Question

Find the unit tangent vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+2 t \mathbf{j}, \quad t=1 $$

Short answer

The unit tangent vector to the curve at \(t=1\) is \(\frac{1}{\sqrt{2}}(1, 1)\).

Step-by-step solution

Calculate derivative

Calculate the derivative of the vector function \(\mathbf{r}(t)\). This is done by taking the derivatives of the coefficient functions: \(r'(t) = 2t\mathbf{i} + 2\mathbf{j}\)

Normalize the derivative

This is done by computing the magnitude of \(\mathbf{r}'(t)\), using the formula for the magnitude of a vector, and then dividing \(\mathbf{r}'(t)\) by its magnitude. The magnitude of \(\mathbf{r}'(t)\) at point \(t=1\) is: \(||r'(1)|| = \sqrt{(2(1))^2 + (2)^2} = 2\sqrt{2}\)

Find the unit tangent vector

The unit tangent vector \(T(t)\) is given by: \(T(t) = \frac{\mathbf{r}'(t)}{ ||\mathbf{r}'(t)|| }\). Thus, the unit tangent vector at \(t=1\) is: \(T(1) = \frac{r'(1)}{||r'(1)||} = \frac{(2,2)}{2\sqrt{2}} = \frac{1}{\sqrt{2}}(1, 1)\)