Gradients and Directional Derivatives

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Objectives:

1.: Understand the definition of the gradient and how to compute it.
2.: Know the properties of the gradient.
3.: Understand what a directional derivative is geometrically, and also how to compute it algebraically.
4.: Understand how the formula for the directional derivative follows from the chain rule.
5.: Be able to find the tangent plane to a surface defined either implicitly or explicitly.

Recap Video

You can watch watch the following video which recaps the ideas of the section.

Test your understanding with the following questions.

The gradient is a scalarvector .

The directional derivative is a scalarvector .

If $\textbf {u}$ is a nonzero vector, $f(x,y)$ is a continuous function, and $P(a,b)$ is a point which of the following is equal to the directional derivative $D_{\textbf {u}}f(P)$ ? Select all that apply.

$\nabla f(P) \cdot \textbf {u}$ $\| \nabla f(P) \| \cos (\theta )$ , where $\theta$ is the angle between $\nabla f(P)$ and $\textbf {u}$ $\nabla f(P) \cdot \hat {\textbf {u}}$

Examples

Below is a video showing an example of computing a tangent plane.

Find the equation of the tangent plane to $xz+x\sin (y) = e^{z-2}+x^2+y-\pi$ at the point $P(1,\pi ,2)$ .

Formula and Properties Recap

Given a function $z = f(x,y)$ , the gradient of $f(x,y)$ , denoted $\nabla f$ is the vector $\nabla f(x,y) = \left \langle f_x(x,y),f_y(x,y) \right \rangle .$ (In three dimensions, just add a $z$ -component).
Given a function $z = f(x,y)$ , a point $P(x_0,y_0)$ , and a nonzero vector $\textbf {u}$ , the directional derivative of $f$ at $P$ in the direction of $\textbf {u}$ , denoted $D_{\textbf {u}}f(x_0,y_0)$ , is $D_{\textbf {u}}f(x_0,y_0) = \nabla f(x_0,y_0) \cdot \hat {\textbf {u}}.$ (Similar in 3D.)

1.: The gradient of $f$ at a point $P(a,b)$ , namely $\nabla f(a,b)$ , is perpendicular to the level curve of the surface $z = f(x,y)$ which passes through $(a,b)$ .
2.: The gradient $\nabla f(a,b)$ gives the direction of the maximum rate of increase of the function $f(x,y)$ at the point $P$ .
3.: The magnitude of the gradient is the maximum rate of increase of the function $f(x,y)$ at the point $P$ .

The tangent plane to a surface $z = f(x,y)$ at the point $(x_0,y_0,z_0)$ has normal vector $\left \langle f_x(x_0,y_0),f_y(x_0,y_0),-1 \right \rangle$ .
The tangent plane to a surface given implicitly as $F(x,y,z) = c$ at a point $P(x_0,y_0,z_0)$ , where $c$ is a number, has normal vector $\nabla F(x_0,y_0,z_0)$ .

Problems

If $f(x,y) = 2x+x^2y+y^2$ , then $\nabla f(x,y) = \left \langle \answer {2+2xy},\answer {x^2+2y} \right \rangle .$

If $f(x,y) = \sin \left (\frac {x}{y}\right ) + 2xy$ , then:

$\nabla f(x,y) = \left \langle \answer {2y+\frac {1}{y}\cos \left (\frac {x}{y}\right )},\answer {2x-\frac {x}{y^2}\cos \left (\frac {x}{y}\right )}\right \rangle$ .
$\nabla f(2\pi ,1) = \left \langle \answer {3},\answer {2\pi } \right \rangle .$

If $f(x,y) = 2x+x^2y+y^2$ and $\textbf {u} = \left \langle 3,4 \right \rangle$ , evaluate $D_{\textbf {u}} f(1,-1)$ .

The formula is $D_{\textbf {u}}f(1,-1) = \nabla f(1,-1) \cdot \hat {\textbf {u}}$ . Steps:

$\nabla f(x,y) = \left \langle \answer {2+2xy},\answer {x^2+2y}\right \rangle$
$\nabla f(1,-1) = \left \langle \answer {0},\answer {-1} \right \rangle$ .
The vector $\hat {\textbf {u}} = \frac {1}{\|\hat {\textbf {u}}\|}\textbf {u} = \left \langle \answer {\frac {3}{5}},\answer {\frac {4}{5}} \right \rangle$ .
The directional derivative is $\nabla f(1,-1) \cdot \hat {\textbf {u}} = \answer {-4/5}$

Consider the function $f(x,y,z) = x^2yz$ and the point $P(1,2,0)$ .

The gradient of $f(x,y,z)$ at $P$ is $\nabla f(P) = \left \langle \answer {0},\answer {0},\answer {2} \right \rangle$ .
The rate of change of $f(x,y,z)$ at $P$ in the direction of $Q(2,0,2)$ is $\answer {\frac {4}{3}}$ .
Find a directional derivative with $\textbf {u} = \overrightarrow {PQ}$ .
Is $f(x,y,z)$ increasing or decreasing at $P$ in the direction of $Q$ ?
Increasing Decreasing
The maximum possible value of $D_{\textbf {u}}f(P)$ (among all possible vectors $\textbf {u}$ ) is $\answer {2}$ .
Is there a vector $\textbf {u}$ such that $D_{\textbf {u}}f(P)=-3$ ?
Yes No

Consider the function $f(x,y) = xy$ , and let $P$ be the point $P(2,3)$ .

The gradient of $f(x,y)$ at $P$ is $\nabla f(2,3) = \left \langle \answer {3},\answer {2} \right \rangle$ .
The level curve of $z = f(x,y)$ which goes through the point $P$ is the one occurring at $z = \answer {6}$ .
Sketch both the level curve and the gradient vector at $P$ . The answer is in the hint.
The level curve for part (c).
A vector which is tangent to the level curve through $P$ is: $\left \langle -2,\answer {3} \right \rangle$ .
The maximum rate of increase of the function $f(x,y)$ at the point $P$ is: $\answer {\sqrt {13}}$ .
The direction of the maximum rate of increase of the function $f(x,y)$ at the point $P$ is (as a unit vector): $\left \langle \answer {\frac {3}{\sqrt {13}}},\answer {\frac {2}{\sqrt {13}}} \right \rangle$ .

You are standing on the hill whose height at any point $(x,y)$ is given by $h(x,y) = 10-x^2-2y^2$ . Your projection onto the $xy$ -plane is $(1,1)$ .

You are currently at height: $\answer {7}$ .
If you walk in the direction of the point $(2,0)$ , is your height increasing or decreasing?
Increasing Decreasing
The direction (as a unit vector) you should start walking in to descend the hill quickest is $\left \langle \answer {\frac {2}{\sqrt {20}}},\answer {\frac {4}{\sqrt {20}}} \right \rangle$ .
Is there any direction you can walk in so that you walk down the hill at a rate of $5$ ?
The maximum rate of decrease is $- \| \nabla f(1,1)\|$ .

Yes No

Find the tangent plane to the surface $z = x^2+y^2$ at the point $P(1,-1,2)$ :

There are two ways to approach this problem.

Approach 1: Let $f(x,y) = x^2+y^2$ . Then the tangent plane has normal vector $\left \langle f_x(1,-1), f_y(1,-1), -1 \right \rangle$ . Computing the partials, we get a normal vector of $\left \langle \answer {2},\answer {-2},-1 \right \rangle .$ We now have a point and a normal vector, so we can say that the tangent plane has equation (in the form $ax+by+cz=d$ ): $\answer {2}x+\answer {-2}y-z = \answer {2}.$
Approach 2: Move the $z$ over to get $x^2+y^2-z=0$ . Now treat this as an expression of the form $F(x,y,z) = c$ . In this case, $F(x,y,z) = x^2+y^2-z$ . and $c=0$ . We know the tangent plane has normal vector $\nabla F(P) = \left \langle \answer {2},\answer {-2},\answer {-1} \right \rangle .$ (Notice this is the same normal vector as in approach 1!) Therefore, the equation of the plane in the form $ax+by+cz=d$ is $\answer {2}x+\answer {-2}y-z = \answer {2}.$

Find the tangent plane to the surface given by $x^2+y^2+z^2 = 3xyz$ at the point $(1,1,1)$

Move all the terms to one side, so define $F(x,y,z) = x^2+y^2+z^2-3xyz$ . Steps:

The gradient of $F$ is $\nabla F(x,y,z) = \left \langle \answer {2x-3yz},\answer {2y-3xz},\answer {2z-3xy} \right \rangle$ .
The gradient at the point $(1,1,1)$ is $\left \langle \answer {-1},\answer {-1},\answer {-1} \right \rangle$ . This is the normal vector for the tangent plane.
The equation of the tangent plane is (in the form $ax+by+cz=d$ ): $\answer {-1}x+\answer {-1}y+\answer {-1}z = -3.$