Answer to Question #124674 in Calculus for Rebecca Kemunto

Question

Find the unit tangent vector to any point on the curve
r=(t^2+1)I+(4t-3)j+(2t^2-6t)k

Answer 1

Given $r =(1+t^2)\hat{i} + (4t-3)\hat{j} + (2t^2-6t)\hat{k}$

differentiating it with respect to t,

$r' = 2t\hat{i} + 4\hat{j} + (4t-6)\hat{k}$

unit tangent vector,

$\hat{r'} = \frac{\vec{r'}}{|r'|} = \frac{2t\hat{i} + 4\hat{j} + (4t-6)\hat{k}}{\sqrt{4t^2 + 16 + (4t-6)^2}}$ $=\frac{2t\hat{i} + 4\hat{j} + (4t-6)\hat{k}}{\sqrt{(20t^2-12t+52)}} = \frac{t\hat{i} + 2\hat{j} + (2t-3)\hat{k}}{\sqrt{(5t^2-3t+13)}}$