Velocity Using Parametric Equations

AI Thread Summary
The discussion revolves around finding the magnitude of the velocity of an object moving according to the parametric equations x(t) = 2t^2 + 3t and y(t) = 4cos(t) at t = 3 seconds. The derivatives dx/dt and dy/dt were calculated as 4t + 3 and -4sin(t), respectively. The user initially attempted to find the velocity using dy/dx but later clarified that the magnitude of the velocity vector should be calculated using the components v_x and v_y. The correct approach involves evaluating the derivatives at t = 3, squaring them, summing these squares, and taking the square root. This method will yield the total velocity magnitude at the specified time.
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Homework Statement



An object moves in two dimensions according to the parametric equations x(t) = At^2 + Bt and y(t) = D cos(Et). The constants A, B, D, and E are A = 2 m/s^2, B = 3 m/s, D = 4 m, and E = 1 rad/s. What is the magnitude of the total velocity of the object at t = 3 s?

Homework Equations





The Attempt at a Solution



I'm not sure if I did this problem right. I plugged back in the constants

x(t) = 2 t^2 + 3 t
y(t) = 4 cos(t)

dx/dt = 4 t + 3
dy/dt = -4 sin(t)

dy/dx = dy/dt dt/dx = [-4 sin(t)]/[4 t + 3]

I thought that this was the velocity?

I then plugged in 3 for t and then plugged this into my calculator
[-4 sin(3)]/(12+3) and got about - .038 m/s but sense it said magnitude only I ignored the negative sign and put .038 m/s

I have the feeling I did this problem wrong. This is for my physics 2 course and is suppose to be a introductory physics course after taking physics 1 (non calculus based) and this is just suppose to be like calculus I based but parametric equations is a calculus 2 topic (in most american schools) and I'm in calculus 2 at the moment and haven't covered the topic yet and only have a brief understanding of it so I'm not sure
 
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What is the magnitude of the vector \vec{v}=[v_x,v_y]?
Or, how do you evaluate the magnitude of a vector knowing its components?

And, \vec{v}=[v_x,v_y]=[\frac{dx}{dt},\frac{dy}{dt}]
 
Ah I thought so, so I plug in 3 into each of the derivatives square both of these values sum these squared values and then take the square root of the whole thing?
 
Yep
|\vec{v}_{t=3}|=\sqrt{\left[\left.\frac{dx}{dt}\right|_{t=3}\right]^2+\left[\left.\frac{dy}{dt}\right|_{t=3}\right]^2}
 
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