Find parametric equations for the tangent line

[Yae Miteo]

Homework Statement

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

Homework Equations

$$x = 1+2 \sqrt{t}, \quad y = t^3 - t, \quad z = t^3 + t, \quad (3, 0, 2)$$

The Attempt at a Solution

I began by re-writing this as $$\vec{r}(t) = <1+2 \sqrt{t}, \quad t^3 - t, \quad t^3 + t >$$
and then taking the derivative to find the normal vector: $$\vec{r}(t) = <1/ \sqrt{t}, \quad 3t^2 - 1, \quad 3t^2 + 1>$$. From here, I tried plugging in (3, 0, 2) into each of the components for the derivative, but that hasn't worked. I think I need to find a value to use for "t" and then put that into the normal vector, but I am not sure how.

 

[LCKurtz]

Homework Statement

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

Homework Equations

$$x = 1+2 \sqrt{t}, \quad y = t^3 - t, \quad z = t^3 + t, \quad (3, 0, 2)$$

The Attempt at a Solution

I began by re-writing this as $$\vec{r}(t) = <1+2 \sqrt{t}, \quad t^3 - t, \quad t^3 + t >$$
and then taking the derivative to find the normal vector: $$\vec{r}(t) = <1/ \sqrt{t}, \quad 3t^2 - 1, \quad 3t^2 + 1>$$. From here, I tried plugging in (3, 0, 2) into each of the components for the derivative, but that hasn't worked. I think I need to find a value to use for "t" and then put that into the normal vector, but I am not sure how.

Part of your problem may be you are calling both the original equation and its derivative $r(t)$. Find the $t$ that makes the original equation pass through $(3,0,2)$.

 

[Ray Vickson]

Homework Statement

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

Homework Equations

$$x = 1+2 \sqrt{t}, \quad y = t^3 - t, \quad z = t^3 + t, \quad (3, 0, 2)$$

The Attempt at a Solution

I began by re-writing this as $$\vec{r}(t) = <1+2 \sqrt{t}, \quad t^3 - t, \quad t^3 + t >$$
and then taking the derivative to find the normal vector: $$\vec{r}(t) = <1/ \sqrt{t}, \quad 3t^2 - 1, \quad 3t^2 + 1>$$. From here, I tried plugging in (3, 0, 2) into each of the components for the derivative, but that hasn't worked. I think I need to find a value to use for "t" and then put that into the normal vector, but I am not sure how.

Please distinguish between $\vec{r}(t)$ and $\vec{r}'(t)$, which is your second one. Anyway $\vec{r}'(t)$ is NOT a "normal vector"! Can you describe what it actually IS?

 

[Yae Miteo]

I made a mistake. I meant to say tangent vector.