Final answer:
The rate of change of temperature as Captain Ralph proceeds in the direction of velocity e^9 is determined by the gradient of the temperature function T and is equal to -12e^3 degrees per second.
Step-by-step explanation:
To determine how fast the temperature decreases as Captain Ralph proceeds in a particular direction, we need to calculate the rate of change of the temperature function T (x, y, z) = e(−x²− 2y²− 3z²)
At the point (1, 1, 1), we would compute the partial derivatives of the temperature function concerning x, y, and z, and evaluate them at this point.
However, since Captain Ralph is moving in the specific direction of the velocity v = e9 meters per second, which we can assume corresponds to the vector (1, 1, 1) to calculate the rate of change, we can use the gradient of T and the dot product to find how fast the temperature changes.
The gradient of T is given by ∇T = (−2x*e(−x²−2y²−3z²), −4y*e(−x²−2y²−3z²), −6z*e(−x²−2y²−3z²)).
Evaluating the gradient at (1, 1, 1) gives us ∇T(1, 1, 1) = (−2e−6, −4e−6, −6e−6).
To find the rate of change in the direction of Captain Ralph's velocity, we take the dot product of ∇T with the velocity vector. Since the problem does not specify the direction, we can simplify the calculation by noting that v is a scalar multiple of the position vector (1, 1, 1).
Thus, the rate of change is the scalar product e9 * (magnitude of ∇T at (1, 1, 1)) * cos(θ), where θ is the angle between the gradient vector and the velocity.
Since the vectors have the same direction, θ = 0, and the cosine term is 1.
So, the rate of change of temperature is simply e9 * −2e−6 + −4e−6 + −6e−6,
which simplifies to −12e3 degrees per second (since e9 × e−6 = e3).