Respuesta :
Answer:
Therefore the unit tangent vector is
[tex]T(t)=\frac{2t \hat i + \hat j}{\sqrt{4t^2+1} }[/tex]
Therefore the parametric form is
[tex]x=16 +8t[/tex]
[tex]y=4+t[/tex]
[tex]z=\frac{16}{3}[/tex]
Step-by-step explanation:
Given [tex]r(t) = t^2 \hat i+t\hat j+\frac{16}{3} \hat k[/tex]
To find the unit tangent we need to find r'(t) and |r'(t)|.
The unit tangent vector [tex]T(t) =\frac{r'(t)}{|r'(t)|}[/tex]
[tex]r(t) = t^2 \hat i+t\hat j+\frac{16}{3} \hat k[/tex]
Differentiate with respect to t
[tex]r'(t)=2t\hat i+\hat j[/tex]
[tex]|r'(t)|=\sqrt{(2t)^2+1^2} =\sqrt{4t^2+1}[/tex]
Therefore the unit tangent vector is
[tex]T(t)=\frac{2t \hat i + \hat j}{\sqrt{4t^2+1} }[/tex] .......(1)
The parametrization equation
x(t) = t² , y(t)=t and z(t) [tex]=\frac{16}{3}[/tex]
Given point is [tex]P(16,4, \frac{16}{3} )[/tex]
Therefore,
[tex]16= t^2[/tex] , [tex]4=t[/tex] and [tex]\frac{16}{3}= \frac{16}{3}[/tex]
Then t =4.
The unit tangent vector at t= 4
[Putting t = 4 in equation (1)]
[tex]T(4)=\frac{2.4 \hat i+\hat j}{\sqrt{4.4^2+1}}[/tex]
[tex]=\frac{1}{65}(8 \hat i+ \hat j)[/tex]
The parametric equation is
x= x₁+at
y =y₁+bt
z=z₁+ct
Here a=8 , b=1 and c=0
[tex]x_1= 16[/tex] , [tex]y_1=4[/tex] and [tex]z_1= \frac{16}{3}[/tex]
Therefore the parametric form is
[tex]x=16 +8t[/tex]
[tex]y=4+1.t =4+t[/tex]
[tex]z=\frac{16}{3} + 0.t =\frac{16}{3}[/tex]
