For the function $f(x,y)=5e^{-4x}\sin (y)$, 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 $\frac{\partial f}{\partial x}=-20e^{-4x}\sin (y)$.

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

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

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

Step 7 ：The tangent vector to the level curve is perpendicular to the gradient. We can find a vector perpendicular to $\mathrm{\nabla }f(1,1)$ 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 $\mathrm{\nabla }f(1,1)$ is $(5e^{-4}\cos (1),20e^{-4}\sin (1))$.

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

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

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