When will the velocity vector make an angle of 45 degrees?

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To determine when the velocity vector makes a 45-degree angle, the relationship between the vertical and horizontal components of the velocity must be established. The velocity vector is derived from the position vector r = bt^2i + ct^3j by taking its derivative, resulting in v = (2bt)i + (3ct^2)j. For the angle to be 45 degrees, the ratio of the vertical component (V_y) to the horizontal component (V_x) must equal 1, indicating that V_y = V_x. This leads to the equation 3ct^2 = 2bt, which can be solved for t to find the specific time when the angle condition is met. Understanding this relationship is crucial for solving the problem effectively.
bharp24
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i have been given that vector r = bt^2i + ct^3j where b and c are positive constants, and i need to know when the velocity vector will make an angle of 45 degrees.

i have been trying to figure it out and am just confused. i have taken the derivative of vec r, because i know that it becomes velocity but i don't know what to do from there. i cannot figure out the components of what makes up the velocity vector. i feel like i am going in circles. please someone help!
 
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Hint: When the velocity vector makes an angle of 45 degrees with respect to the horizontal, what can you say about V_y/V_x?
 
it will = 1 because the sin and cos of 45 degrees are the same...right?
 
Right!
 

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