Answer:
Since the equation was missing, I solved it with another equation and got an answer of T(0) = <3j / 5 + 4k / 5>.
Please see my explanation. I hope this helps
Step-by-step explanation:
The question asked us to find out unit tangent vector.
Recall unit vector = vector / magnitude of vector
Since the question is missing with an equation. I suppose an equation.
r(t)=Cost i, 3t j, 2Sin(2t) k at t=0
Lets take out differentiation
r'(t) = <(-Sint), 3, 2(Cos(2t)(2))>
r'(t)= <-Sint, 3, 4Cos(2t)>
Now substitute t=0 in the differentiate found above.
r'(0)= <-Sin(0), 3, 4Cos(2*0)>
r'(0)= <0, 3, 4(1)>
r'(0)= <0,3,4>
vector r'(0)=<0i, 3j, 4k>
Now lets find out magnitude of vector
|r'(0)| = [tex]\sqrt{0^{2}+3^{2}+4^{2} }[/tex]
|r'(0)| = [tex]\sqrt{0+9+16}[/tex]
|r'(0)| = [tex]\sqrt{25}[/tex]
|r'(0)| = 5
Unit Tangent Vector
T(0) = <0, 3, 4> / 5
T(0) = <3j / 5 + 4K / 5>
