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 _______________
Show Hint
The direction of steepest increase of a scalar field is its gradient; find the magnitude of grad phi at P(3,4,1).
Step 1: Understanding the Concept:
The steepest climb of a scalar field at any point points along its gradient vector, and the size of that steepest climb equals the magnitude of the gradient there. So the question is really asking for $|\nabla\phi|$ at P$(3,4,1)$.
Step 2: Key Formula or Approach:
\[ \nabla\phi = \phi_x\,\hat{i} + \phi_y\,\hat{j} + \phi_z\,\hat{k}, \qquad |\nabla\phi| = \sqrt{\phi_x^2+\phi_y^2+\phi_z^2} \]
Step 3: Detailed Explanation.
Differentiate $\phi=x^2-yz$ term by term. Treating $y$ and $z$ as constants while differentiating with respect to $x$ gives $\phi_x=2x$. Treating $x$ and $z$ as constants while differentiating with respect to $y$ gives $\phi_y=-z$, since $-yz$ drops its $y$ and leaves behind $-z$. In the same way, $\phi_z=-y$.
Now plug in the point $P(3,4,1)$ directly:
\[ \phi_x = 2(3) = 6, \qquad \phi_y = -(1) = -1, \qquad \phi_z = -(4) = -4 \]
The gradient at P is the vector $(6,-1,-4)$. Its length is found by summing the squares of its components and taking the square root:
\[ |\nabla\phi| = \sqrt{6^2+1^2+4^2} = \sqrt{36+1+16} = \sqrt{53} \]
\[ \sqrt{53} \approx 7.28 \]
Step 4: Final Answer:
The magnitude of the fastest-increase direction vector at P is about $7.28$.
\[ \boxed{7.28} \]
