Question

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

Hint:

Always compute partial derivatives carefully, substitute the point values, and then calculate the magnitude using the square root of the sum of squares.

Answer 1

Step 1: Write the given function.
\[ T(x,y) = \tfrac{1}{3}xy(x+y) \] Step 2: Compute gradient components.
The gradient is given by: \[ \nabla T = \left(\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}\right) \] Differentiate with respect to $x$: \[ \frac{\partial T}{\partial x} = \tfrac{1}{3} \left[ y(x+y) + xy(1) \right] = \tfrac{1}{3}( yx + y^2 + xy ) = \tfrac{1}{3}(2xy + y^2) \] Differentiate with respect to $y$: \[ \frac{\partial T}{\partial y} = \tfrac{1}{3} \left[ x(x+y) + xy(1) \right] = \tfrac{1}{3}( x^2 + xy + xy ) = \tfrac{1}{3}(x^2 + 2xy) \] Step 3: Evaluate at the point (1,1).
\[ \frac{\partial T}{\partial x}\Big|_{(1,1)} = \tfrac{1}{3}(2\cdot1\cdot1 + 1^2) = \tfrac{1}{3}(3) = 1 \] \[ \frac{\partial T}{\partial y}\Big|_{(1,1)} = \tfrac{1}{3}(1^2 + 2\cdot1\cdot1) = \tfrac{1}{3}(3) = 1 \] Step 4: Magnitude of gradient.
\[ |\nabla T| = \sqrt{1^2 + 1^2} = \sqrt{2} \approx 1.41 \] \[ \boxed{\text{Magnitude of gradient = 1.41}} \]