Final answer:
The maximum rate of change of the function teˢᵗ at the point (0,2) is found by computing the gradient and its magnitude, resulting in √17. This is accomplished by evaluating the partial derivatives at that point.
Step-by-step explanation:
To find the maximum rate of change of the function f(s,t) = tes at the point (0,2), we need to compute the gradient of the function and evaluate its magnitude at that point. The gradient (∇f) is a vector that points in the direction of the steepest ascent of a function. We find the partial derivatives with respect to both variables s and t:
- ∇f/ ∇s = t2es
- ∇f/ ∇t = (1 + st)est
Evaluating these partial derivatives at the point (0,2) gives:
- ∇f/ ∇s |_0,2) = (2)2e(0)×(2) = 4
- ∇f/ ∇t |_0,2) = (1 + (0)×(2))e(0)×(2) = 1
The gradient at (0,2) is then ∇f = <4, 1>. The magnitude of the gradient gives us the maximum rate of change at that point, which is √(42 + 12) = √17.