Question

For a two-dimensional field described by \( T(x, y) = \frac{1}{3} xy(x + y) \), the magnitude of its gradient at the point \( (1, 1) \) is .......... (rounded off to two decimal places).

Hint:

When calculating the gradient of a two-dimensional field, remember that the gradient is the vector of partial derivatives with respect to \( x \) and \( y \), and its magnitude is the square root of the sum of the squares of these derivatives.

Answer 1

To find the gradient of the field \( T(x, y) \), we first compute the partial derivatives with respect to \( x \) and \( y \). The gradient \( \nabla T(x, y) \) is given by: \[ \nabla T(x, y) = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right) \] The partial derivatives are: \[ \frac{\partial T}{\partial x} = \frac{1}{3} y(x + y) + \frac{1}{3} xy \] \[ \frac{\partial T}{\partial y} = \frac{1}{3} x(x + y) + \frac{1}{3} xy \] Now, evaluate the gradient at the point \( (1, 1) \): \[ \frac{\partial T}{\partial x} = \frac{1}{3} \cdot 1(1 + 1) + \frac{1}{3} \cdot 1 \cdot 1 = \frac{1}{3} \cdot 2 + \frac{1}{3} = 1 \] \[ \frac{\partial T}{\partial y} = \frac{1}{3} \cdot 1(1 + 1) + \frac{1}{3} \cdot 1 \cdot 1 = \frac{1}{3} \cdot 2 + \frac{1}{3} = 1 \] The magnitude of the gradient is: \[ |\nabla T| = \sqrt{1^2 + 1^2} = \sqrt{2} \approx 1.41 \] Thus, the magnitude of the gradient is approximately 1.40 to 1.42.