Question

Find the unit tangent vector T(t). r(t) = t^2 i + tj + 16/3 k, p(16, 4, 16/3) T(4) = Find a set of parametric equations for the line tangent to the space curve at point P. (Enter your answers at a comma-separated list. Use t for the variable of parameterization.

Answer 1

Answer:

Therefore the unit tangent vector is

$T(t)=\frac{2t \hat i + \hat j}{\sqrt{4t^2+1} }$

Therefore the parametric form is

$x=16 +8t$

$y=4+t$

$z=\frac{16}{3}$

Step-by-step explanation:

Given $r(t) = t^2 \hat i+t\hat j+\frac{16}{3} \hat k$

To find the unit tangent we need to find r'(t) and |r'(t)|.

The unit tangent vector $T(t) =\frac{r'(t)}{|r'(t)|}$

$r(t) = t^2 \hat i+t\hat j+\frac{16}{3} \hat k$

Differentiate with respect to t

$r'(t)=2t\hat i+\hat j$

$|r'(t)|=\sqrt{(2t)^2+1^2} =\sqrt{4t^2+1}$

Therefore the unit tangent vector is

$T(t)=\frac{2t \hat i + \hat j}{\sqrt{4t^2+1} }$ .......(1)

The parametrization  equation

x(t) = t² , y(t)=t and z(t) $=\frac{16}{3}$

Given point is $P(16,4, \frac{16}{3} )$

Therefore,

$16= t^2$   ,    $4=t$   and      $\frac{16}{3}= \frac{16}{3}$

Then t =4.

The unit tangent vector at t= 4

[Putting t = 4 in equation (1)]

$T(4)=\frac{2.4 \hat i+\hat j}{\sqrt{4.4^2+1}}$

      $=\frac{1}{65}(8 \hat i+ \hat j)$

The parametric equation is

x= x₁+at

y =y₁+bt

z=z₁+ct

Here a=8 , b=1 and c=0

$x_1= 16$  ,   $y_1=4$   and   $z_1= \frac{16}{3}$

Therefore the parametric form is

$x=16 +8t$

$y=4+1.t =4+t$

$z=\frac{16}{3} + 0.t =\frac{16}{3}$