\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Applications of The Gradient

Coordinate Systems

Vectors

Using Vectors

Applications

Vector Functions

Partial Derivatives

Partial Integrals

Double Integrals - 2Int

Triple Integrals - 3Int

Practice

Vector Fields

Calculus Tools

Learning Tools

Articles

This page covers applications of gradients and directional derivatives.

There are several applications where the gradient is used and has additional meanings based on what the function represents. Here are three of them.

Temperature

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

Practice

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 video

video by Dr Chris Tisdell

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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

Practice

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 video

video by Dr Chris Tisdell

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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 video

video by Dr Chris Tisdell

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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 video

video by Krista King Math

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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 video

video by MIP4U

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You CAN Ace Calculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Temperature - Practice

Tangent Plane - Practice

How to Ace Calculus: The Streetwise Guide

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