2-dimentional movement of a point

[SimpliciusH]

Ok, I know this post is not up to specs but the real problem with this is my trouble with calculus and derivatives (I'm actually worse at that than at English ;) ) in general. I only need help with this step of the homework (basic math help).

This describes the movement of the point:
x= 4*x0*cos (Angular velocity*time)
y= x0*sin (2*angular velocity*time)

Angular velocity is given as is x0.

At first I just thought dx/dt = v, and went for it by using the rule (dsink*k/dx=k*cos(k*x)) got:

Vx=-4x0*angular velocity* sin (angular velocity*t)
Vy=x0*2*angular velocity*cos (2*angular velocity*t)

I know this is wrong. Since I realized as soon as I wrote it that, angular velocity is v/r so d(v/r)/dt, trying to get throught that using the rule for d(f(x)/g(x))/dx it got quickly complicated since looking at the XY graph (I got something that looks like a 8) for movment leads me to believe r is not constant.


Anyway the more basic problem is that I don't know how to solve sin (f(x)*x)/dx or the second one for cos :(

I can solve d(f(x)*x)/dx and dsin(x)/dx (since both are covered by the simple equations) but putting them together confuses the heck out of me.


You guys where very patient and helpful with my previous problem, I hope I can learn from you again. :) Again, thaks for taking the time to help a newbie with something (for you) this trivial. :)




PS I hope angular velocity is the proper english word the derivative for what I mean is dAngle/dTime its related to velocity (v= r * angular velocity).

 

[kuruman]

I know this is wrong. Since I realized as soon as I wrote it that, angular velocity is v/r...

It is not wrong. You took the derivatives correctly so the velocity components are what they are. Here you do not have circular motion, so ω=v/r does not apply. If you had circular motion then

x²+y² = constant. This is not the case here.

 

[kerimek]

As a scientist, my first suggestion would be to review the fundamental principles and equations of calculus and derivatives. It is important to have a strong foundation in these concepts in order to solve more complex problems.

That being said, let's take a look at the movement of the point described in the post. From the given equations, we can see that x and y are both functions of time. This means that we can use the chain rule to find the derivatives of x and y with respect to time.

To find the derivative of x with respect to time, we can use the chain rule as follows:

dx/dt = d/dt[4*x0*cos(angular velocity*time)]
= 4*x0*(-sin(angular velocity*time))*d/dt[angular velocity*time]
= 4*x0*(-sin(angular velocity*time))*angular velocity
= -4*x0*angular velocity*sin(angular velocity*time)

Similarly, the derivative of y with respect to time can be found using the chain rule:

dy/dt = d/dt[x0*sin(2*angular velocity*time)]
= x0*cos(2*angular velocity*time)*d/dt[2*angular velocity*time]
= x0*cos(2*angular velocity*time)*2*angular velocity
= 2*x0*angular velocity*cos(2*angular velocity*time)

Now, we can see that the velocity in the x-direction (Vx) is equal to the derivative of x with respect to time, and the velocity in the y-direction (Vy) is equal to the derivative of y with respect to time. So, we can write the velocities as:

Vx = -4*x0*angular velocity*sin(angular velocity*time)
Vy = 2*x0*angular velocity*cos(2*angular velocity*time)

I hope this helps with your understanding of derivatives and solving this problem. Remember to always go back to the fundamental principles and equations when facing a new problem, and don't hesitate to seek help when needed. Good luck with your homework!