If r =bt^2i+ct^3j , 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

To find when the velocity vector makes an angle of 45 degrees with the x- and y-axes, we need to find the values of t that satisfy this condition.

Let's start by finding the magnitude of the velocity vector:

|v| = √((bt^2)^2 + (ct^3)^2)
= √(b^2t^4 + c^2t^6)
= √(b^2t^4(1 + c^2t^2))

Next, we can find the slope of the velocity vector using the formula:

tan(45 degrees) = |v|y / |v|x

Since we want the velocity vector to make a 45 degree angle with both the x- and y-axes, the slope of the velocity vector must be 1:

1 = (ct^3) / (bt^2)
=> c = b

Now we can substitute c = b into the magnitude equation:

|v| = √(b^2t^4(1 + b^2t^2))
= bt^2√(1 + b^2t^2)

To make it more convenient, let's denote z = b^2t^2:

|v| = b√(z(1 + z))

Now we need to find the values of t (or z) that make |v| = 1:

1 = b√(z(1 + z))
1/b = √(z(1 + z))
(1/b)^2 = z(1 + z)
z^2 + z - (1/b)^2 = 0

This is a quadratic equation in z. We can solve it by using the quadratic formula:

z = (-1 ± √(1^2 - 4(1)(-(1/b)^2))) / (2(1))
z = (-1 ± √(1 + 4/b^2)) / 2
z = (-1 ± √((b^2 + 4) / b^2)) / 2

Since b and c are positive constants, z must also be positive. Therefore, we take the positive square root:

z = (-1 + √((b^2 + 4) / b^2)) / 2

Finally, we can solve for t by substituting back z = b^2t^2:

(-1 + √((b^2 + 4) / b^2)) / 2 = b^2t^2
t^2 = (-1 + √((b^2 + 4) / b^2)) / 2b^2
t = ±√((-1 + √((b^2 + 4) / b^2)) / (2b^2))

Therefore, the velocity vector makes an angle of 45 degrees with the x- and y-axes at the values of t given by t = ±√((-1 + √((b^2 + 4) / b^2)) / (2b^2)).