17Calculus - Gradients and Directional Derivatives

Using Vectors

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Partial Integrals

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The directional derivative and gradient are very closely related. So we cover both on this page. First, a couple of quick definitions.

Directional Derivative - - You know from single variable calculus that the derivative is just the slope of the tangent line. If you are given a point, you can find the slope of the specific tangent line at that specific point. Similarly, the directional derivative is the slope of a tangent line but it applies to three dimensions where you are given a point and a unit vector. This unit vector tells you which specific tangent line you are talking about at that point. In summary, a directional derivative is the slope of a tangent line in a specific direction.

Gradient - - The gradient is a vector (which is why its also called the gradient vector) calculated using the partial derivative. It contains information about the slope at every point. We can use the gradient to get the directional derivative.

Remember that in multi-variable calculus most concepts use special notation. The directional derivative and gradient are no exception.

Here is a great video explaining the directional derivative and gradient in detail and the associated notation, including some examples. It is a long video, so we separate it into video clips on this page as the discussion continues.

Dr Chris Tisdell - Gradient and Directional Derivative (Part 1) [19mins-21secs]

video by Dr Chris Tisdell

$$\displaystyle{ \nabla g(x,y) = \frac{\partial g}{\partial x} \hat{i} + \frac{\partial g}{ \partial y}\hat{j} }$$ where $$\displaystyle{ \nabla }$$ is the 'del' operator $$\displaystyle{ \nabla = \frac{\partial }{\partial x}\hat{i} + \frac{\partial }{ \partial y}\hat{j} }$$

The gradient can also be indicated by grad(g) and can be called the grad of g. Notice that we do not use vector notation on $$\nabla g(x,y)$$. We could write $$\vec{ \nabla } g(x,y)$$ to clearly indicate that the result is a vector and some instructors do this. However, many of them don't and we won't either. So you just need to remember that $$\nabla g(x,y)$$ is a vector (or, more precisely, a vector field).

Functions of More Than Two Variables

You can easily extend the definition of the gradient and directional derivative to functions of 3 or more variables. However, you lose the physical interpretation of slope. For three dimensions, the gradient looks like you would expect.

 $$\displaystyle{ \nabla h(x,y,z) = \frac{\partial h}{\partial x} \hat{i} + \frac{\partial h}{ \partial y}\hat{j} + \frac{\partial h}{ \partial z}\hat{k} }$$

Before we go on, let's make sure we understand this by working a few practice problems.

Unless otherwise instructed, find the gradient of these functions.

$$f(x,y) = x\sin y + \cos y$$

Problem Statement

Find the gradient of $$f(x,y) = x\sin y + \cos y$$

$$\nabla f(x,y) = (\sin y) \vhat{i} + (x\cos y - \sin y)\vhat{j}$$

Problem Statement

Find the gradient of $$f(x,y) = x\sin y + \cos y$$

Solution

 $$\displaystyle{ \nabla f(x,y) = \frac{\partial f}{\partial x} \vhat{i} + \frac{\partial f}{\partial y} \vhat{j} }$$ $$= (\sin y + 0) \vhat{i} + (x\cos y - \sin y)\vhat{j}$$ $$= (\sin y) \vhat{i} + (x\cos y - \sin y) \vhat{j}$$

$$\nabla f(x,y) = (\sin y) \vhat{i} + (x\cos y - \sin y)\vhat{j}$$

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If $$f(x,y,z) = x^3+y+z^2$$, calculate $$\nabla f$$ and $$\nabla f(1,2,3)$$.

Problem Statement

If $$f(x,y,z) = x^3+y+z^2$$, calculate $$\nabla f$$ and $$\nabla f(1,2,3)$$.

Solution

Dr Chris Tisdell - 804 video solution

video by Dr Chris Tisdell

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Calculate $$\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }$$ at $$(\tau, \kappa) = (1,5)$$

Problem Statement

Calculate $$\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }$$ at $$(\tau, \kappa) = (1,5)$$

Hint

This problem uses unusual variables $$\tau$$ and $$\kappa$$. However, they work just as if you had t and x.

Problem Statement

Calculate $$\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }$$ at $$(\tau, \kappa) = (1,5)$$

Hint

This problem uses unusual variables $$\tau$$ and $$\kappa$$. However, they work just as if you had t and x.

Solution

World Wide Center of Mathematics - 2293 video solution

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Calculate the gradient vector field of $$f(x,y) = \ln(2x+5y)$$.

Problem Statement

Calculate the gradient vector field of $$f(x,y) = \ln(2x+5y)$$.

Solution

MIP4U - 2294 video solution

video by MIP4U

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Given $$f(x,y) = xy$$, determine $$\nabla f(-3,2)$$.

Problem Statement

Given $$f(x,y) = xy$$, determine $$\nabla f(-3,2)$$.

Solution

MIP4U - 2295 video solution

video by MIP4U

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$$f(x,y,z) = z e^{-xy}$$

Problem Statement

Find the gradient of $$f(x,y,z) = z e^{-xy}$$

$$\nabla f(x,y,z) = -yze^{-xy} \vhat{i} - xz e^{-xy} \vhat{j} + e^{-xy} \vhat{k}$$

Problem Statement

Find the gradient of $$f(x,y,z) = z e^{-xy}$$

Solution

 $$\displaystyle{ \nabla f(x,y,z) = \frac{\partial f}{\partial x} \vhat{i} + \frac{\partial f}{\partial y} \vhat{j} + \frac{\partial f}{\partial z} \vhat{k} }$$ $$\displaystyle{ z(-y)e^{-xy} \vhat{i} + z (-x) e^{-xy} \vhat{j} + e^{-xy} \vhat{k} }$$

$$\nabla f(x,y,z) = -yze^{-xy} \vhat{i} - xz e^{-xy} \vhat{j} + e^{-xy} \vhat{k}$$

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Directional Derivative

$$\displaystyle{ D_{\vec{u}}f = \nabla f \cdot \vec{u} }$$
We read this as the directional derivative of f in the direction of $$\vec{u}$$.

The directional derivative can be thought of as the magnitude of the gradient at a certain point in a specific direction. To calculate the directional derivative, we take the dot product of the gradient with the unit vector ( $$\vec{u}$$ ) in the direction we are interested in. The equation looks like this.

Here is a video going through a proof showing that the directional derivative is the dot product of the gradient with a direction vector. This video is not required for understanding how to use the dot product version of the directional derivative but it will help you understand it. So it is a good video to watch.

Dr Chris Tisdell - Proof: Directional Derivative is a Dot Product [8mins-6secs]

video by Dr Chris Tisdell

Here is a video clip discussing the properties of the directional derivative based on the use of the dot product.

Dr Chris Tisdell - Gradient and Directional Derivative (Part 2) [13mins-23secs]

video by Dr Chris Tisdell

From the last video clip, you now know that the direction of maximum increase of a function f is the gradient $$\nabla f$$ and the direction of maximum decrease is $$-\nabla f$$. ( Remember that these are vectors. )

Since the gradient is just a partial derivative, the basic rules of algebra are followed as well as the chain rule. If you would like to confirm these rules, here is a quick video clip to watch.

Dr Chris Tisdell - Gradient and Directional Derivative (Part 3) [2mins-25secs]

video by Dr Chris Tisdell

Practice - Directional Derivative

Unless otherwise instructed, find the directional derivative of these functions.

Calculate the derivative of $$f(x,y) = xe^y + \cos(xy)$$ at the point $$(2,0)$$ in the direction of $$\vec{v} = 3\hat{i} - 4\hat{j}$$.

Problem Statement

Calculate the derivative of $$f(x,y) = xe^y + \cos(xy)$$ at the point $$(2,0)$$ in the direction of $$\vec{v} = 3\hat{i} - 4\hat{j}$$.

Solution

Dr Chris Tisdell - 799 video solution

video by Dr Chris Tisdell

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At the point $$(1,1)$$, determine the directions in which $$f(x,y) = x^2/2 + y^2/2$$ (a) increases most rapidly; (b) decreases most rapidly; (c) has zero change

Problem Statement

At the point $$(1,1)$$, determine the directions in which $$f(x,y) = x^2/2 + y^2/2$$ (a) increases most rapidly; (b) decreases most rapidly; (c) has zero change

Solution

Dr Chris Tisdell - 800 video solution

video by Dr Chris Tisdell

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Calculate the derivative of $$f(x,y,z) = x^3 - xy^2-z$$ at the point $$(1,1,0)$$ in the direction $$\vec{v} = 2\hat{i} - 3\hat{j} + 6\hat{k}$$.

Problem Statement

Calculate the derivative of $$f(x,y,z) = x^3 - xy^2-z$$ at the point $$(1,1,0)$$ in the direction $$\vec{v} = 2\hat{i} - 3\hat{j} + 6\hat{k}$$.

Solution

Dr Chris Tisdell - 801 video solution

video by Dr Chris Tisdell

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Calculate the directional derivative of $$f(x,y) = 1 - x^2/2 - y^4/4$$ in the direction $$\hat{i} + \hat{j}$$ at the point $$(1,2)$$.

Problem Statement

Calculate the directional derivative of $$f(x,y) = 1 - x^2/2 - y^4/4$$ in the direction $$\hat{i} + \hat{j}$$ at the point $$(1,2)$$.

Solution

Dr Chris Tisdell - 803 video solution

video by Dr Chris Tisdell

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Calculate the directional derivative of $$f(x,y) = x^2 + 2xy + 3y^2$$ at the point $$(2,1)$$ in the direction $$\hat{i}+\hat{j}$$.

Problem Statement

Calculate the directional derivative of $$f(x,y) = x^2 + 2xy + 3y^2$$ at the point $$(2,1)$$ in the direction $$\hat{i}+\hat{j}$$.

Solution

Krista King Math - 807 video solution

video by Krista King Math

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Calculate the maximum directional derivative and its direction for $$f(x,y) = 2x^2 + 3xy + 4y^2$$ at $$(1,1)$$.

Problem Statement

Calculate the maximum directional derivative and its direction for $$f(x,y) = 2x^2 + 3xy + 4y^2$$ at $$(1,1)$$.

Solution

Krista King Math - 808 video solution

video by Krista King Math

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Calculate the directional derivative of $$f(x,y) = x^2y^3 - y^4$$ at $$(2,1)$$ in the direction given by the angle $$\theta = \pi/4$$.

Problem Statement

Calculate the directional derivative of $$f(x,y) = x^2y^3 - y^4$$ at $$(2,1)$$ in the direction given by the angle $$\theta = \pi/4$$.

Solution

PatrickJMT - 810 video solution

video by PatrickJMT

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Calculate the value of the maximum rate of change of the function $$f(x,y) = \sin(xy)$$ at the point $$(1,0)$$.

Problem Statement

Calculate the value of the maximum rate of change of the function $$f(x,y) = \sin(xy)$$ at the point $$(1,0)$$.

Solution

Krista King Math - 1522 video solution

video by Krista King Math

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Calculate the direction of the maximum rate of change of the scalar field $$\phi(x,y,z) = -2xy + x\ln(y+z)$$ at $$P(1,3,-2)$$ and the maximum value of the directional derivative at P.

Problem Statement

Calculate the direction of the maximum rate of change of the scalar field $$\phi(x,y,z) = -2xy + x\ln(y+z)$$ at $$P(1,3,-2)$$ and the maximum value of the directional derivative at P.

Solution

Dr Chris Tisdell - 1776 video solution

video by Dr Chris Tisdell

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