Okay so I just recently started learning calculus with more than one variable and whilst I'm coming to grips with many of the ideas and stuff I'm finding it difficult to visualize certain things for example, what does a directional derivative look like in 3D and stuff like this.
Is there a good site or something where I can get a visualization to try and help cement my understanding?
Thank you.
On
In 3D you have the graph of a function of two variables $(x,y,z)=(x,y,f(x,y))$. Fix a point $(x_0,y_0)$ in your domain. For the sake of concreteness, suppose $(x_0, y_0)=(2,3)$ and $f(x,y)=e^{xy}$. Imagine slicing the graph by the plane $x=2$, that is, all points of the form $(x,y,z)=(2,y,z)$, where $y,z$ range through all real numbers. This plane is parallel to the $yz$-plane. Now focus your attention on the part of the graph that is in this plane $(x,y,z)=(2,y,z)$. That is, consider the points $(x,y,z)=(2,y,f(2,y))=(2,y,e^{2y})$ for all real numbers $y$. This is a curve, call it $C_1$, in the plane $x=2$ that, in particular, contains the point $(2,3,e^{6})$.
Since $x$ is constant on $C_1$, we can temporally ignore this component. We are then looking at the graph of a function of one variable, say $z=g(y)=e^{2y}$. We know that its derivative $g'(y)=2e^{2y}$ and that, in particular, $g'(3)=2e^{6}.$ This value represents the slope of the tangent line to the curve $z=g(y)=e^{2y}$ at the point $(y,z)=(3,g(3))=(3,e^{6}).$ For $y$ close to $3$, $g(y)$ is close to $g(3)+g'(3)(y-3)=e^6+2e^6(y-3).$ In 3D, this value represents the slope of the tangent line to the curve $(x,y,z)=(2,y,f(2,y))=(2,y,e^{2y})$ at the point $(x,y,z)=(2,3,f(2,3))=(2,3,e^{6})$.
Let's go back to considering all of the graph of $z=f(x,y)=e^{xy}$. If we stand on the surface formed by the graph and move in just the y direction from the given point, $(2,3,e^6)$, we are remaining in the $x=2$ plane and following the curve $C_1$ discussed above. It follows that our altitude (the $z=f(x,y)$ value) changes almost in accordance with the rule $$g(3)+g'(3)(y-3)=e^6+2e^6(y-3)=f(2,3)+2e^6(y-3)$$ when $y$ is close to $3$.
This is the meaning of the directional derivative when the direction is the direction of the positive $y$ axis. Of course, you know it as the partial derivative of $f$ with respect to $y$. However, there is nothing in our reasoning above that prevents us from doing the same for any other direction. Suppose, for example, that we want the directional derivative in the direction pointing away from the origin of the xy plane at an angle of $\theta$ radians from the positive $x$ axis. Now note that in polar coordinates a vector of length $|t|$ with direction specified by $\theta$ is given by $$\begin{align} (x,y)&=t (\cos\theta,\sin\theta) =t\left(\cos\theta,\left(\frac{\cos\theta}{\cos\theta}\right)\sin\theta\right)= t (\cos\theta,\cos\theta \tan\theta)\\ &=(t, t \tan\theta). \end{align} $$ Let's consider again the example $f(x,y)=e^{xy}$ with $(x_0,y_0)=(2,3)$, and $\theta=\pi/4$. The line in the xy plane with our specified direction and through the point $(2,3)$ is given by $(2,3)+(t, t \tan\pi/4)=(2+t, 3+t)$. This time we slice the graph of $f$ by the plane $(x(t), y(t), z(t))=(2+t, 3+t, z)$. The part of the graph of $f$ that is in this plane is the curve $C_2$ given by $(x(t), y(t), z(t))=(2+t, 3+t, f(2+t,3+t))=(2+t, 3+t, e^{(2+t)(3+t)}).$ When $t=0$, we are at the point on the graph $(2,3, e^6)$. If we vary $t$ we trace the curve $C_2$. If $t$ is close to zero, then $f(2+t,3+t,e^{(2+t)(3+t)})=z(t)$ is close to $e^6+\frac{dz}{dt}t$, where $\frac{dz}{dt}$ is the directional derivative. It is the rate at which $f$ changes as we move from $(2,3)$ along the line $x+2=y+3$.
The basic idea is that, given a function of two variables and a point in its domain, we can turn it into a function of one variable by ignoring all but one of its directions through that point. It seems that people like to teach the directional derivative as a generalisation of partial derivatives. I prefer to think the other way: partial derivatives are just special cases of the directional derivative. If we know how to differentiate functions of one variable, we can apply the same procedures to take partial derivatives. This ease of calculation makes them seem more intuitive than the general directional derivative but in fact one must understand the directional derivative in order to understand partial derivatives.
The Math Insight website at http://mathinsight.org/ is replete with diagrams and visualizations of concepts from multivariable and vector calculus.
For the directional derivative, see: http://mathinsight.org/directional_derivative_gradient_introduction.