Final answer:
The rate of change in T at the point (0, 0) in the direction of (-1, -2) is -1. The direction in which T increases most rapidly from point (0, 1) is (-1/2, -1), and the rate of change in T from point (0, 1) in this direction is √(5)/2.
Step-by-step explanation:
To determine the rate of change in T at the point (0, 0) in the direction of (-1, -2), we can use the gradient vector. The gradient vector is given by (partial derivative of T with respect to x, partial derivative of T with respect to y).
Taking the partial derivatives of T(x, y) = ln(eˣ + y²), we get:
a. Tx(x, y) = 1 / (eˣ + y²)
Ty(x, y) = 2y / (eˣ + y²)
Substituting the values for x and y in the gradient vector:
Gradient vector = (Tx(0, 0), Ty(0, 0)) = (1 / 1, 0) = (1, 0)
The rate of change in T at the point (0, 0) in the direction of (-1, -2) is given by the dot product of the gradient vector and the direction vector:
Rate of change = Gradient vector dot Direction vector
Rate of change = (1, 0) dot (-1, -2) = 1 × (-1) + 0 × (-2) = -1
Therefore, the rate of change in T at the point (0, 0) in the direction of (-1, -2) is -1.
b. To determine the direction in which T increases most rapidly from point (0, 1), we can again use the gradient vector. Evaluating the gradient vector at the point (0, 1):
Gradient vector = (Tx(0, 1), Ty(0, 1))
Gradient vector = (1 / (e⁰ + 1²), 2(1) / (e⁰ + 1²))
Gradient vector = (1/2, 1)
The direction opposite to the gradient vector is the direction in which T increases most rapidly. So the direction is (-1/2, -1). The magnitude (length) of the gradient vector represents the rate of change.
Therefore, the rate of change in T from point (0, 1) in the direction (-1/2, -1) is the magnitude of the gradient vector:
Rate of change = |Gradient vector| = |(1/2, 1)| = √((1/2)² + 1²) = √(1/4 + 1) = √(5/4) = √(5)/2
Therefore, the rate of change in T from point (0, 1) in the direction (-1/2, -1) is √(5)/2.