Velocity Using Parametric Equations

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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}
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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