now, let u = √t. Therefore, our position function can be written as x(u) = cos(u).

Now, differentiate x(u) with respect to u:

dx/du = -sin(u).

But u = √t, so we need to differentiate u with respect to t:

du/dt = (1/2√t).

Applying the chain rule, we can find dx/dt:

dx/dt = (dx/du) * (du/dt) = -sin(u) * (1/2√t) = -(sin(√t))/(2√t).

Now, we have the derivative of the position function, x(t), with respect to time, t.

To find the velocity of the particle at the first instance it is at the origin, we need to find the value of t when x(t) = 0.

Setting x(t) = 0, we get cos(√t) = 0.

This means, at √t = (2n + 1)*(π/2), for any integer n.

Solving for t, we get t = [(2n + 1)*(π/2)]^2.

So, the particle is at the origin at t = [(2n + 1)*(π/2)]^2.

However, when t = 0, the expression for velocity, dx/dt = -(sin(√t))/(2√t), is undefined due to the presence of √t in the denominator.

This means that at t = 0, the velocity of the particle is undefined.

In the context of this problem, if the velocity is undefined, it suggests that the particle abruptly changes its speed or direction at t = 0. It could be interpreted as a point of discontinuity in the motion of the particle.

x=cos(sqrt(t))

v=-sin(sqrt(t))*1/2sqrtt

wo when is position zero?
0=cos(sqrt(t))
sqrt(t)=PI/2
t=PI^2/4

v(PI^2/4)=-sin(PI/2)*1/2sqrtPI/2)

but sin(PI/2)=1
v(PI^2/4)=1/sqrt(PI/2)= 0.797884561

So I dont see where the undefined comes from.