1. considere el siguiente ejercicio:\n\ncompruebe la identidad $f_{xx} + f_{yy} = f_t$ para la…
1. considere el siguiente ejercicio:\n\ncompruebe la identidad $f_{xx} + f_{yy} = f_t$ para la función:\n$$f(x,y,t) = e^{-t}(x^2 + y^2)$$
Answer
Explanation:
Step1: Calculate the first partial derivative with respect to $x$
$$f_x = \frac{\partial}{\partial x} [e^{-t}(x^2 + y^2)] = 2xe^{-t}$$
Step2: Calculate the second partial derivative with respect to $x$
$$f_{xx} = \frac{\partial}{\partial x} [2xe^{-t}] = 2e^{-t}$$
Step3: Calculate the first partial derivative with respect to $y$
$$f_y = \frac{\partial}{\partial y} [e^{-t}(x^2 + y^2)] = 2ye^{-t}$$
Step4: Calculate the second partial derivative with respect to $y$
$$f_{yy} = \frac{\partial}{\partial y} [2ye^{-t}] = 2e^{-t}$$
Step5: Calculate the partial derivative with respect to $t$
$$f_t = \frac{\partial}{\partial t} [e^{-t}(x^2 + y^2)] = -e^{-t}(x^2 + y^2)$$
Step6: Sum the second spatial derivatives
$$f_{xx} + f_{yy} = 2e^{-t} + 2e^{-t} = 4e^{-t}$$
Step7: Compare the sum with the time derivative
$$4e^{-t} \neq -e^{-t}(x^2 + y^2)$$
Answer:
The identity $f_{xx} + f_{yy} = f_t$ does not hold for the function $f(x,y,t) = e^{-t}(x^2 + y^2)$ because $4e^{-t} \neq -e^{-t}(x^2 + y^2)$.