Question

If r⃗ =bt2i^+ct3j^, where b and c are positive constants, when does the velocity vector make an angle of 45.0∘ with the x- and y-axes?

Answer 1

The velocity vector for r⟶ = bt^2î + ct^3ï makes an angle of 45 degrees with the x- and y-axes when the time t equals 2b/(3c), found by setting the magnitudes of the x and y components of the velocity vector equal to each other.

Step-by-step explanation:

To determine when the velocity vector makes an angle of 45 degrees with the x- and y-axes for the given position vector r⟶ = bt^2î + ct^3ï, where b and c are positive constants, we first need to find the velocity vector by differentiating the position vector with respect to time.

The velocity vector v⟶ is given by: v⟶ = dr⟶/dt = 2btî + 3ct^2ï. An angle of 45 degrees between the velocity vector and the axes means the components along the x- and y-axes must be equal. This occurs when 2bt = 3ct^2, which simplifies to t = 2b/(3c). Therefore, the velocity vector makes a 45-degree angle with the axes at this specific time t.

Answer 2

The displacement function is given by : r = bt2i + ct3j meters
The velocity function is the derivative of the displacement:
v = r' = 2bti + 3ct2j meter / second

Now, for the velocity vector make an angle of 45.0∘ with the x- and y-axes, the i and j components have to be equal:
2bt = 3ct2 (Now solve for t)
2bt = t (3ct)
either t = zero (rejected)
or t = 2b / 3c seconds (accepted)