Question

Find the unit tangent vector \(\mathbf{T}(t)\) and find a set of parametric equations for the line tangent to the space curve at point \(\boldsymbol{P}\) $$ \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t \mathbf{k}, \quad P(0,0,0) $$

Short answer

Unit tangent vector at \(P(0,0,0)\) is \(\mathbf{T}(0) = \frac{\mathbf{i} + \mathbf{k}}{\sqrt{2}}\) and parametric equations for the tangent line at \(P(0,0,0)\) are \(x=t/\sqrt{2}\), \(y=0\), \(z=t/\sqrt{2}\)

Step-by-step solution

Find the derivative of \(\mathbf{r}(t)\)

First, find the derivative of \(\mathbf{r}(t)\) which is \(\frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t\mathbf{i}+t^{2} \mathbf{j} + t\mathbf{k}) = (\mathbf{i} + 2t\mathbf{j} + \mathbf{k})\). This gives us the tangent vector.

Calculate the magnitude of the derivative

Next, calculate the magnitude of \(\frac{d\mathbf{r}}{dt}\) at point \(P(0,0,0)\). So, \(\left\| \frac{d\mathbf{r}}{dt}\right\| _{t=0}=\sqrt{(1)^2+(2\cdot0)^2+(1)^2}=\sqrt{2}\)

Find the unit tangent vector \(\mathbf{T}(t)\)

To find the unit tangent vector \(\mathbf{T}(t)\), divide the derivative vector by its magnitude at point \(P\) to get \(\mathbf{T}(t) = \frac{\frac{d\mathbf{r}}{dt}}{\left\|\frac{d\mathbf{r}}{dt}\right\|} = \frac{(\mathbf{i} + 2t\mathbf{j} + \mathbf{k})}{\sqrt{2}}\) at \(t=0\) we find \(\mathbf{T}(0) = \frac{\mathbf{i} + \mathbf{k}}{\sqrt{2}}\)

Find a set of parametric equations

Finally, use the unit tangent vector to find a set of parametric equations for the line tangent to the curve at \(P(0,0,0)\). This will give \(x=0+t/\sqrt{2}\), \(y=0\) and \(z=0+t/\sqrt{2}\)