You CAN Ace Calculus

 partial derivatives basics of vectors

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

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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).

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

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

### 804 solution video

video by Dr Chris Tisdell

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

### 2293 solution video

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

### 2294 solution video

video by MIP4U

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

Problem Statement

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

Solution

### 2295 solution video

video by MIP4U

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

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

### 799 solution video

video by Dr Chris Tisdell

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

### 800 solution video

video by Dr Chris Tisdell

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

### 801 solution video

video by Dr Chris Tisdell

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

### 803 solution video

video by Dr Chris Tisdell

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

### 807 solution video

video by Krista King Math

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

### 808 solution video

video by Krista King Math

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

### 810 solution video

video by PatrickJMT

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

### 1522 solution video

video by Krista King Math

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

### 1776 solution video

video by Dr Chris Tisdell

Applications

There are several applications where the gradient is used and has additional meanings. Here are three of them.

 temperature application

When the function f is a temperature, the gradient can be interpreted as the change in temperature.

If we have a material where the heat flow (from hot to cold) is given by $$T = x^3 - 3xy^2$$, determine the direction of maximum decrease of temperature at the point $$(1,2)$$.

Problem Statement

If we have a material where the heat flow (from hot to cold) is given by $$T = x^3 - 3xy^2$$, determine the direction of maximum decrease of temperature at the point $$(1,2)$$.

Solution

### 802 solution video

video by Dr Chris Tisdell

 calculating a tangent plane

To determine the equation of a plane, we need one point and a normal vector. Since the gradient is normal to the level curves (and also to the tangent plane), we have a normal vector and we are usually given a point in the problem statement. This is enough information to determine the equation of the tangent plane.

Here is a video clip with a great explanation of this idea. He also explains how using the gradient notation simplifies the notation significantly.

### Dr Chris Tisdell - Gradient and Directional Derivative (Part 4) [5mins-39secs]

video by Dr Chris Tisdell

Calculate the tangent plane and the normal line to the surface $$x^2 + y^2 + z = 9$$ at the point $$(1,2,4)$$.

Problem Statement

Calculate the tangent plane and the normal line to the surface $$x^2 + y^2 + z = 9$$ at the point $$(1,2,4)$$.

Solution

### 805 solution video

video by Dr Chris Tisdell

Calculate the gradient, directional derivative and equation of the tangent plane of $$g(x,y,z) = \sin(xyz)$$ at the point $$(\pi, 1/2, 1/2)$$ in the direction $$\vec{v} = \langle \sqrt{3},-2,3 \rangle$$.

Problem Statement

Calculate the gradient, directional derivative and equation of the tangent plane of $$g(x,y,z) = \sin(xyz)$$ at the point $$(\pi, 1/2, 1/2)$$ in the direction $$\vec{v} = \langle \sqrt{3},-2,3 \rangle$$.

Solution

### 806 solution video

video by Dr Chris Tisdell

Calculate the equation of the tangent plane to $$x^4 + xy + y^2 = 19$$ at the point $$(2,-3)$$.

Problem Statement

Calculate the equation of the tangent plane to $$x^4 + xy + y^2 = 19$$ at the point $$(2,-3)$$.

Solution

### 809 solution video

video by Krista King Math

Calculate the equation of the tangent plane to the surface $$f(x,y)=2y\cos(5x-3y)$$ at the point $$(3,5,10)$$.

Problem Statement

Calculate the equation of the tangent plane to the surface $$f(x,y)=2y\cos(5x-3y)$$ at the point $$(3,5,10)$$.

Solution

### 2201 solution video

video by MIP4U

These types of problems will often be worked in the opposite direction than what we have discussed so far, i.e. you may be given a gradient and then expected to come up with the potential function. These are not hard but require some changes in your thinking to solve. The idea is that you are given a gradient and you need to come up with the original function. Here is a great video, with examples, showing how this works.

### Dr Chris Tisdell - Potential Function Example (Gradient) [5mins-19secs]

video by Dr Chris Tisdell

Calculate the potential function for $$\nabla f = (2x+yz)\hat{i} + xz\hat{j} + xy\hat{k}$$.

Problem Statement

Calculate the potential function for $$\nabla f = (2x+yz)\hat{i} + xz\hat{j} + xy\hat{k}$$.

Solution

### 811 solution video

video by PatrickJMT

 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.

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