Applications of implicit differentiation
Summary
TLDRThis 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
📚 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.
📐 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.
Mindmap
Keywords
💡Implicit Differentiation
💡Slope of the Tangent
💡Derivative
💡Product Rule
💡dy/dx
💡Equation of Tangent Line
💡Factor
💡Plug in the Point
💡Simplify
💡Recap
💡Homework
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
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