Answer :
Answer:
a.
[tex]T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))[/tex]
b.
[tex]T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)[/tex]
[tex]N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)[/tex]
Step-by-step explanation:
Remember that for any curve [tex]r(t)[/tex]
The tangent vector is given by
[tex]T(t) = \frac{r'(t) }{| r'(t)| }[/tex]
And the normal vector is given by
[tex]N(t) = \frac{T'(t)}{|T'(t)|}[/tex]
a.
For this case, using the chain rule
[tex]r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\[/tex]
And also remember that
[tex]|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t[/tex]
Therefore
[tex]T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )[/tex]
Similarly, using the quotient rule and the chain rule
[tex]T'(t) = ( -2t cos(t^2) , -2t sin(t^2))[/tex]
And also
[tex]|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t[/tex]
Therefore
[tex]N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))[/tex]
Notice that
1. [tex]|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1[/tex]
2. [tex]N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0[/tex]
b.
Simlarly
[tex]r'(t) = (2t,-6,0) \\[/tex]
and
[tex]|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}[/tex]
Therefore
[tex]T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)[/tex]
Then
[tex]T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)[/tex]
and also
[tex]|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )[/tex]
And since
[tex]N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)[/tex]