Question

a curve is given by X=t²+1,y=4t-3,z=2t²-6t. tangent vectors to the curve at t=1 and t=2 are​

Answer 1

Answer:

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Answer 2

Answer:

tangent at t =1  is  $2\vec{i}+4\vec{j}-2\vec{k}$

and tangent at t= 2 is$4\vec{i} +4\vec{j}+2\vec{k}$

Step-by-step explanation:

Explanation :

Given , X = $t^{2}+1$

           Y = $4t - 3$

and    Z = $t^{2}- 6t$.

We know that , formula of a tangent vector is $\frac{d\vec{r}}{dt}$

Step 1:

let $\vec{r}=x\vec{i} + y\vec{j}+z\vec{k}$

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

∴ $\frac{d\vec{r}}{dt} =2t\vec{i} +4\vec{j}+ (4t-6)\vec{k}$ .............(i)

Step2:

At t = 1

put the value of t=1 in the equation (i)

we get ,

$\frac{d\vec{r}}{dt} ] = 2\vec{i}+4\vec{j}-2\vec{k}$

Step 3:

at t=2

put the value of t=2 in the equation (i)

we get,

$\frac{d\vec{r}}{dt} ] = 4\vec{i}+4\vec{j}+2\vec{k}$

Final answer:

Hence , the tangent vector to the curve at t =1 and t=2 is ($2\vec{i}+4\vec{j}-2\vec{k}$)

and ($4\vec{i}+4\vec{j}+2\vec{k}$)