Answer:
Therefore the unit tangent vector is
![T(t)=\frac{2t \hat i + \hat j}{\sqrt{4t^2+1} }](https://tex.z-dn.net/?f=T%28t%29%3D%5Cfrac%7B2t%20%5Chat%20i%20%2B%20%5Chat%20j%7D%7B%5Csqrt%7B4t%5E2%2B1%7D%20%7D)
Therefore the parametric form is
![x=16 +8t](https://tex.z-dn.net/?f=x%3D16%20%2B8t)
![y=4+t](https://tex.z-dn.net/?f=y%3D4%2Bt)
![z=\frac{16}{3}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B16%7D%7B3%7D)
Step-by-step explanation:
Given ![r(t) = t^2 \hat i+t\hat j+\frac{16}{3} \hat k](https://tex.z-dn.net/?f=r%28t%29%20%3D%20t%5E2%20%5Chat%20i%2Bt%5Chat%20j%2B%5Cfrac%7B16%7D%7B3%7D%20%5Chat%20k)
To find the unit tangent we need to find r'(t) and |r'(t)|.
The unit tangent vector ![T(t) =\frac{r'(t)}{|r'(t)|}](https://tex.z-dn.net/?f=T%28t%29%20%3D%5Cfrac%7Br%27%28t%29%7D%7B%7Cr%27%28t%29%7C%7D)
![r(t) = t^2 \hat i+t\hat j+\frac{16}{3} \hat k](https://tex.z-dn.net/?f=r%28t%29%20%3D%20t%5E2%20%5Chat%20i%2Bt%5Chat%20j%2B%5Cfrac%7B16%7D%7B3%7D%20%5Chat%20k)
Differentiate with respect to t
![r'(t)=2t\hat i+\hat j](https://tex.z-dn.net/?f=r%27%28t%29%3D2t%5Chat%20i%2B%5Chat%20j)
![|r'(t)|=\sqrt{(2t)^2+1^2} =\sqrt{4t^2+1}](https://tex.z-dn.net/?f=%7Cr%27%28t%29%7C%3D%5Csqrt%7B%282t%29%5E2%2B1%5E2%7D%20%20%3D%5Csqrt%7B4t%5E2%2B1%7D)
Therefore the unit tangent vector is
.......(1)
The parametrization equation
x(t) = t² , y(t)=t and z(t) ![=\frac{16}{3}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B16%7D%7B3%7D)
Given point is ![P(16,4, \frac{16}{3} )](https://tex.z-dn.net/?f=P%2816%2C4%2C%20%5Cfrac%7B16%7D%7B3%7D%20%29)
Therefore,
,
and ![\frac{16}{3}= \frac{16}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B16%7D%7B3%7D%3D%20%5Cfrac%7B16%7D%7B3%7D)
Then t =4.
The unit tangent vector at t= 4
[Putting t = 4 in equation (1)]
![T(4)=\frac{2.4 \hat i+\hat j}{\sqrt{4.4^2+1}}](https://tex.z-dn.net/?f=T%284%29%3D%5Cfrac%7B2.4%20%5Chat%20i%2B%5Chat%20%20j%7D%7B%5Csqrt%7B4.4%5E2%2B1%7D%7D)
![=\frac{1}{65}(8 \hat i+ \hat j)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B65%7D%288%20%5Chat%20i%2B%20%5Chat%20j%29)
The parametric equation is
x= x₁+at
y =y₁+bt
z=z₁+ct
Here a=8 , b=1 and c=0
,
and ![z_1= \frac{16}{3}](https://tex.z-dn.net/?f=z_1%3D%20%5Cfrac%7B16%7D%7B3%7D)
Therefore the parametric form is
![x=16 +8t](https://tex.z-dn.net/?f=x%3D16%20%2B8t)
![y=4+1.t =4+t](https://tex.z-dn.net/?f=y%3D4%2B1.t%20%3D4%2Bt)
![z=\frac{16}{3} + 0.t =\frac{16}{3}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B16%7D%7B3%7D%20%2B%200.t%20%3D%5Cfrac%7B16%7D%7B3%7D)