For the function $f(x,y)=3e^{-4x}\sin (y)$, find a unit tangent vector to the level curve at the point $(-5,2)$ that has a positive $x$ component. Present your answer with three decimal places of accuracy.
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Step 1 ：The level curve of a function $f(x,y)$ is defined by the equation $f(x,y)=c$ for some constant $c$. The gradient of $f$ at a point $(x,y)$ is perpendicular to the level curve at that point. Therefore, the tangent to the level curve at $(x,y)$ is perpendicular to the gradient of $f$ at $(x,y)$.

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

Step 3 ：The partial derivative of $f$ with respect to $x$ is $\frac{\partial f}{\partial x}=-12e^{-4x}\sin (y)$. Evaluating this at the point $(-5,2)$ gives $\frac{\partial f}{\partial x}(-5,2)=-12e^{20}\sin (2)$.

Step 4 ：The partial derivative of $f$ with respect to $y$ is $\frac{\partial f}{\partial y}=3e^{-4x}\cos (y)$. Evaluating this at the point $(-5,2)$ gives $\frac{\partial f}{\partial y}(-5,2)=3e^{20}\cos (2)$.

Step 5 ：Therefore, the gradient of $f$ at the point $(-5,2)$ is $\mathrm{\nabla }f(-5,2)=(-12e^{20}\sin (2),3e^{20}\cos (2))$.

Step 6 ：The tangent vector to the level curve at $(-5,2)$ is perpendicular to the gradient vector. Therefore, we can find the tangent vector by rotating the gradient vector by 90 degrees. This can be done by swapping the components of the gradient vector and changing the sign of one of them. We want the tangent vector to have a positive $x$ component, so we change the sign of the $y$ component. This gives the tangent vector $(3e^{20}\cos (2),12e^{20}\sin (2))$.

Step 7 ：Finally, we need to normalize the tangent vector to make it a unit vector. The length of the vector is $\sqrt{(3e^{20}\cos (2){)}^2+(12e^{20}\sin (2){)}^2}=15e^{20}$. Therefore, the unit tangent vector is $(\frac{3e^{20}\cos (2)}{15e^{20}},\frac{12e^{20}\sin (2)}{15e^{20}})=(0.2\cos (2),0.8\sin (2))$.

Step 8 ：Rounding to three decimal places, the unit tangent vector is $(0.176,0.713)$.