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Section 2.4 Directional Derivatives

Subsection 2.4.1 Motivation

We have calculated derivatives on surfaces by facing either in the direction of the x-axis or the direction of the y-axis. We could do these by treating the other axis as a constant (because on the curve it is). We have also calculated the derivative on the surface facing other directions by considering curves like \(p(t)=(t,3t)\text{.}\) In this section we develop the idea of the derivative on a surface being calculated in a direction.

Activity 35.

The goal of this activity is to connect the idea of derivative on a surface coming from derivative of a curve to the idea of direction.
Suppose Guido is again romping on the surface \(z=x^2+y^2\text{.}\)
(a)
If Guido is walking along this surface following the path \(p_1(t)=(t+2,1),\) what curve \(P_1(t)=(x,y,z)\) is he following?
(b)
Calculate the instantaneous rate of change of his altitude (\(z\)) at \(t=0\text{.}\)
(c)
If Guido is walking along this surface following the path \(p_2(t)=(2,t+1),\) what curve \(P_2(t)=(x,y,z)\) is he following?
(d)
Calculate the instantaneous rate of change of his altitude (\(z\)) at \(t=0\text{.}\)
(e)
If Guido is walking along this surface following the path \(p_3(t)=(t+2,t+1),\) what curve \(P_3(t)=(x,y,z)\) is he following?
(f)
Calculate the instantaneous rate of change of his altitude (\(z\)) at \(t=0\text{.}\)
(g)
What is the instantaneous rate of change of Guido’s altitude (\(z\)) at \((2,1)\) if he is facing \(\vec{d}_1=\langle 1,0 \rangle \text{,}\) \(\vec{d}_2=\langle 0,1 \rangle \text{,}\) and \(\vec{d}_3=\langle 1,1 \rangle \text{?}\)
Note, you should not be doing any calculations in this task.
(h)
What is the instantaneous rate of change of Guido’s altitude (\(z\)) at \((2,1)\) if he is facing \(\vec{d}_1=(1,1),\) \(\vec{d}_2=(2,2),\) and \(\vec{d}_3=(\sqrt{2},\sqrt{2})\text{?}\)
Once again, you should not be doing any calculations in this task.

Subsection 2.4.2 Notation

Definition 2.4.1. Gradient.

For a function \(f:\R^n \to \R\) the vector of all partial derivatives is called the gradient and is denoted \(\nabla f = \left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \ldots \frac{\partial f}{\partial x_n} \right).\)

Checkpoint 2.4.2.

Calculate the gradients for the following functions.
(a)
\(f(x,y)=x^2+xy+y^2.\)
(b)
\(g(x,y,z)=xyz.\)

Subsection 2.4.3 Derivation

Activity 36.

We have already extended the definition for single variable derivative to a definition for a multi-variable derivative on space curves (vector valued function). In this activity we connect the single variable definition of derivative with the concept of directional derivative on a surface.
(a)
Copy the following definition for derivatives for single variable functions.
\begin{equation*} f^\prime(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}. \end{equation*}
(b)
Review how the definition for derivative of vector valued functions is constructed in Subsection 1.4.2.
(c)
Modify the definition of derivative to work with surface functions \(f(x,y)\text{.}\) Note this is different from the vector valued function defintion (do not copy that one).
The following can be proved and is helpful for calculating directional derivatives.

Checkpoint 2.4.4.

Calculate the directional derivative of \(f(x,y)=\sin(x+y)\) in the following directions.
(a)
\(\vec{u}=\langle 1,0 \rangle\)
(b)
\(\vec{u}=\langle 1/\sqrt{2},1/\sqrt{2} \rangle\)
(c)
\(\vec{u}=\langle 1/\sqrt{5},2/\sqrt{5} \rangle\)
(d)
\(\vec{u}=\langle 2,1 \rangle\)

Subsection 2.4.4 Maximum Gradient

Activity 37.

In this activity we ask in which direction the directional derivative will be largest.
Use the function \(f(x,y)=x^2+xy+y^2\) for the following.
(a)
Calculate \(\nabla f(1,3)\text{.}\)
(b)
Calculate the directional derivative of \(f\) at \((1,3)\) in the directions \(\langle -7,5 \rangle\text{,}\) \(\langle 5,6 \rangle\text{,}\) \(\langle 5,7 \rangle\text{.}\)
(c)
On a plane, graph \(\nabla f(1,3)\) and the three directions. Label the magnitude for each directional derivative.
(d)
In which direction the directional derivative will be greatest for any function \(f(x,y)\text{?}\)
Figure 2.4.5. Comapring the Gradient to the Surface