# Proving grad operator yields perpendicular vector to contour

• I
• henry wang
In summary, the conversation discusses the concept of a function f(x,y,z) and a constant c, and how the set of points satisfying f(x,y,z) = c forms a two-dimensional surface. The conversation also explains how the derivative of f(x,y,z) with respect to time (dg/dt) is equal to 0 when the path \vec{r}(t) stays on the surface f(x,y,z)=c. This is proven through the definition of g(t).
henry wang
I saw a link on MIT open courseware proving grad operator yields perpendicular vector to contour, but I can't make sense of how dg/dt=0.
Can someone explain to me please.

henry wang said:
I saw a link on MIT open courseware proving grad operator yields perpendicular vector to contour, but I can't make sense of how dg/dt=0.
Can someone explain to me please.

If you have a function $f(x,y,z)$, and you have a constant $c$, then the set of points satisfying $f(x,y,z) = c$ is a two-dimensional surface. So now we let $\vec{r}(t)$ be any path that stays on that surface. In terms of components, if we write $\vec{r}(t) = (x(t), y(t), z(t))$, then to say that $\vec{r}(t)$ stays on the surface $f(x,y,z)=c$ just means that for all $t$, we have:

$f(x(t), y(t), z(t)) = c$

So far, this is just true by assumption. We're assuming that $x(t), y(t), z(t)$ are three functions such that $f(x(t), y(t), z(t)) = c$.

Since $f(x(t), y(t), z(t)) = c$, then it immediately follows that:

$\frac{d}{dt} f(x(t), y(t), z(t)) = \frac{dc}{dt} = 0$

All they're doing is defining $f(x(t), y(t), z(t)) = g(t)$. That's just the definition of $g$. So it immediately follows that $\frac{d}{dt} g(t) = 0$.

henry wang
stevendaryl said:
If you have a function $f(x,y,z)$, and you have a constant $c$, then the set of points satisfying $f(x,y,z) = c$ is a two-dimensional surface. So now we let $\vec{r}(t)$ be any path that stays on that surface. In terms of components, if we write $\vec{r}(t) = (x(t), y(t), z(t))$, then to say that $\vec{r}(t)$ stays on the surface $f(x,y,z)=c$ just means that for all $t$, we have:

$f(x(t), y(t), z(t)) = c$

So far, this is just true by assumption. We're assuming that $x(t), y(t), z(t)$ are three functions such that $f(x(t), y(t), z(t)) = c$.

Since $f(x(t), y(t), z(t)) = c$, then it immediately follows that:

$\frac{d}{dt} f(x(t), y(t), z(t)) = \frac{dc}{dt} = 0$

All they're doing is defining $f(x(t), y(t), z(t)) = g(t)$. That's just the definition of $g$. So it immediately follows that $\frac{d}{dt} g(t) = 0$.
Thank you very much, that make a lot of sense, I understand now!

## 1. What is a gradient operator?

The gradient operator, denoted as ∇ (nabla), is a mathematical operator that calculates the directional derivative of a scalar-valued function in three-dimensional space. It is commonly used in vector calculus to determine the rate and direction of change of a function.

## 2. How is the gradient operator used to determine perpendicularity to a contour?

The gradient operator is used to determine perpendicularity to a contour by taking the dot product of the gradient vector and the unit tangent vector to the contour. If the dot product is zero, then the gradient vector is perpendicular to the contour, indicating that the contour is a level curve of the function.

## 3. Can the gradient operator be used for any type of function?

Yes, the gradient operator can be used for any scalar-valued function in three-dimensional space. However, it is important to note that the function must be differentiable at the point of interest for the gradient operator to yield meaningful results.

## 4. How is the perpendicular vector to a contour determined using the gradient operator?

The perpendicular vector to a contour is determined by taking the gradient of the function at the point of interest and then rotating it 90 degrees counterclockwise. This vector will be perpendicular to the contour at that point.

## 5. What is the significance of the gradient operator yielding a perpendicular vector to a contour?

The fact that the gradient operator yields a perpendicular vector to a contour is significant because it allows us to determine the direction of steepest ascent or descent for a function. This can be useful in many applications, such as optimization problems or understanding the behavior of a physical system.

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