If position vector r=bt²i + ct³j, where c are positive constants, when does the velocity vector make an angle of 45° with the x and y axes?

We want to find the value of t such that the velocity vector makes an angle of 45° with both axes.

We found that:

t = (2/3)*(b/c)

The velocity vector makes an angle of 45° with the x and y axes.

We know that the position vector is:

r = r=b*t²i + c*t³j

Remember that the versor "i" corresponds to the x-component, and the versor "j" corresponds to the y-component, then:

r = r=b*t²i + c*t³j  = (b*t², c*t³)

The velocity vector is the vector that we get when we differentiate the position one, remember that if:

f(x) = a*x^n

then

f'(x) = n*a*x^(n - 1)

Using this, we can find that the velocity vector is:

v = (2*b*t,  3*c*t²)

Now we want to know, when does the velocity vector make an angle of 45° with the x and y axes.

Let's think of the vector as the hypotenuse of a triangle rectangle, where the x-component is the adjacent cathetus, and the y-component is the opposite cathetus. (so the angle is measured counterclockwise from the x-axis)

We have the trigonometric equation:

tan(a) = (opposite cathetus)/(adjacent cathetus)

So now we can replace these things with the known ones:

a = 45°

opposite cathetus = y-component = 3*c*t²

adjacent cathetus = x-component = 2*b*t

So we will get:

tan(45°)  =  (3*c*t²)/( 2*b*t)

1 = (3/2)*(c/b)*t

Now we can solve this for the variable, t.

1*(2/3)*(b/c) = t

t = (2/3)*(b/c)

We can conclude that at the time:

t = (2/3)*(b/c)

The velocity vector makes an angle of 45° with the x and y axes.

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