Given the vector-valued function $r\left(t\right)=(t^2,\frac{2}{3}{t}^3,t)$, find the unit tangent vector $T$, the unit normal vector $N$, and the unit binormal vector $B$ at the point $(4,\frac{-16}{3},-2)$. Which of the following correctly gives the unit tangent vector $T$ at that point?

• A. $(0,1,0)$
• B. $(1,0,0)$
• C. $(\frac{2}{3},\frac{-8}{9},\frac{-1}{3})$
• D. $(\frac{1}{3},\frac{-2}{3},\frac{-2}{3})$

Answer: (C)

Verified step by step guidance

1. Step 1: Compute the derivative of the vector-valued function r(t) to find the velocity vector v(t). The derivative is calculated component-wise: v(t) = (d/dt[t^2], d/dt[(2/3)t^3], d/dt[t]).
2. Step 2: Evaluate v(t) at the given point (4, -16/3, -2). To do this, determine the value of t that corresponds to the given point by solving r(t) = (4, -16/3, -2). Substitute this value of t into v(t).
3. Step 3: Normalize the velocity vector v(t) to find the unit tangent vector T. The formula for normalization is T = v(t)/||v(t)||, where ||v(t)|| is the magnitude of v(t). Compute the magnitude using ||v(t)|| = sqrt(v1^2 + v2^2 + v3^2), where v1, v2, and v3 are the components of v(t).
4. Step 4: Compute the derivative of the unit tangent vector T with respect to t to find the normal vector N. Normalize this derivative to obtain the unit normal vector N = T'(t)/||T'(t)||.
5. Step 5: Compute the cross product of the unit tangent vector T and the unit normal vector N to find the unit binormal vector B. The formula for the cross product is B = T × N.