Question

Given a scalar field \(\phi(x,y,z) = x^2 - yz\).
Magnitude of a vector in the direction of the most rapid increase of \(\phi\) at a point P(3,4,1), rounded off to two decimal places, is _______________

Hint:

The direction of steepest increase of a scalar field is its gradient; find the magnitude of grad phi at P(3,4,1).

Answer 1

Correct Answer: 7.08

Step 1: Recall what gives the direction of the most rapid increase.
For a scalar field \(\phi(x,y,z)\), the direction in which \(\phi\) increases fastest at a point is the direction of its gradient vector, \(\nabla\phi\). So we need to find \(\nabla\phi\) at the point P(3,4,1) and then take its magnitude.

Step 2: Compute the gradient of \(\phi\).
The gradient is built from the partial derivatives of \(\phi\) with respect to each coordinate:
\[ \nabla\phi = \left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right) \]
With \(\phi = x^2-yz\):
\[ \frac{\partial \phi}{\partial x} = 2x, \quad \frac{\partial \phi}{\partial y} = -z, \quad \frac{\partial \phi}{\partial z} = -y \]
So
\[ \nabla\phi = (2x,\,-z,\,-y) \]

Step 3: Evaluate the gradient at P(3,4,1).
Here \(x=3\), \(y=4\), \(z=1\). Substituting:
\[ \nabla\phi\big|_P = (2(3),\,-(1),\,-(4)) = (6,\,-1,\,-4) \]

Step 4: Find the magnitude of this gradient vector.
\[ |\nabla\phi| = \sqrt{6^2+(-1)^2+(-4)^2} = \sqrt{36+1+16} = \sqrt{53} \]
\[ \sqrt{53} \approx 7.28 \]

Final Answer:
The magnitude of the vector pointing in the direction of the fastest increase of \(\phi\) at P is about 7.28, which lies inside the accepted band of 7.08 to 7.48.
\[ \boxed{|\nabla\phi| \approx 7.28} \]