Applications of implicit differentiation

Summary

This video script offers a comprehensive tutorial on implicit differentiation, emphasizing practical applications. It guides through finding slopes of tangents at specific points using derivatives. The script illustrates step-by-step processes for equations like 2x^2 - 2xy - 2y^2 = 1 and x^2 + y^2 = 10, detailing how to isolate dy/dx and calculate tangent lines. It encourages practice and provides homework guidance, fostering a deeper understanding of calculus concepts.

Takeaways

• 📘 Implicit differentiation is used when you cannot solve for y explicitly.
• 🔍 The slope of the tangent line at a point on a curve is found by differentiating both sides of the equation implicitly.
• ✏️ When differentiating, apply the product rule where necessary and remember that the derivative of y with respect to x is dy/dx.
• 📐 The derivative of a constant is zero, which simplifies the equation when finding dy/dx.
• 🔄 Isolate dy/dx to solve for it, often by moving terms to the other side of the equation.
• 📉 Factor and simplify the equation to find the expression for dy/dx.
• 📌 Substitute the x and y coordinates of the given point into the expression for dy/dx to find the slope at that point.
• 📐 The equation of the tangent line is y = mx + b, where m is the slope and b is found by substituting the point into the equation.
• 🔢 To find b, use the coordinates of the given point and the known slope to solve for the y-intercept.
• 📑 The process involves taking derivatives, applying algebraic manipulations, and substituting values to find both the slope and the equation of the tangent line.

Q & A

What is the purpose of learning implicit differentiation?

The purpose of learning implicit differentiation is to apply the concept to solve problems where explicit differentiation is not straightforward or when the relationship between variables is given implicitly.

What is the first example given in the script for applying implicit differentiation?

The first example is to determine the slope of the tangent line to the curve defined by the equation 2x^2 - 2xy - 2y^2 = 1 at the point (-3, 1).

How is the derivative of y^2 with respect to x calculated?

The derivative of y^2 with respect to x is calculated by treating y as a function of x and applying the chain rule, resulting in 2y(dy/dx).

What does the term dy/dx represent in the context of implicit differentiation?

In implicit differentiation, dy/dx represents the derivative of y with respect to x, which is the slope of the tangent line to the curve at any given point.

How do you isolate dy/dx when applying implicit differentiation?

To isolate dy/dx, you collect all terms containing dy/dx on one side of the equation and simplify the equation to solve for dy/dx.

What is the slope of the tangent line at the point (-3, 1) for the given example?

The slope of the tangent line at the point (-3, 1) is 2/5.

What is the process for finding the equation of the tangent line at a given point?

The process involves finding the slope (dy/dx) using implicit differentiation, then using the point-slope form of a line (y - y1 = m(x - x1)) to find the equation of the tangent line.

What is the significance of the negative signs in the derivative calculations?

The negative signs in the derivative calculations indicate the direction of the change in the function. They are crucial for correctly determining the slope of the tangent line.

How do you find the value of B in the equation of the tangent line y = mx + B?

To find the value of B, you substitute the given point (x1, y1) into the equation y = mx + B and solve for B.

What is the equation of the tangent line at the point (-2, 1) for the curve x^2 + y^2 = 10?

The equation of the tangent line at the point (-2, 1) is y = (1/3)x + 10/3.

Where can students find the homework assignments mentioned in the script?

Students can find the homework assignments in the class notebook under Chapter 3, where the related worksheets are located.

Outlines

00:00

📚 Implicit Differentiation Application

This paragraph introduces the concept of implicit differentiation with a focus on its applications. The speaker emphasizes the importance of understanding the purpose behind learning new mathematical concepts and applies this to finding the slope of a tangent line at a specific point. The example provided involves differentiating the equation 2x^2 - 2xy - 2y^2 = 1 implicitly to find the slope at the point (-3, 1). The process involves taking the derivative of both sides of the equation, applying the product rule, and isolating dy/dx. The final step is to substitute the coordinates of the given point into the derived expression to find the slope of the tangent line. The speaker concludes by summarizing the steps involved in finding the slope and encourages the audience to practice these skills.

05:04

📐 Finding Tangent Line Equations

The second paragraph continues the mathematical theme by focusing on finding the equation of a tangent line at a given point. The speaker outlines the steps to derive the slope (M) using implicit differentiation and then to find the equation of the tangent line in the form y = MX + B. An example is given where the equation x^2 + y^2 = 10 is differentiated to find the slope at the point (1, 3). The speaker guides through the process of isolating dy/dx, simplifying the expression, and then substituting the point's coordinates to find the slope. Once the slope is determined, the next step is to find the y-intercept (B) by using the given point and the derived slope. The speaker concludes by summarizing the process and encourages the audience to practice the problem-solving steps on their own, offering assistance if needed.

Keywords

💡Implicit Differentiation

Implicit differentiation is a method used in calculus to find the derivative of an equation that cannot be easily solved for one variable explicitly. It involves differentiating both sides of an equation with respect to a variable, treating the equation as a whole. In the video, implicit differentiation is used to find the slope of the tangent line to a curve at a given point without needing to solve for y explicitly.

💡Slope of the Tangent

The slope of the tangent line to a curve at a specific point is the rate of change of the function at that point, which can be found using derivatives. The video uses implicit differentiation to calculate the slope at a given point on a curve defined by an equation.

💡Derivative

A derivative in calculus represents the rate at which a function is changing at a certain point. It is the primary tool used in the video to find the slope of tangent lines. The video demonstrates how to take derivatives of both sides of an equation to apply implicit differentiation.

💡Product Rule

The product rule is a derivative rule that states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. The video mentions the product rule when differentiating terms involving products, such as '2xy'.

💡dy/dx

dy/dx represents the derivative of y with respect to x, which is the slope of the tangent line to the curve at any point. The video script focuses on isolating dy/dx to find the slope of the tangent line at specific points on the curve.

💡Equation of Tangent Line

The equation of a tangent line is a linear equation that represents the tangent to a curve at a specific point. The video script explains how to derive this equation using the slope found from implicit differentiation and a given point on the curve.

💡Factor

Factoring is a method of expressing a polynomial as the product of its factors. In the video, factoring is used as a step in simplifying the expression obtained after implicit differentiation to isolate dy/dx.

💡Plug in the Point

Plugging in the point refers to substituting the coordinates of a specific point into an equation to find an unknown value, such as the y-intercept in the equation of a tangent line. The video demonstrates this process when finding the equation of the tangent line at a given point.

💡Simplify

Simplification in calculus often involves reducing complex expressions to a more manageable form. The video script mentions simplifying expressions after differentiation to isolate terms and solve for dy/dx.

💡Recap

A recap is a summary or review of the main points covered. The video script includes a recap section to review the steps and concepts covered, such as finding the slope of the tangent and the equation of the tangent line.

💡Homework

Homework in the context of the video refers to the practice problems assigned to reinforce the concepts learned. The video script directs students to find and complete homework related to the topics discussed, such as implicit differentiation and finding tangent lines.

Highlights

Introduction to applications of implicit differentiation.

Purpose of learning is to apply what we've learned.

Starting with basic applications of implicit differentiation.

First example involves finding the slope of a tangent line.

Equation given: 2x^2 - 2xy - 2y^2 = 1.

Derivative of x^2 is 2x, using the power rule.

Product rule applied to derivative of 2xy.

Derivative of 2y^2 involves chain rule.

Isolating dy/dx to solve for the slope.

Simplifying to find dy/dx = (x - y) / (x + 2y).

Finding the slope at the point (-3, 1).

Slope of the tangent at (-3, 1) is 2/5.

Recap on finding the slope of a tangent line.

Finding the equation of the tangent line y = mx + b.

Derivative of y = mx + b to find the slope.

Isolating dy/dx and simplifying the equation.

Finding the slope at the point (-2, 1).

Equation of the tangent line is y = 1/2x + 2.

Second example involves finding the equation of a tangent line at a given point.

Equation given: x^2 + y^2 = 10.

Derivative of x^2 is 2x, derivative of y^2 is 2y(dy/dx).

Isolating dy/dx to get -x/y.

Finding the slope at the point (1, 3).

Equation of the tangent line is y = -1/3x + 10/3.

Guidance on finding homework exercises.

Location of homework exercises in class notebook.

Transcripts

applications of implicit differentiation

every time you learn something new there

should be a purpose so we're going to

learn to apply what we've learned

there's some truly magnificent

applications coming out we're going to

start with the basic ones today

first example determine the slope of the

tangent 2x squared minus 2xy minus 2 y

squared equals 1 at the point minus 3 1

slope means find the derivative

we have a Y we have a y squared we do

not want to sell for y explicitly so

we're going to do an implicit derivative

that means taking the derivative of the

left hand side of the equation and the

right hand side of the equation

I'll take the derivative of x squared

with respect to X derivative of 2x Y

with respect to X derivative of 2y

squared with respect to X we revel it

with 1 with respect to X derivative of x

squared we know that's 2x this is a

product rule

derivative of first times second

first derivative of second remembering

derivative of y is dy/dx they're both

negative because of the negative sign in

front derivative of 2y squared will do

that with respect to Y first giving us

for Y and then multiply by dy/dx you'll

notice that there's a dy/dx term every

time you take a derivative of Y

derivative of a number with no X's is

zero our goal is to solve for dy/dx

there's all sorts of negative signs here

well take those terms to the other side

trying to get dy/dx by itself we'll

factor x 2x plus 4 or y and then we'll

divide both sides by the 2x plus 4y

simplify that as much as possible

there's a common term in there so we'll

divide everything by 2 getting is X

minus y over X plus 2y now our goal was

to find the slope at the point negative

3 1 so negative 3 will go in for X

negative 1 and for y

add those together and a better answer

than negative 2 over negative 5 is

positive 2 over caught is positive 5

so the slope of the tangent at negative

3 negative 1 is 2 over 5

quick recap slope of tangent find

derivative derivative of both sides of

the equation

watch out for your product rule and

remember that anytime you find the

derivative of the Y there will be a dy d

extra get dy/dx by itself so we'll

isolate all the terms with new ID ax by

putting them on one side of the equation

factor simplify reduce plug in your pool

find the equation of the tangent line at

the given point now other than the fact

that we have X and Y's together you know

how to find the equation of the tangent

line

we've done that before think about y

equals MX plus B equation of a lie

you're welcome to try this one on your

own put me on pause try it see if you're

right for those who want to keep working

with me let's keep going

equation of tangent line y equals MX

plus B we'll find the slope first which

means taking the derivative of both

sides here's another product rule

derivative of first times second first

times derivative a second and anytime we

take derivative of Y we have a dy/dx

derivative of a number with no zero with

no X's is zero trying to isolate the

term with dy/dx so we'll take the 2y to

the other side of the equation

/ 2x and then reduce get rid of the

common term of two

on equation by the slope at the point

negative 2 1 first so negative 2 goes in

for X 1 goes in for y so now we have a

slope but we can plug in for M so Y is

1/2 X plus B the only thing we do not

know is B but we do know X and we do

know Y X negative 2 and Y is 1

1/2 of negative 2 is negative 1 bring

the negative 1 over we now know what B

is that was the only thing missing in

this part of the equation so we'll put

the B in and we'll find out what the

equation is so we'll have y equals 1/2 X

plus 2 for the equation of the tangent

line y equals 1/2 X plus 2

quick recap on that one equation of a

line y equals MX plus B M slope finder

written

isolate dy/dx simplify plug in the point

and that goes in for your slope B is the

only thing we don't know but we do know

X and we do know why get x and y from

the point simplify and find out what B

is and then plug that in and there's our

equation

we'll do a second example of that one

find the equation of the tangent line at

the given point x squared plus y squared

equals ten at the point one three try

this one on your L put me on hold

do it yourself and then check the answer

time to check your answer what you

should have done was taken the

derivative of both sides of the equation

and that will give us a slope derivative

of x squared is 2x derivative of Y

squared take it with respect to Y first

and then multiply by dy/dx derivative of

10 is 0

you'll notice that I'm starting to skip

steps only do the steps that you need in

order to solve the question we're going

to get dy/dx by itself so we'll keep

this term on one side and put the 2x

over

/ - lie reduce your expression - x over

y and we'll put in 1 for X and 3 for y

so we have the slope of our tangent

looking for y equals MX plus B and we'll

put the slope and immediately B's the

only thing we don't know but we do know

X and we do know why

negative one third of one is negative

one third pull that to the other side

and add it to the 3 3 would be equal to

9 over 3 plus 1 over 3 so B is 10 thirds

that gives us an equation now y equals

negative 1/3 X plus y or B and there's

our equation I'd love to ask if you and

how many got it right but we'll just

keep looking what you need to get

working on now

if you did not do any yesterday because

you didn't know where to find it well

now you know where to find it or I'll

show you where to find so today question

number two page one one two five and

yesterday you should have done question

number one so if you can't find the

homework into thinking there's homework

let's go look for it shall we

and then the class notebook I'm in

Chapter three because that's one we're

working one and it says homework now the

file is a little too big to put here so

I told you where to find it go to files

if you accidentally went to file you

know there's no homework there I don't

have to do anything let's read that

again files highlighted next to class

notebook

here in all the class material

we're doing Chapter three

here are the worksheets

related rates chapter work

and there is the exercise for question

one all your answers as usual in

question two they've got lots of work to

get done please get yourself to work and

if you have questions just let me know

at any point in time and I'll help you