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.
You can read more about vectors in:
brainly.com/question/10841907
