# Vector grad as a normal vector

i have done a little on multivariable calculus in school -partial derivates and tangent planes to f(x,y), but now we have moved on to functions of more than 2 variables

my teacher, however, doesnt really teach us - she gives us equations and tells us to do questions and i usually find myself having to read the textbook to convince myself that i understand the equations. shes only part time so she is never about to help me at lunch time, so if you guys could help me that would be great

my textbook wasnt able to convince me why the vector grad is $$\nabla _{g}$$ is normal to the surface - surely if it has the same direction as the surface in x, y and z then it cant be going away from the surface at an angle of pi/2 rad? :S

i know that:

$$\nabla _{g}$$ = < $$\frac{\partial{g}}{\partial{x}}$$ , $$\frac{\partial{g}}{\partial{y}}$$ , $$\frac{\partial{g}}{\partial{z}}$$ >

but its not clear to me why this should be perpendicular to the surface. would you be able to help justify why it is? i dont know anything about vector spaces and stuff like that, because a level further maths is pretty limited.

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arildno
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g(x,y,z) is no surface, but g(x,y,z)=CONSTANT is a surface!

And, the gradient of g at a point on this surface (called a level surface), indicates the direction along which g increases most swiftly.
Clearly, the direction that g increases most swiftly must be orthogonal to the direction(s) along which g retains its value (i.e, along the level surface g=CONSTANT).

Thus, the gradient of g is orthogonal to the surface g=constant.

g(x,y,z) is no surface, but g(x,y,z)=CONSTANT is a surface!

And, the gradient of g at a point on this surface (called a level surface), indicates the direction along which g increases most swiftly.
Clearly, the direction that g increases most swiftly must be orthogonal to the direction(s) along which g retains its value (i.e, along the level surface g=CONSTANT).

Thus, the gradient of g is orthogonal to the surface g=constant.

oh yeah sorry i meant that it was = constant.

im not sure i understand why it represents the direction along which the gradient increases most swiftly.. surely it just represents a vector which is travelling in the same direction as the surface in the x y z directions, if it has the same slope in all 3 directions?

arildno
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surely it just represents a vector which is travelling in the same direction as the surface in the x y z directions, if it has the same slope in all 3 directions?
Nope.
At every point (x,y,z) (incluing, therefore, every point ON a particular surface), the gradient tells us the direction in which g(x,y,z) increases most swiftly.

This is easily seen as follows.
Let h(t) be the function:
$$h(t)=g(\vec{x}_{0}+t*\vec{n}), \vec{x}_{0}=(x_{0},y_{0},z_{0}), \vec{n}=(n_{x},n_{y},n_{z}),||\vec{n}||=1$$
where $\vec{x}_{0}$ is some arbitrary fixed point, and $\vec{n}$ is some unit vector in some fixed direction.
h(t) is therefore an evaluation of g along a straight line going through $\vec{x}_{0}$.

Now, we have, by the chain rule (evaluated at t=0):
$$\frac{dh}{dt}_{t=0}=\nabla{g}_{\vec{x}_{0}}\cdot\vec{n}=||\nabla{g}_{\vec{x}_{0}}||\cos\theta_{n}$$
where the angle is between the line we are following and the direction of the gradient at our point.
This expression is maximized when $\vec{n}$ is parallell to the gradient of g.

If you, at some point walks along a level surface, the value of g won't change at all.
Thus, the directions along the level surface cannot coincide with the direction along which g increases most strongly; they have to be orthogonal to the gradient direction.

lavinia
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df is zero on tangent vectors to the surface because f is constant on the surface. The directional derivative of f along the surface is zero. If you want to express this with an inner product with a vector then this vector is by definition perpendicular to the surface. If you use the standard metric on Euclidean space, this vector is the gradient.

LCKurtz
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I'll throw in 2 cents worth too. Your surface is g(x,y,z) = c. Suppose you have a smooth curve on the surface parameterized as

$$\vec R(t) = \langle x(t),y(t),z(t)\rangle$$

Then g(x(t),y(t),z(t)) ≡ 0 as a function of t. Therefore its derivative with respect to t must be identically 0. We have, by the chain rule,

$$g'(t)=g_x\frac{dx}{dt}+ g_y\frac{dy}{dt}+g_z\frac{dz}{dt}$$
$$= \langle g_x,g_y,g_z\rangle \cdot \langle \frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\rangle=\nabla g \cdot \frac{d\vec R}{dt} = 0$$

This tells you that the gradient of g is perpendicular to R'(t). But R'(t) is tangent to the curve R(t), which lies on the surface. Think of all the curves on the surface passing through a particular point P on the surface. The gradient at P is perpendicular to all these curves. The only way that could happen is if it is perpendicular to the surface itself.