Question

Determine the unit tangeDetermine the unit tangent vector to the following curve at t = 1 r = 2t2ˆi + (t3 − 4t) ˆj+ (5t − t2 )kˆnt vector to the following curve at t = 1 r = 2t2ˆi + (t3 − 4t) ˆj+ (5t − t2 )kˆ

Answer 1

Let's first write the question properly for a better understanding.

$\textbf{QUESTION}\\\boxed{\boxed{\begin{minipage}{15em}\sf\bold{\textsf{Determine the tangent unit vector to the following curve at t=1 :}}\\ \\ \\ \vec{\sf r}=\sf 2t^2\hat{\imath}+(t^3-4t)\hat{\jmath}+(5t-t^2)\hat{k} \end{minipage}}}$

Here, we will use the simple concept that the derivative of a function gives the tangent vector to the curve of the function at that point.

We have:

$\vec{r}(t)=2t^2\hat{\imath}+(t^3-4t)\hat{\jmath}+(5t-t^2)\hat{k}\\\\\\ \implies \vec{r'}(t)=\dfrac{d\vec{r}(t)}{dt} = 4t\hat{\imath}+(3t^2-4)\hat{\jmath}+(5-2t)\hat{k}$

By just putting t=1, we get the value of tangent vector.

Let's name this Tangent Vector as $\vec{\tau}$

$\vec{\tau} = \vec{r'}(1) = 4(1)\hat{\imath}+(3(1)^2-4)\hat{\jmath}+(5-2(1))\hat{k} \\\\\\ \implies \vec{\tau} = 4\hat{\imath}-\hat{\jmath}+3\hat{k}$

But, we need the Unit Vector.

To get a unit vector, we must divide the vector by its magnitude.

Hence:

$\hat{\tau} = \dfrac{\vec{\tau}}{|\vec{\tau}|} = \dfrac{4\hat{\imath}-\hat{\jmath}+3\hat{k}}{\sqrt{4^2+(-1)^2+3^2}} \\\\\\\\ \implies \huge\boxed{\hat{\tau} = \dfrac{4\hat{\imath}-\hat{\jmath}+3\hat{k}}{\sqrt{26}}}$

Thus, we found the unit tangent vector to the curve at the given point.

Answer 2

Answer:

Step-by-step explanation: