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In 2D geometry where $y=f(x)$ then $f'(a)$ means slope of the tangent line at ($a$, $f(a)$). It means the angle made with the positive $X$ axis Now extending to 3D geometry let's say $z$=$f(x, y)$ so ∂z/∂x at let's say ($A$, $B$, $C$) gives us the slope of the tangent line at that point. But what's the definition of slope here? Is it the angle made with the positive $X$ axis or $Z$ axis or something totally different altogether?

Spectre
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Nigel Goveas
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  • The partial derivatives give the slope of the surface in the direction of the axis that you're taking the partial with respect to. So, if you take a surface, and intersect it with a plane parallel to the x-z plane, $\partial z/\partial x$ tells you the familiar slope of the curve. The slope is generalized using the gradient, $\nabla f=(\partial z/\partial x,\partial z/\partial y)$, which (as in 2D) points in the direction of greatest increase in a function. – George Jul 30 '17 at 01:01

1 Answers1

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For a two dimensional surface, you can define the directional derivative.

First, let me give you an intuitive idea of what that is.

Suppose you are standing on a horizontal surface. Your feet make a right angle with your legs. The slope of your feet is $0$.

Now suppose you are standing on a curved surface, such as the side of a hill. Depending upon which compass direction you are facing, your feet may have positive slope (toes up) or negative slope (toes down) or $0$ slope.

That is what is meant by a directional derivative at a point on a surface. The direction is given by a two dimensional vector rather than a compass heading. The partial derivatives of the surface function

$$ \frac{\partial}{\partial x}f(x,y),\quad\frac{\partial }{\partial y}f(x,y)$$

are the components of the gradient vector

$$\nabla f(x,y)=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$$

and the directional derivative in the direction of the unit vector $\mathbf{u}$ is given by the formula

$$ D_\mathbf{u}f=\nabla f\cdot \mathbf{u} $$

As an example, consider $f(x,y)=x^2y-x$ and we wish to find the directional derivative at the point $(2,1)$ when facing in the direction of the unit vector $\mathbf{u}=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$

$$\nabla f(2,1)=\left(2xy-1,x^2\right){\huge\vert}_{(2,1)}=(3,4)$$

So the directional derivative in the direction $\mathbf{u}$ is

$$ D_\mathbf{u}f(2,1)=(3,4)\cdot\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\frac{3+4\sqrt{3}}{2} $$

John Wayland Bales
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  • I have contact information at https://jwbales.us/precal/ but I am unable to take on tutoring responsibilities. – John Wayland Bales Jul 30 '17 at 15:22
  • Suppose I take ∂z/∂x does it mean that it's the slope of the tangent line with the surface at a particular point? – Nigel Goveas Jul 31 '17 at 06:31
  • Yes, the unit vector in the direction of the $x$-axis is $\mathbf{i}=(1,0)$, so the directional derivative $D_{\mathbf{i}}f=\nabla f\cdot\mathbf{i}=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)\cdot(1,0)=\frac{\partial f}{\partial x}$. Likewise, $\frac{\partial f}{\partial y}$ is the directional derivative in the direction of the $y$-axis. – John Wayland Bales Jul 31 '17 at 12:49
  • @Frobenius-my up vote doesn't get considered by this site since my reputation is less than 15 – Nigel Goveas Aug 25 '17 at 21:14