In this exercise we need to use the concept of derivative. In each item, we will get two vectors, Thus, we will have four unit vectors in all. Therefore, let's solve this problem step by step:
a.1) First vector
We know that the derivative is the slope of the tangent line of a function at a certain point. First of all, we need to parameterize the function. To do this, just say that [tex]x=t[/tex], therefore:
[tex]\vec{v}(t)=\left[\begin{array}{c}
t\\
f(t)
\end{array}\right] = \left[\begin{array}{c}
t\\
8sin(t) \end{array}\right][/tex]
Therefore. let's name [tex]\vec{A}[/tex] to the first vector:
[tex]\vec{A}=\frac{d\vec{v}(t)}{dt}=\left[\begin{array}{c}
\frac{d}{dt}(t)\\ \\
\frac{d}{dt}(8sint)
\end{array}\right] \rightarrow \vec{A}=\left[\begin{array}{c}
1\\ 8cos(t)\end{array}\right][/tex]
By evaluating this derivative at the point at the point (π/6, 4), that is, [tex]t=\pi /6[/tex] we have:
[tex]\vec{A}=\frac{d\vec{v}(\pi /6)}{dt}=\left[\begin{array}{c}
1\\ 8cos(\pi /6)\end{array}\right]=\left[\begin{array}{c}
1\\ 4\sqrt{3}\end{array}\right][/tex]
Finally, the unit vector is found by dividing this vector by its norm:
Norm:
[tex]\left|\vec{A}\right|=\sqrt{1^2+(4\sqrt{3})^2}=7[/tex]
Therefore:
[tex]\boxed{\vec{UT}_{A}=(\frac{1}{7},\frac{4\sqrt{3}}{7})}[/tex]
a.1) Second vector
This vector can be found by multiplying the previous one by -1, therefore:
[tex]\boxed{\vec{UT}_{A'}=(-\frac{1}{7},-\frac{4\sqrt{3}}{7})}[/tex]
b.1) First vector
This vector is found by rotating one of the tangent unit vector 90 degrees. To do this, we need to interchange the components of the vector and set up one of them negative, say:
[tex](x,y) \rightarrow (-y,x)[/tex]
For this purpose, let's take the vector [tex]\vec{UT}_{A}[/tex], so the first unit normal vector is:
[tex]\boxed{\vec{UT}_{B}=\left(-\frac{4\sqrt{3}}{7},\frac{1}{7}\right)}[/tex]
b.2) Second vector
This vector is also found as in the item a.2), therefore:
[tex]\boxed{\vec{UT}_{B'}=\left(\frac{4\sqrt{3}}{7},-\frac{1}{7}\right)}[/tex]
