Final answer:
To find the unit tangent vector T and the curvature for the parameterized curve r(t) = (-6t, -6ln(cos t)), we need to find the velocity vector v(t) and normalize it to obtain T. Then, we find the acceleration vector a(t) and use it to calculate the curvature k.
Step-by-step explanation:
To find the unit tangent vector T and the curvature for the parameterized curve r(t) = (-6t, -6ln(cos t)), we will first find the velocity vector v(t) and then normalize it to obtain T. Next, we will find the acceleration vector a(t) and use it to calculate the curvature k using the formula k = ||a(t)||/||v(t)||^2.
Step 1: Find the velocity vector v(t)
The velocity vector is the derivative of the position vector, so v(t) = r'(t). Differentiating each component of r(t), we get v(t) = (-6, 6tan(t)).
Step 2: Find the unit tangent vector T
The unit tangent vector is obtained by normalizing the velocity vector, so T = v(t)/||v(t)||. Using the magnitude of v(t) given by ||v(t)|| = sqrt((-6)^2 + (6tan(t))^2), we can calculate the unit tangent vector.
Step 3: Find the acceleration vector a(t)
The acceleration vector is the derivative of the velocity vector, so a(t) = v'(t). Differentiating each component of v(t), we get a(t) = (0, 6sec^2(t)).
Step 4: Find the curvature k
The curvature is given by the formula k = ||a(t)||/||v(t)||^2. Plugging in the magnitudes of a(t) and v(t), we can calculate the curvature.