For the function $f(x, y)=5 e^{-4 x} \sin (y)$, find a unit tangent vector to the level curve at the point $(1,1)$ that has a positive $x$ component. Present your answer with three decimal places of accuracy.

Step 1 ：The level curve of a function $f(x, y)$ is defined by $f(x, y) = c$ for some constant $c$. The gradient of $f$ is perpendicular to the level curve, so the tangent vector to the level curve is perpendicular to the gradient.

Step 2 ：First, we need to find the gradient of $f$. The gradient is a vector whose components are the partial derivatives of $f$ with respect to $x$ and $y$. So we need to compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

Step 3 ：The partial derivative of $f$ with respect to $x$ is $\frac{\partial f}{\partial x} = -20 e^{-4 x} \sin(y)$.

Step 4 ：The partial derivative of $f$ with respect to $y$ is $\frac{\partial f}{\partial y} = 5 e^{-4 x} \cos(y)$.

Step 5 ：So the gradient of $f$ is $\nabla f = \left(-20 e^{-4 x} \sin(y), 5 e^{-4 x} \cos(y)\right)$.

Step 6 ：We evaluate the gradient at the point $(1,1)$ to get $\nabla f(1,1) = \left(-20 e^{-4} \sin(1), 5 e^{-4} \cos(1)\right)$.

Step 7 ：The tangent vector to the level curve is perpendicular to the gradient. We can find a vector perpendicular to $\nabla f(1,1)$ by swapping the components and negating one of them. We choose to negate the second component to ensure that the $x$ component is positive.

Step 8 ：So a vector perpendicular to $\nabla f(1,1)$ is $\left(5 e^{-4} \cos(1), 20 e^{-4} \sin(1)\right)$.

Step 9 ：We want a unit vector, so we need to normalize this vector. The length of the vector is $\sqrt{(5 e^{-4} \cos(1))^2 + (20 e^{-4} \sin(1))^2}$.

Step 10 ：Dividing the vector by its length gives the unit vector $\left(\frac{5 e^{-4} \cos(1)}{\sqrt{(5 e^{-4} \cos(1))^2 + (20 e^{-4} \sin(1))^2}}, \frac{20 e^{-4} \sin(1)}{\sqrt{(5 e^{-4} \cos(1))^2 + (20 e^{-4} \sin(1))^2}}\right)$.

Step 11 ：Evaluating this expression to three decimal places gives the unit tangent vector $\boxed{(0.196, 0.981)}$.