Question

Find the unit tangent vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=e^{t} \cos t \mathbf{i}+e^{t} \mathbf{j}, \quad t=0 $$

Short answer

The unit tangent vector to the curve at t=0 is \( \frac{1}{\sqrt{2}} (\mathbf{i} + \mathbf{j}) \)\).

Step-by-step solution

Compute the Derivative

Derive the given vector function component by component\n\n\(\mathbf{r}'(t) = \frac {d}{dt}(e^{t} \cos t \mathbf{i}+e^{t} \mathbf{j}) = (e^{t} \cos t - e^{t} \sin t) \mathbf{i} + e^{t} \mathbf{j} \)

Evaluate the derivative at the specified value of t

Substitute t=0 into \(\mathbf{r}'(t)\) to find the tangent vector at t=0\n\n\(\mathbf{r}'(0) = (e^{0} \cos 0 - e^{0} \sin 0) \mathbf{i} + e^{0} \mathbf{j} = (1*1 - 1*0) \mathbf{i} + 1 * \mathbf{j} = \mathbf{i} + \mathbf{j}\)

Normalize the tangent vector

Normalize the tangent vector by dividing by its magnitude to obtain the unit vector, where magnitude of a vector v = \(sqrt{(v_{x})^2+ (v_{y})^2)}\)\n\nMagnitude \(\mathbf{r}'(0) = sqrt{((1)^2+(1)^2)} = sqrt{2}\)\n\nUnit tangent vector = \(\mathbf{T} = \frac{\mathbf{r}'(0)}{\|r'(0)\|} = \frac{1}{sqrt{2}} (\mathbf{i} + \mathbf{j})\)