Final answer:
To find the unit tangent and unit normal vectors of the vector function r(t) = 9t, 4 cos t, 4 sin t, we first need to find the velocity vector v(t). Taking the derivative of r(t) with respect to t gives us v(t) = 9, -4 sin t, 4 cos t. We can then find the magnitude of v(t) to obtain the speed of the particle. The unit tangent vector t(t) is obtained by dividing v(t) by its magnitude, and the unit normal vector n(t) is obtained by taking the derivative of t(t) with respect to t.
Step-by-step explanation:
To find the unit tangent and unit normal vectors t(t) and n(t), we need to find the velocity vector v(t) and the acceleration vector a(t) first.
Given the vector function r(t) = 9t, 4 cos t, 4 sin t, the velocity vector v(t) is obtained by taking the derivative of r(t) with respect to t.
Therefore, v(t) = 9, -4 sin t, 4 cos t.
Next, we can find the magnitude of v(t) which is also the speed of the particle:
|v(t)| = √(9² + (-4sin t)² + (4cos t)²) = √(81 + 16sin² t + 16cos² t) = √(97 + 16sin² t).
The unit tangent vector t(t) is given by:
t(t) = v(t) / |v(t)| = (9 / √(97 + 16sin² t))i + (-4sin t / √(97 + 16sin² t))j + (4cos t / √(97 + 16sin² t))k.
The unit normal vector n(t) is obtained by taking the derivative of the unit tangent vector t(t) with respect to t:
n(t) = (-36sint / (97 + 16sin² t)^(3/2))i + (9cos t / (97 + 16sin² t)^(3/2))j + (4sintcost / (97 + 16sin² t)^(3/2))k.