Question:easy

Let \( \varphi \) be a scalar function. Then, \( \nabla \varphi \) is

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Recall that the gradient of a scalar field always points normal to its level surface.
Updated On: Jul 27, 2026
  • always perpendicular to the surface of constant \( \varphi \)
  • always parallel to the surface of constant \( \varphi \)
  • the minimum rate of change of scalar \( \varphi \)
  • always zero
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Take a curve that stays on the level surface.
Let \( \mathbf{r}(t) \) be any curve lying entirely on the surface \( \varphi = \text{constant} \), so \( \varphi(\mathbf{r}(t)) = c \) for all \( t \).

Step 2: Differentiate along the curve.
By the chain rule, \( \dfrac{d}{dt}\varphi(\mathbf{r}(t)) = \nabla \varphi \cdot \mathbf{r}'(t) = 0 \), since \( \varphi \) does not change along the curve.
Here \( \mathbf{r}'(t) \) is a tangent vector to the surface at that point, and it can point along any direction within the surface, since the curve can be chosen freely.

Step 3: Read off the geometric meaning.
Since \( \nabla \varphi \) has a zero dot product with every possible tangent vector of the surface, it cannot have any component lying inside the surface.

Final Answer:
\( \nabla \varphi \) has to be entirely normal to the surface, which rules out the parallel, zero, and minimum rate options directly. \[ \nabla \varphi \perp \{\varphi = \text{constant}\} \]
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