The position vector r of a particle moving in an xy plane is given by [r = (6t^3-3t)i + (4-1t^4)j] with r in meters and t in seconds. What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 3 s? Give your answer in the range of (-180 degrees; 180 degrees].
To find the angle between the positive x-axis and the line tangent to the particle's path at a specific value of t, we need to calculate the vector tangent to the path at that moment and then find the angle between that vector and the positive x-axis.
Let's first calculate the tangent vector by finding the derivative of the position vector with respect to time (t).
Given: r = (6t^3 - 3t)i + (4 - t^4)j
Taking the derivative of r with respect to t, we get:
r' = d/dt[(6t^3 - 3t)i + (4 - t^4)j]
r' = (18t^2 - 3)i - (4t^3)j
Substituting t = 3 into r', we get:
r'(3) = (18(3)^2 - 3)i - (4(3)^3)j
= (162 - 3)i - (108)j
= 159i - 108j
Now, we have the tangent vector. To find the angle between this vector and the positive x-axis, we can use the dot product and the arctangent function.
The dot product of two vectors A = (A1, A2) and B = (B1, B2) is given by:
A · B = A1B1 + A2B2
Let A = (x1, y1) be the unit vector in the positive x-direction (1, 0), and B = (159, -108) be the tangent vector we calculated above.
A · B = (1)(159) + (0)(-108)
= 159
The magnitude of A is given by ||A|| = sqrt(A1^2 + A2^2) = sqrt(1^2 + 0^2) = 1.
The magnitude of B is given by ||B|| = sqrt(159^2 + (-108)^2) = sqrt(25281 + 11664) = sqrt(36945) = 192.202
Using the dot product formula and the magnitudes of A and B:
cos(theta) = (A · B) / (||A|| * ||B||)
= 159 / (1 * 192.202)
= 0.826
To find the angle theta, we can take the arccosine (inverse cosine) of the value obtained:
theta = arccos(0.826)
≈ 33.815 degrees
Since the angle of rotation is measured counterclockwise, the angle between the positive x-axis and the tangent line is approximately 33.815 degrees in the range (-180 degrees, 180 degrees].