Final answer:
To find a unit tangent vector to the level curve of f(x, y) at (-5,4) with a positive x component, compute the gradient of f, evaluate at the point, find a perpendicular vector with a positive x component, normalize it, and round the components to four decimal places. The unit tangent vector T is approximately <0.8897, -0.4563>.
Step-by-step explanation:
To find a unit tangent vector to the level curve of the function f(x, y) = -4x² + 4xy + 3y² + 5x - 5y - 3 at the point (-5,4) with a positive x component, we first need to find the gradient of f, which is given by the partial derivatives, namely ∇f = . Calculating the derivatives we get fx = -8x + 4y + 5 and fy = 4x + 6y - 5. Evaluating at (-5,4) gives us ∇f(-5,4) = <20,39>. This gradient vector points perpendicularly to the level curve.
The tangent vector should be perpendicular to the gradient, so we swap the components and change one of their signs. To ensure a positive x component, we take the gradient vector <20,39> and turn it into the tangent vector <39,-20>. To normalize this vector to a unit vector, we divide by its magnitude, which is calculated as √(39² + (-20)²).
The magnitude of our tangent vector is √(1521+400) = √1921, which equals 43.8406 (rounded to four decimal places). So, the unit tangent vector T is <39/43.8406, -20/43.8406> rounded to four decimal places gives us T ≈ <0.8897, -0.4563>.