maths.free › Multivariable Calculus › 4. Differentiation of Functions of Several Variables › Directional Derivatives and the Gradient
Directional Derivatives and the Gradient
Determine the directional derivative in a given direction for a function of two variables.
Directional Derivatives and the Gradient
This section relies on many of the tools we have developed throughout , most importantly we will use the linearization from . The chain rule introduced in is used to define an efficient calculation of the directional derivative but is not critical for this section.
You will likely want to cover this section over several classes since these ideas are important to many other partner fields. The gradient is defined after the development of the directional derivative because the interpretations of the gradient are dependent on the directional derivative. This presentation emphasizes the geometric measurement of the directional derivative more than the algebraic form of the gradient.
Introduction
The partial derivatives of a multivariable function tell us the instantaneous rate at which the function's output changes as we hold all but one input variable constant (allowing the remaining input variable to change). It is natural to wonder how we can measure the rate at which a function changes in a direction other than parallel to a coordinate axis. In this section, we investigate this question and will connect the rates of change in other directions to the rates of change given by the partial derivatives. The Preview Activity investigates these concepts in terms of a contour map representing elevation in terms of location.
We can use cardinal directions to specify the direction of displacement vectors. These directions can be described by a compass rose. The compass rose given in is an example of a sixteen point compass rose. Directions of the form ESE are read as east-southeast and point in the direction halfway between east and southeast.
The Preview Activity demonstrates how visual information such as a contour plot or a surface plot can be helpful in determining information about how quickly the output of a function changes as its inputs change in different directions. In this section, we will focus on how to calculate this kind of directional derivative using algebraic or numerical representations of functions and relate this calculation to other ways we have described change for a function.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Directional Derivatives
We begin by determining how to measure the rate of change for a function of two variables in a direction that is not parallel to one of the coordinate directions. We use the classic calculus approach to set up this measurement: approximate the measurement, quantify how the approximation changes on a smaller scale, and finally use a limit to find the value. To measure the rate of change in the output of a function, we use a difference quotient: \[\begin{aligned}\end{aligned}\].
Let \(f\) be a function of the variables \(x\) and \(y\). We wish to measure the rate of change of \(f\) at the base point \((x_0,y_0)\) in the direction of the vector \(\langle a,b\rangle\). To do this, we write the difference quotient using the points \((x_0+a,y_0+b)\) and \((x_0,y_0\)): \[\begin{aligned}\end{aligned}\]. The denominator comes from using the distance formula to measure the change in the input as distance in the \(xy\)-plane. shows the plot of a surface given by \(z=f(x,y)\) with the points used in the difference quotient labeled.
The difference quotient completes is the first step in the classic calculus approach and provides a good start to approximating the rate of change for the output of \(f\) in the direction \(\langle a,b\rangle\). However, it is not clear how this approximation changes when we look at smaller scales. Specifically, we need to look at what happens to this approximation as we shrink the step size \(\sqrt{a^2+b^2}\) to \(0\) while maintaining the same direction of change. In order to maintain our direction and look at smaller step sizes, we must separate the length of the vector \(\langle a,b \rangle\) from its direction. To do this, we use the unit vector in the direction of \(\langle a,b \rangle\).
Let \(\vu=\langle u_1,u_2\rangle\) be the unit vector in the same direction as \(\langle a,b \rangle\). Vectors in the same direction as \(\langle a,b \rangle\) can be written in the form \(t \langle u_1,u_2\rangle\), where \(t \gt 0\) is the step size and \(\vu=\langle u_1,u_2\rangle\) is the unit vector. Taking a step of length \(t\) in the direction \(\langle u_1,u_2 \rangle\), the difference quotient can be written as \[\begin{aligned}\end{aligned}\]. We will look at how this difference quotient changes at smaller scales, i.e., when the step size \(t\) gets smaller. This accomplishes the second step of the classic calculus approach because it quantifies how the approximation works on smaller scales. We can now take the limit as \(t \to 0\) to define the directional derivative.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Efficiently Computing the Directional Derivative
We want to find a way to evaluate directional derivatives without resorting to evaluating the limit definition in Equation every time. We will look at this algebraic approach from two perspectives. First, we will use the chain rule from to define a simple algebraic rule which will allow us to efficiently calculate directional derivatives. Second, we will use the linearization of a locally linear/differentiable function to compute directional derivatives.
We are interested in the instantaneous rate of change of \(f\) at a point \((x_0,y_0)\) in the direction of a unit vector \(\vu = \langle u_1, u_2 \rangle\). In particular, the input variables \(x\) and \(y\) are changing according to \[\begin{aligned}\end{aligned}\]. A trace along the surface in the direction of \(\langle u_1,u_2\rangle\) is given by \(\vr(t)=\langle x_0+ u_1 t, y_0 + u_2 t , f(x_0+ u_1 t, y_0 + u_2 t) \rangle\). Observe that \(\frac{dx}{dt} = u_1\) and \(\frac{dy}{dt} = u_2\) for all values of \(t\). Since \(\vu\) is a unit vector in the \(xy\)-plane, a unit change in the parameter \(t\) corresponds to moving one unit in the \(\vu\) direction. This allows us to use the multivariable chain rule to calculate the directional derivative as a measure of the instantaneous rate of change of \(f\) in the direction \(\vu\).
The output of \(f\) along the given direction can be written as a composition of functions, \(f(t)=f(\vr(t))=f(x(t), y(t))\), which means we can apply the multivariable chain rule: \[\begin{aligned}D_{\vu}f(x_0,y_0) \amp = \frac{d}{dt}\left[ f(\vr(t)) \right] \\ \amp = f_x(x_0,y_0)\frac{dx}{dt} + f_y(x_0,y_0)\frac{dy}{dt} \\ \amp = f_x(x_0,y_0) u_1 + f_y(x_0,y_0) u_2\end{aligned}\] This allows us to compute the directional derivative at an arbitrary point according to the following formula.
To use equation, we must have a unit vector \(\vu = \langle u_1, u_2 \rangle\) in the direction of motion. In the event that we have a direction prescribed by a non-unit vector, we must first rescale the vector to have length 1.
In the following activity, we use algebraic techniques to calculate and interpret the of directional derivative of a function.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
The Gradient
We have used the chain rule and linearization to see see that the instantaneous rate of change a function \(f = f(x,y)\) in the direction of a unit vector \(\vu = \langle u_1, u_2 \rangle\) is given by \[\begin{aligned}\end{aligned}\]. You may recognize the form of as being similar to a dot product. We can also view this equation as stating that the rate of change along the linearization in a given direction can be expressed as a linear combination of the rates of change in the coordinate directions by using partial derivatives with the weight of each partial derivative specified by the components of the unit vector \(\vu\).
We can think about Equation in a way that will have geometric meaning related to the dot product. The directional derivative \(D_{\vu}f(x_0,y_0)\) is the dot product of \(\left\langle f_x(x_0,y_0), f_y(x_0,y_0) \right\rangle\) and \(\vu=\langle u_1,u_2\rangle\). Notice that the vector \(\left\langle f_x(x_0,y_0), f_y(x_0,y_0) \right\rangle\) comes from simple calculations that tell us how the function \(f\) is changing near the input \((x_0,y_0)\).
We read \(\nabla f\) as the gradient of \(f\), grad \(f\) or del \(f\).The symbol \(\nabla\) is called nabla, which comes from a Greek word for a certain type of harp that has a similar shape. Notice that \(\nabla f\) varies from point to point, and also provides an alternate formulation of the directional derivative.
In the following activity, we investigate some of what the gradient tells us about the behavior of a function \(f\).
Because \(\nabla f(x_0,y_0)\) is a vector, we can consider its direction and length separately. The next subsection discusses the information conveyed by these geometric aspects about the behavior of \(f\) near \((x_0,y_0)\).
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Vector Properties of the Gradient
shows how we can separate the directional derivative of a function of two variables into two separate parts: the gradient vector evaluated at the point of interest and the unit vector in the direction we want to change the inputs of the function. Recall from equation that the dot product of two vectors depends on the lengths of the vectors and the angle between the vectors. If \(\theta\) is the angle between \(\nabla f(x_0,y_0)\) and \(\vu\) (where \(\vu\) is a unit vector), then combining and gives \[\begin{aligned}D_{\vu}f(x_0,y_0) \amp= \nabla f(x_0,y_0)\cdot\vu \\ \amp= \vecmag{\nabla f(x_0,y_0)} \vecmag{\vu} \cos(\theta)\end{aligned}\]. Remember that \(\vecmag{\vu}=1\) because \(\vu\) is a unit vector. Hence, the directional derivative is the length of the gradient vector times the cosine of the angle between \(\nabla f(x_0,y_0)\) and \(\vu\): \[\begin{aligned}\end{aligned}\] Equation will be extremely useful in interpreting the gradient geometrically.
Because the magnitude of a vector is always non-negative, the sign of a directional derivative depends on \(\cos(\theta)\). graphically shows examples of the following statements (from left to right):
- \(\vu\)
- \(\vu\)
- \(\vu\)
By a parallel argument, the smallest (or most negative) value the directional derivative can take is when \(\theta=\pi\). In this case, the direction vector is in the opposite direction of the gradient vector. Thus if \(\theta=\pi\), then \[\begin{aligned}\end{aligned}\] because \(\cos(\pi)=-1\).
We summarize our most recent work by stating important facts about the gradient.
Note that the third and fifth bullets above are algebraic statements of the second and fourth bullets, respectively.
Exercises look at how the directional derivative changes as a function of the direction used or of the point where the directional derivative is being evaluated.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Applications
The gradient has many natural applications. For example, situations often arise where we are interested in knowing the direction in which a function is increasing or decreasing most rapidly. This type of question is natural when constructing a road through the mountains or planning the flow of water across a landscape. In the next activity, we examine how the gradient can help with navigation to the top of a mountain in foggy conditions.
The technique described in the previous activity has many applications related to maximizing or minimizing functions. For example, consider a two-dimensional version of how a heat-seeking missile might work. (This application is borrowed from United States Air Force Academy Department of Mathematical Sciences.) Suppose that the temperature surrounding a fighter jet can be modeled by the function \(T\) defined by \[\begin{aligned}\end{aligned}\] where \((x,y)\) is a point in the plane of the fighter jet and \(T(x,y)\) is measured in degrees Celsius. Some contours and gradients \(\nabla T\) are shown on the left in Figure.
A heat-seeking missile will always travel in the direction in which the temperature increases most rapidly; that is, it will always travel in the direction of the gradient \(\nabla T\). If a missile is fired from the point \((2,4)\), then its path will be that shown on the right in Figure.
This type of strategy is sometimes called gradient ascent. The gradient decent method is used to find relative minimums for functions by taking steps in the direction opposite the gradient. Gradient decent methods are used extensively in economics and machine learning to find where the difference between predictions from a model and data are as small as possible (minimize error).
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Directional Derivatives
We start with the graph of a surface defined by the equation \(z=f(x,y).\) Given a point \((a,b)\) in the domain of \(f,\) we choose a direction to travel from that point. We measure the direction using an angle \(\theta ,\) which is measured counterclockwise in the x, y-plane, starting at zero from the positive x-axis (). The distance we travel is \(h\) and the direction we travel is given by the unit vector \(u=(\text{cos}\ \theta )i+(\text{sin}\ \theta )j.\) Therefore, the z-coordinate of the second point on the graph is given by \(z=f(a+h\ \text{cos}\ \theta ,b+h\ \text{sin}\ \theta ).\)
We can calculate the slope of the secant line by dividing the difference in \(z\text{-values}\) by the length of the line segment connecting the two points in the domain. The length of the line segment is \(h.\) Therefore, the slope of the secant line is
\[{m}_{\text{sec}}=\frac{f(a+h\ \text{cos}\ \theta ,b+h\ \text{sin}\ \theta )-f(a,b)}{h}.\]To find the slope of the tangent line in the same direction, we take the limit as \(h\) approaches zero.
provides a formal definition of the directional derivative that can be used in many cases to calculate a directional derivative.
Another approach to calculating a directional derivative involves partial derivatives, as outlined in the following theorem.
Condensed — the full section is in OpenStax Calculus Volume 3.
Gradient
The right-hand side of is equal to \({f}_{x}(x,y)\text{cos}\ \theta +{f}_{y}(x,y)\text{sin}\ \theta ,\) which can be written as the dot product of two vectors. Define the first vector as \(∇f(x,y)={f}_{x}(x,y)\text{i}+{f}_{y}(x,y)\text{j}\) and the second vector as \(u=(\text{cos}\ \theta )i+(\text{sin}\ \theta )j.\) Then the right-hand side of the equation can be written as the dot product of these two vectors:
\[{D}_{u}f(x,y)=∇f(x,y)\cdot u.\]The first vector in has a special name: the gradient of the function \(f.\) The symbol \(∇\) is called nabla and the vector \(∇f\) is read \(\text{“del}\ f\text{.”}\)
If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. For example, if we wished to find the directional derivative of the function in in the direction of the vector \(〈-5,12〉,\) we would first divide by its magnitude to get \(u.\) This gives us \(u=〈\text{-}(5\text{/}13),12\text{/}13〉.\) Then
\[\begin{array}{ll}{D}_{u}f(x,y) & =∇f(x,y)\cdot u \\ & =-\frac{5}{13}(2x-\frac{77}{12}y)+\frac{12}{13}(\text{-}x+6y) \\ & =-\frac{22}{13}x+\frac{17}{13}y.\end{array}\]Example
Try it.
Find the gradient \(∇f(x,y)\) of each of the following functions:
- \(f(x,y)={x}^{2}-xy+3{y}^{2}\)
- \(f(x,y)=\text{sin}\ 3x\ \text{cos}\ 3y\)
Solution
For both parts a. and b., we first calculate the partial derivatives \({f}_{x}\) and \({f}_{y},\) then use .
\(\begin{array}{lll}{f}_{x}(x,y) & = & 2x-y\ \text{and}\ {f}_{y}(x,y)=\text{-}x+6y,\ \text{so} \\ ∇f(x,y) & = & {f}_{x}(x,y)i+{f}_{y}(x,y)j \\ & = & (2x-y)i+(\text{-}x+6y)j.\end{array}\)
\(\begin{array}{lll}{f}_{x}(x,y) & = & 3\ \text{cos}\ 3x\ \text{cos}\ 3y\ \text{and}\ {f}_{y}(x,y)=-3\ \text{sin}\ 3x\ \text{sin}\ 3y,\ \text{so} \\ ∇f(x,y) & = & {f}_{x}(x,y)i+{f}_{y}(x,y)j \\ & = & (3\ \text{cos}\ 3x\ \text{cos}\ 3y)i-(3\ \text{sin}\ 3x\ \text{sin}\ 3y)j.\end{array}\)
Condensed — the full section is in OpenStax Calculus Volume 3.
Gradients and Level Curves
Recall that if a curve is defined parametrically by the function pair \((x(t),y(t)),\) then the vector \({x}^{'}(t)i+{y}^{'}(t)j\) is tangent to the curve for every value of \(t\) in the domain. Now let’s assume \(z=f(x,y)\) is a differentiable function of \(x\ \text{and}\ y,\) and \(({x}_{0},{y}_{0})\) is in its domain. Let’s suppose further that \({x}_{0}=x({t}_{0})\) and \({y}_{0}=y({t}_{0})\) for some value of \(t,\) and consider the level curve \(f(x,y)=k.\) Define \(g(t)=f(x(t),y(t))\) and calculate \({g}^{'}(t)\) on the level curve. By the chain Rule,
\[{g}^{'}(t)={f}_{x}(x(t),y(t)){x}^{'}(t)+{f}_{y}(x(t),y(t)){y}^{'}(t).\]But \({g}^{'}(t)=0\) because \(g(t)=k\) for all \(t.\) Therefore, on the one hand,
\[{f}_{x}(x(t),y(t)){x}^{'}(t)+{f}_{y}(x(t),y(t)){y}^{'}(t)=0;\]on the other hand,
\[{f}_{x}(x(t),y(t)){x}^{'}(t)+{f}_{y}(x(t),y(t)){y}^{'}(t)=∇f(x,y)\cdot 〈{x}^{'}(t),{y}^{'}(t)〉\text{.}\]Therefore,
\[∇f(x,y)\cdot 〈{x}^{'}(t),{y}^{'}(t)〉=0.\]Thus, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector is tangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the following theorem.
We can use this theorem to find tangent and normal vectors to level curves of a function.
Example
Try it.
For the function \(f(x,y)=2{x}^{2}-3xy+8{y}^{2}+2x-4y+4,\) find a tangent vector to the level curve at point \((-2,1).\) Graph the level curve corresponding to \(f(x,y)=18\) and draw in \(∇f(-2,1)\) and a tangent vector.
Solution
First, we must calculate \(∇f(x,y)\text{:}\)
\[{f}_{x}(x,y)=4x-3y+2\ \text{and}\ {f}_{y}=-3x+16y-4\ \text{so}\ ∇f(x,y)=(4x-3y+2)i+(-3x+16y-4)j.\]Next, we evaluate \(∇f(x,y)\) at \((-2,1)\text{:}\)
\[∇f(-2,1)=(4(-2)-3(1)+2)i+(-3(-2)+16(1)-4)j=-9i+18j.\]This vector is orthogonal to the curve at point \((-2,1).\) We can obtain a tangent vector by reversing the components and multiplying either one by \(-1.\) Thus, for example, \(-18i-9j\) is a tangent vector (see the following graph).
Three-Dimensional Gradients and Directional Derivatives
The definition of a gradient can be extended to functions of more than two variables.
Calculating the gradient of a function in three variables is very similar to calculating the gradient of a function in two variables. First, we calculate the partial derivatives \({f}_{x},{f}_{y},\) and \({f}_{z},\) and then we use .
Example
Try it.
Find the gradient \(∇f(x,y,z)\) of each of the following functions:
- \(f(x,y,z)=5{x}^{2}-2xy+{y}^{2}-4yz+{z}^{2}+3xz\)
- \(f(x,y,z)={e}^{-2z}\text{sin}\ 2x\ \text{cos}\ 2y\)
Solution
For both parts a. and b., we first calculate the partial derivatives \({f}_{x},{f}_{y},\) and \({f}_{z},\) then use .
\(\begin{array}{lll}{f}_{x}(x,y,z) & = & 10x-2y+3z,\ {f}_{y}(x,y,z)=-2x+2y-4z\ \text{and}\ {f}_{z}(x,y,z)=3x-4y+2z,\ \text{so} \\ ∇f(x,y,z) & = & {f}_{x}(x,y,z)i+{f}_{y}(x,y,z)j+{f}_{z}(x,y,z)k \\ & = & (10x-2y+3z)i+(-2x+2y-4z)j+(3x-4y+2z)k.\end{array}\)
\(\begin{array}{lll}{f}_{x}(x,y,z) & = & 2{e}^{-2z}\text{cos}\ 2x\ \text{cos}\ 2y,\ {f}_{y}(x,y,z)=-2{e}^{-2z}\text{sin}\ 2x\ \text{sin}\ 2y\ \text{and} \\ {f}_{z}(x,y,z) & = & -2{e}^{-2z}\text{sin}\ 2x\ \text{cos}\ 2y,\ \text{so} \\ ∇f(x,y,z) & = & {f}_{x}(x,y,z)i+{f}_{y}(x,y,z)j+{f}_{z}(x,y,z)k \\ & = & (2{e}^{-2z}\text{cos}\ 2x\ \text{cos}\ 2y)i+(-2{e}^{-2z}\sin 2x\sin 2y)j+(-2{e}^{-2z}\sin 2x\cos 2y)k \\ & = & 2{e}^{-2z}(\text{cos}\ 2x\ \text{cos}\ 2y\ \text{i}-\text{sin}\ 2x\ \text{sin}\ 2y\ \text{j}-\text{sin}\ 2x\ \text{cos}\ 2y\ \text{k}).\end{array}\)
Condensed — the full section is in OpenStax Calculus Volume 3.
Key Concepts
- A directional derivative represents a rate of change of a function in any given direction.
- The gradient can be used in a formula to calculate the directional derivative.
- The gradient indicates the direction of greatest change of a function of more than one variable.
Key Equations
| directional derivative (two dimensions) | \({D}_{u}f(a,b)=\underset{h\to 0}{\text{lim}}\frac{f(a+h\ \text{cos}\ \theta ,b+h\ \text{sin}\ \theta )-f(a,b)}{h}\) or \({D}_{u}f(x,y)={f}_{x}(x,y)\text{cos}\ \theta +{f}_{y}(x,y)\text{sin}\ \theta\) |
| gradient (two dimensions) | \(∇f(x,y)={f}_{x}(x,y)i+{f}_{y}(x,y)j\) |
| gradient (three dimensions) | \(∇f(x,y,z)={f}_{x}(x,y,z)i+{f}_{y}(x,y,z)j+{f}_{z}(x,y,z)k\) |
| directional derivative (three dimensions) | \(\begin{array}{ll}{D}_{u}f(x,y,z) & =∇f(x,y,z)\cdot u \\ & ={f}_{x}(x,y,z)\text{cos}\ \alpha +{f}_{y}(x,y,z)\text{cos}\ \beta +{f}_{x}(x,y,z)\text{cos}\ \gamma \end{array}\) |
Directional Derivatives and the Gradient
For the following exercises, find the directional derivative using the limit definition only.
For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(u\) or \(v\) as appropriate.
For the following exercises, find the directional derivative of the function in the direction of the unit vector \(u=\text{cos}\ \theta i+\text{sin}\ \theta j.\)
For the following exercises, find the gradient.
For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(Q.\)
For the following exercises, find the derivative of the function at \(P\) in the direction of \(u.\)
For the following exercises, find the gradient vector at the indicated point.
- the tangent plane and
- the normal line to the given surface at the given point.
Condensed — the full section is in OpenStax Calculus Volume 3.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Find all directions in which the directional derivative of \(f(x,y) = ye^{-xy}\) is 1 at the point \((0,2)\).
Révèle la réponse
We have \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\], so \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. For a unit vector \(\vu = \langle a, b \rangle\), \[\begin{aligned}\end{aligned}\]. We want \(D_{\vu} f(0, 2) = 1\), so we want a unit direction vector \(\langle a, b \rangle\) with \(-4a + b = 1\). Note that since \(\langle a, b \rangle\) is a unit vector we have \(1 = \sqrt{a^2 + b^2}\). Solving these equations simultaneously yields \(a=0, b = 1\) or \(a = -\frac{8}{17}, b = -\frac{15}{17}\). Thus any positive scalar multiple of \(\langle 0, 1 \rangle\) or \(\langle -\frac{8}{17}, -\frac{15}{17} \rangle\) will give a direction in which the directional derivative is 1.
-
Find, if possible, a function \(f\) such that \[\begin{aligned}\end{aligned}\]. If not possible, explain why.
Révèle la réponse
Recall \[\begin{aligned}\end{aligned}\]. If \(f_x (x, y, z) = \sin(yz)\), we must have \[\begin{aligned}\end{aligned}\], where \(k_1(y, z)\) is a function of \(y\) and \(z\) only. If \(f_y(x, y, z) = \sin(yz) + 2y\), we must have \[\begin{aligned}\end{aligned}\], where \(k_2(x, z)\) is a function of \(x\) and \(z\) only. Finally, if \(f_z (x, y, z) = xy \cos(yz) + \frac{5}{z}\), we must have \[\begin{aligned}\end{aligned}\], where \(k_3(x, y)\) is a function of \(x\) and \(y\) only. A function that satisfies all three of these possibilities is \[\begin{aligned}\end{aligned}\], where \(c\) is a constant (and \(x \neq 0\)).
-
Let \(\theta =\text{arccos}(3\text{/}5).\) Find the directional derivative \({D}_{u}f(x,y)\) of \(f(x,y)={x}^{2}-xy+3{y}^{2}\) in the direction of \(\text{u}=(\text{cos}\ \theta )i+(\text{sin}\ \theta )j.\) What is \({D}_{u}f(-1,2)?\)
Révèle la réponse
First of all, since \(\text{cos}\ \theta =3\text{/}5\) and \(\theta\) is acute, this implies
\[\text{sin}\ \theta =\sqrt{1-{(\frac{3}{5})}^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}.\]Using \(f(x,y)={x}^{2}-xy+3{y}^{2},\) we first calculate \(f(x+h\ \text{cos}\ \theta ,y+h\ \text{sin}\ \theta )\text{:}\)
\[\begin{array}{ll}f(x+h\ \text{cos}\ \theta ,y+h\ \text{sin}\ \theta ) & ={(x+h\ \text{cos}\ \theta )}^{2}-(x+h\ \text{cos}\ \theta )(y+h\ \text{sin}\ \theta )+3{(y+h\ \text{sin}\ \theta )}^{2} \\ & ={x}^{2}+2xh\ \text{cos}\ \theta +{h}^{2}{\text{cos}}^{2}\theta -xy-xh\ \text{sin}\ \theta -yh\ \text{cos}\ \theta \\ & \ {\text{-h}}^{2}\text{sin}\ \theta \ \text{cos}\ \theta +3{y}^{2}+6yh\ \text{sin}\ \theta +3{h}^{2}{\text{sin}}^{2}\theta \\ & ={x}^{2}+2xh(\frac{3}{5})+\frac{9{h}^{2}}{25}-xy-\frac{4xh}{5}-\frac{3yh}{5}-\frac{12{h}^{2}}{25}+3{y}^{2} \\ & \ +6yh(\frac{4}{5})+3{h}^{2}(\frac{16}{25}) \\ & ={x}^{2}-xy+3{y}^{2}+\frac{2xh}{5}+\frac{9{h}^{2}}{5}+\frac{21yh}{5}.\end{array}\]We substitute this expression into :
\[\begin{array}{ll}{D}_{u}f(a,b) & =\underset{h\to 0}{\text{lim}}\frac{f(a+h\ \text{cos}\ \theta ,b+h\ \text{sin}\ \theta )-f(a,b)}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{({x}^{2}-xy+3{y}^{2}+\frac{2xh}{5}+\frac{9{h}^{2}}{5}+\frac{21yh}{5})-({x}^{2}-xy+3{y}^{2})}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{\frac{2xh}{5}+\frac{9{h}^{2}}{5}+\frac{21yh}{5}}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{2x}{5}+\frac{9h}{5}+\frac{21y}{5} \\ & =\frac{2x+21y}{5}.\end{array}\]To calculate \({D}_{u}f(-1,2),\) we substitute \(x=-1\) and \(y=2\) into this answer:
\[\begin{array}{ll}{D}_{u}f(-1,2) & =\frac{2(-1)+21(2)}{5} \\ & =\frac{-2+42}{5} \\ & =8.\end{array}\](See the following figure.)
-
Let \(\theta =\text{arccos}(3\text{/}5).\) Find the directional derivative \({D}_{u}f(x,y)\) of \(f(x,y)={x}^{2}-xy+3{y}^{2}\) in the direction of \(u=(\text{cos}\ \theta )i+(\text{sin}\ \theta )j.\) What is \({D}_{u}f(-1,2)?\)
Révèle la réponse
First, we must calculate the partial derivatives of \(f\text{:}\)
\[\begin{array}{l}{f}_{x}=2x-y \\ {f}_{y}=\text{-}x+6y,\end{array}\]Then we use with \(\theta =\text{arccos}(3\text{/}5)\text{:}\)
\[\begin{array}{ll}{D}_{u}f(x,y) & ={f}_{x}(x,y)\text{cos}\ \theta +{f}_{y}(x,y)\text{sin}\ \theta \\ & =(2x-y)\frac{3}{5}+(\text{-}x+6y)\frac{4}{5} \\ & =\frac{6x}{5}-\frac{3y}{5}-\frac{4x}{5}+\frac{24y}{5} \\ & =\frac{2x+21y}{5}.\end{array}\]To calculate \({D}_{u}f(-1,2),\) let \(x=-1\) and \(y=2\text{:}\)
\[{D}_{u}f(-1,2)=\frac{2(-1)+21(2)}{5}=\frac{-2+42}{5}=8.\]This is the same answer obtained in .
-
Find the directional derivative \({D}_{u}f(x,y)\) of \(f(x,y)=3{x}^{2}y-4x{y}^{3}+3{y}^{2}-4x\) in the direction of \(u=(\text{cos}\ \frac{\pi }{3})i+(\text{sin}\ \frac{\pi }{3})j\) using . What is \({D}_{u}f(3,4)?\)
Révèle la réponse
\(\begin{array}{l} \\ \\ {D}_{u}f(x,y)=\frac{(6xy-4{y}^{3}-4)(1)}{2}+\frac{(3{x}^{2}-12x{y}^{2}+6y)\sqrt{3}}{2} \\ {D}_{u}f(3,4)=\frac{72-256-4}{2}+\frac{(27-576+24)\sqrt{3}}{2}=-94-\frac{525\sqrt{3}}{2}\end{array}\)
-
Find the gradient \(∇f(x,y)\) of each of the following functions:
- \(f(x,y)={x}^{2}-xy+3{y}^{2}\)
- \(f(x,y)=\text{sin}\ 3x\ \text{cos}\ 3y\)
Révèle la réponse
For both parts a. and b., we first calculate the partial derivatives \({f}_{x}\) and \({f}_{y},\) then use .
\(\begin{array}{lll}{f}_{x}(x,y) & = & 2x-y\ \text{and}\ {f}_{y}(x,y)=\text{-}x+6y,\ \text{so} \\ ∇f(x,y) & = & {f}_{x}(x,y)i+{f}_{y}(x,y)j \\ & = & (2x-y)i+(\text{-}x+6y)j.\end{array}\)
\(\begin{array}{lll}{f}_{x}(x,y) & = & 3\ \text{cos}\ 3x\ \text{cos}\ 3y\ \text{and}\ {f}_{y}(x,y)=-3\ \text{sin}\ 3x\ \text{sin}\ 3y,\ \text{so} \\ ∇f(x,y) & = & {f}_{x}(x,y)i+{f}_{y}(x,y)j \\ & = & (3\ \text{cos}\ 3x\ \text{cos}\ 3y)i-(3\ \text{sin}\ 3x\ \text{sin}\ 3y)j.\end{array}\)
-
Find the gradient \(∇f(x,y)\) of \(f(x,y)=({x}^{2}-3{y}^{2})\text{/}(2x+y).\)
Révèle la réponse
\(∇f(x,y)=\frac{2{x}^{2}+2xy+6{y}^{2}}{{(2x+y)}^{2}}i-\frac{{x}^{2}+12xy+3{y}^{2}}{{(2x+y)}^{2}}j\)
-
Find the direction for which the directional derivative of \(f(x,y)=3{x}^{2}-4xy+2{y}^{2}\) at \((-2,3)\) is a maximum. What is the maximum value?
Révèle la réponse
The maximum value of the directional derivative occurs when \(∇f\) and the unit vector point in the same direction. Therefore, we start by calculating \(∇f(x,y)\text{:}\)
\[\begin{array}{lll}{f}_{x}(x,y) & = & 6x-4y\ \text{and}\ {f}_{y}(x,y)=-4x+4y,\ \text{so} \\ ∇f(x,y) & = & {f}_{x}(x,y)i+{f}_{y}(x,y)j=(6x-4y)i+(-4x+4y)j.\end{array}\]Next, we evaluate the gradient at \((-2,3)\text{:}\)
\[∇f(-2,3)=(6(-2)-4(3))i+(-4(-2)+4(3))j=-24i+20j.\]We need to find a unit vector that points in the same direction as \(∇f(-2,3),\) so the next step is to divide \(∇f(-2,3)\) by its magnitude, which is \(\sqrt{{(-24)}^{2}+{(20)}^{2}}=\sqrt{976}=4\sqrt{61}.\) Therefore,
\[\frac{∇f(-2,3)}{‖∇f(-2,3)‖}=\frac{-24}{4\sqrt{61}}i+\frac{20}{4\sqrt{61}}j=\frac{-6\sqrt{61}}{61}i+\frac{5\sqrt{61}}{61}j.\]This is the unit vector that points in the same direction as \(∇f(-2,3).\) To find the angle corresponding to this unit vector, we solve the equations
\[\text{cos}\ \theta =\frac{-6\sqrt{61}}{61}\ \text{and}\ \text{sin}\ \theta =\frac{5\sqrt{61}}{61}\]for \(\theta .\) Since cosine is negative and sine is positive, the angle must be in the second quadrant. Therefore, \(\theta =\pi -\text{arcsin}((5\sqrt{61})\text{/}61)\approx 2.45\ \text{rad.}\)
The maximum value of the directional derivative at \((-2,3)\) is \(‖∇f(-2,3)‖=4\sqrt{61}\) (see the following figure).
-
Find the direction for which the directional derivative of \(g(x,y)=4x-xy+2{y}^{2}\) at \((-2,3)\) is a maximum. What is the maximum value?
Révèle la réponse
The gradient of \(g\) at \((-2,3)\) is \(∇g(-2,3)=i+14j.\) The unit vector that points in the same direction as \(∇g(-2,3)\) is \(\frac{∇g(-2,3)}{‖∇g(-2,3)‖}=\frac{1}{\sqrt{197}}i+\frac{14}{\sqrt{197}}j=\frac{\sqrt{197}}{197}i+\frac{14\sqrt{197}}{197}j,\) which gives an angle of \(\theta =\text{arcsin}((14\sqrt{197})\text{/}197)\approx 1.499\ \text{rad}.\) The maximum value of the directional derivative is \(‖∇g(-2,3)‖=\sqrt{197}.\)
-
For the function \(f(x,y)=2{x}^{2}-3xy+8{y}^{2}+2x-4y+4,\) find a tangent vector to the level curve at point \((-2,1).\) Graph the level curve corresponding to \(f(x,y)=18\) and draw in \(∇f(-2,1)\) and a tangent vector.
Révèle la réponse
First, we must calculate \(∇f(x,y)\text{:}\)
\[{f}_{x}(x,y)=4x-3y+2\ \text{and}\ {f}_{y}=-3x+16y-4\ \text{so}\ ∇f(x,y)=(4x-3y+2)i+(-3x+16y-4)j.\]Next, we evaluate \(∇f(x,y)\) at \((-2,1)\text{:}\)
\[∇f(-2,1)=(4(-2)-3(1)+2)i+(-3(-2)+16(1)-4)j=-9i+18j.\]This vector is orthogonal to the curve at point \((-2,1).\) We can obtain a tangent vector by reversing the components and multiplying either one by \(-1.\) Thus, for example, \(-18i-9j\) is a tangent vector (see the following graph).
-
For the function \(f(x,y)={x}^{2}-2xy+5{y}^{2}+3x-2y+4,\) find the tangent to the level curve at point \((1,1).\) Draw the graph of the level curve corresponding to \(f(x,y)=9\) and draw \(∇f(1,1)\) and a tangent vector.
Révèle la réponse
\(∇f(x,y)=(2x-2y+3)i+(-2x+10y-2)j\)
\(∇f(1,1)=3i+6j\)
Tangent vector: \(6i-3j\) or \(-6i+3j\)
-
Find the gradient \(∇f(x,y,z)\) of each of the following functions:
- \(f(x,y,z)=5{x}^{2}-2xy+{y}^{2}-4yz+{z}^{2}+3xz\)
- \(f(x,y,z)={e}^{-2z}\text{sin}\ 2x\ \text{cos}\ 2y\)
Révèle la réponse
For both parts a. and b., we first calculate the partial derivatives \({f}_{x},{f}_{y},\) and \({f}_{z},\) then use .
\(\begin{array}{lll}{f}_{x}(x,y,z) & = & 10x-2y+3z,\ {f}_{y}(x,y,z)=-2x+2y-4z\ \text{and}\ {f}_{z}(x,y,z)=3x-4y+2z,\ \text{so} \\ ∇f(x,y,z) & = & {f}_{x}(x,y,z)i+{f}_{y}(x,y,z)j+{f}_{z}(x,y,z)k \\ & = & (10x-2y+3z)i+(-2x+2y-4z)j+(3x-4y+2z)k.\end{array}\)
\(\begin{array}{lll}{f}_{x}(x,y,z) & = & 2{e}^{-2z}\text{cos}\ 2x\ \text{cos}\ 2y,\ {f}_{y}(x,y,z)=-2{e}^{-2z}\text{sin}\ 2x\ \text{sin}\ 2y\ \text{and} \\ {f}_{z}(x,y,z) & = & -2{e}^{-2z}\text{sin}\ 2x\ \text{cos}\ 2y,\ \text{so} \\ ∇f(x,y,z) & = & {f}_{x}(x,y,z)i+{f}_{y}(x,y,z)j+{f}_{z}(x,y,z)k \\ & = & (2{e}^{-2z}\text{cos}\ 2x\ \text{cos}\ 2y)i+(-2{e}^{-2z}\sin 2x\sin 2y)j+(-2{e}^{-2z}\sin 2x\cos 2y)k \\ & = & 2{e}^{-2z}(\text{cos}\ 2x\ \text{cos}\ 2y\ \text{i}-\text{sin}\ 2x\ \text{sin}\ 2y\ \text{j}-\text{sin}\ 2x\ \text{cos}\ 2y\ \text{k}).\end{array}\)
-
Find the gradient \(∇f(x,y,z)\) of \(f(x,y,z)=\frac{{x}^{2}-3{y}^{2}+{z}^{2}}{2x+y-4z}.\)
Révèle la réponse
\(\begin{array}{ll}∇f(x,y,z) & =\frac{2{x}^{2}+2xy+6{y}^{2}-8xz-2{z}^{2}}{{(2x+y-4z)}^{2}}i-\frac{{x}^{2}+12xy+3{y}^{2}-24yz+{z}^{2}}{{(2x+y-4z)}^{2}}j \\ & +\frac{4{x}^{2}-12{y}^{2}-4{z}^{2}+4xz+2yz}{{(2x+y-4z)}^{2}}k.\end{array}\)
-
Calculate \({D}_{u}f(1,-2,3)\) in the direction of \(\text{v}=\text{-}i+2j+2k\) for the function
\[f(x,y,z)=5{x}^{2}-2xy+{y}^{2}-4yz+{z}^{2}+3xz.\]Révèle la réponse
First, we find the magnitude of \(v\text{:}\)
\[‖v‖=\sqrt{{(-1)}^{2}+{(2)}^{2}+{(2)}^{2}}=3.\]Therefore, \(\frac{v}{‖v‖}=\frac{\text{-}i+2j+2k}{3}=-\frac{1}{3}i+\frac{2}{3}j+\frac{2}{3}k\) is a unit vector in the direction of \(v,\) so \(\text{cos}\ \alpha =-\frac{1}{3},\text{cos}\ \beta =\frac{2}{3},\ \text{and}\ \text{cos}\ \gamma =\frac{2}{3}.\) Next, we calculate the partial derivatives of \(f\text{:}\)
\[\begin{array}{lll}{f}_{x}(x,y,z) & = & 10x-2y+3z \\ {f}_{y}(x,y,z) & = & -2x+2y-4z \\ {f}_{z}(x,y,z) & = & -4y+2z+3x,\end{array}\]then substitute them into :
\[\begin{array}{ll}{D}_{u}f(x,y,z) & ={f}_{x}(x,y,z)\text{cos}\ \alpha +{f}_{y}(x,y,z)\text{cos}\ \beta +{f}_{z}(x,y,z)\text{cos}\ \gamma \\ & =(10x-2y+3z)(-\frac{1}{3})+(-2x+2y-4z)(\frac{2}{3})+(-4y+2z+3x)(\frac{2}{3}) \\ & =-\frac{10x}{3}+\frac{2y}{3}-\frac{3z}{3}-\frac{4x}{3}+\frac{4y}{3}-\frac{8z}{3}-\frac{8y}{3}+\frac{4z}{3}+\frac{6x}{3} \\ & =-\frac{8x}{3}-\frac{2y}{3}-\frac{7z}{3}.\end{array}\]Last, to find \({D}_{u}f(1,-2,3),\) we substitute \(x=1,y=-2,\ \text{and}\ z=3\text{:}\)
\[\begin{array}{ll}{D}_{u}f(1,-2,3) & =-\frac{8(1)}{3}-\frac{2(-2)}{3}-\frac{7(3)}{3} \\ & =-\frac{8}{3}+\frac{4}{3}-\frac{21}{3} \\ & =-\frac{25}{3}.\end{array}\] -
Calculate \({D}_{u}f(x,y,z)\) and \({D}_{u}f(0,-2,5)\) in the direction of \(\text{v}=-3i+12j-4k\) for the function \(f(x,y,z)=3{x}^{2}+xy-2{y}^{2}+4yz-{z}^{2}+2xz.\)
Révèle la réponse
\(\begin{array}{lll}{D}_{u}f(x,y,z) & = & -\frac{3}{13}(6x+y+2z)+\frac{12}{13}(x-4y+4z)-\frac{4}{13}(2x+4y-2z) \\ {D}_{u}f(0,-2,5) & = & \frac{384}{13}\end{array}\)
-
\(f(x,y)=5-2{x}^{2}-\frac{1}{2}{y}^{2}\) at point \(P(3,4)\) in the direction of \(\text{u}=(\text{cos}\ \frac{\pi }{4})i+(\text{sin}\ \frac{\pi }{4})j\)
-
\(f(x,y)={y}^{2}\text{cos}(2x)\) at point \(P(\frac{\pi }{3},2)\) in the direction of \(\text{u}=(\text{cos}\ \frac{\pi }{4})i+(\text{sin}\ \frac{\pi }{4})j\)
Révèle la réponse
\(-2\sqrt{6}-\sqrt{2}\)
-
Find the directional derivative of \(f(x,y)={y}^{2}\text{sin}(2x)\) at point \(P(\frac{\pi }{4},2)\) in the direction of \(u=5i+12j.\)
-
\(f(x,y)=xy,\) \(P(0,-2),\) \(v=\frac{1}{2}i+\frac{\sqrt{3}}{2}j\)
Révèle la réponse
\(-1\)
-
\(h(x,y)={e}^{x}\text{sin}\ y,P(1,\frac{\pi }{2}),v=\text{-}i\)
-
\(h(x,y,z)=xyz,P(2,1,1),v=2i+j-k\)
Révèle la réponse
\(\frac{2}{\sqrt{6}}\)
-
\(f(x,y)=xy,P(1,1),u=〈\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}〉\)
-
\(f(x,y)={x}^{2}-{y}^{2},\begin{array}{ll}u=〈\frac{\sqrt{3}}{2},\frac{1}{2}〉, & P(1,0)\end{array}\)
Révèle la réponse
\(\sqrt{3}\)
-
\(f(x,y)=3x+4y+7,\begin{array}{ll}u=〈\frac{3}{5},\frac{4}{5}〉, & P(0,\frac{\pi }{2})\end{array}\)
-
\(\begin{array}{lll}f(x,y)={e}^{x}\text{cos}\ y, & u=〈0,1〉, & P=(0,\frac{\pi }{2})\end{array}\)
Révèle la réponse
\(-1.0\)
-
\(\begin{array}{lll}f(x,y)={y}^{10}, & u=〈0,-1〉, & P=(1,-1)\end{array}\)
-
\(f(x,y)=\text{ln}({x}^{2}+{y}^{2}),\begin{array}{ll}u=〈\frac{3}{5},\frac{4}{5}〉, & P(1,2)\end{array}\)
Révèle la réponse
\(\frac{22}{25}\)
-
\(f(x,y)={x}^{2}y,\begin{array}{ll}P(-5,5), & v=3i-4j\end{array}\)
-
\(f(x,y,z)={y}^{2}+xz,\begin{array}{ll}P(1,2,2), & v=〈2,-1,2〉\end{array}\)
Révèle la réponse
\(\frac{2}{3}\)
-
\(f(x,y)={x}^{2}+2{y}^{2},\theta =\frac{\pi }{6}\)
-
\(f(x,y)=\frac{y}{x+2y},\theta =-\frac{\pi }{4}\)
Révèle la réponse
\(\frac{\text{-}\sqrt{2}(x+y)}{2{(x+2y)}^{2}}\)
-
\(f(x,y)=\text{cos}(3x+y),\theta =\frac{\pi }{4}\)
-
\(w(x,y)=y{e}^{x},\theta =\frac{\pi }{3}\)
Révèle la réponse
\(\frac{{e}^{x}(y+\sqrt{3})}{2}\)
-
\(\begin{array}{ll}f(x,y)=x\ \text{arctan}(y), & \theta =\frac{\pi }{2}\end{array}\)
-
\(\begin{array}{ll}f(x,y)=\text{ln}(x+2y), & \theta =\frac{\pi }{3}\end{array}\)
Révèle la réponse
\(\frac{1+2\sqrt{3}}{2(x+2y)}\)
-
Find the gradient of \(f(x,y)=\frac{14-{x}^{2}-{y}^{2}}{3}.\) Then, find the gradient at point \(P(1,2).\)
-
Find the gradient of \(f(x,y,z)=xy+yz+xz\) at point \(P(1,2,3).\)
Révèle la réponse
\(〈5,4,3〉\)
-
Find the gradient of \(f(x,y,z)\) at \(P\) and the directional derivative in the direction of \(u\text{:}\) \(f(x,y,z)=\text{ln}({x}^{2}+2{y}^{2}+3{z}^{2}),\begin{array}{ll}P(2,1,4), & u=\frac{-3}{13}i-\frac{4}{13}j-\frac{12}{13}k\end{array}.\)
-
\(f(x,y,z)=4{x}^{5}{y}^{2}{z}^{3},\begin{array}{ll}P(2,-1,1), & u=\frac{1}{3}i+\frac{2}{3}j-\frac{2}{3}k\end{array}\)
Révèle la réponse
\(-320\)
-
\(f(x,y)={x}^{2}+3{y}^{2},\begin{array}{ll}P(1,1), & Q(4,5)\end{array}\)
Symbols used here
Vector of partial derivatives; points uphill.
The non-negative number whose square (n-th power) is x.
Ratio of a circle's circumference to its diameter, 3.14159…
The usual name for an angle.
Ratios of sides in a right triangle; coordinates on the unit circle.
Chance of A; chance of A given that B happened.
Instantaneous rate of change; slope of the graph.
2.71828…, the base whose exponential is its own derivative.
i² = −1.
Inequalities that allow equality; < and > exclude it.
Derivative with respect to x, holding the other variables fixed.
Antiderivative (indefinite) or signed area from a to b (definite).
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Directional Derivatives and the Gradient
- Determine the directional derivative in a given direction for a function of two variables.
- Determine the gradient vector of a given real-valued function.
- Explain the significance of the gradient vector with regard to direction of change along a surface.
- Use the gradient to find the tangent to a level curve of a given function.
- Calculate directional derivatives and gradients in three dimensions.
- If
- If
- If
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Essayez votre propre
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
Plus en Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems