Welcome to a lesson on tangent planes.
The goal of this video is to determine the equation of a tangent plane to a surface.
Let's start by viewing a tangent plane to a surface.
Here this yellow plane is tangent to the surface
at the purple point.
Let's animate this so you get a better feel for this tangent plane.
Again this yellow plane
is tangent to the surface
at the purple point.
So the goal of our lesson is to determine
the equation of this tangent plane
at a given point.
I like to start
by reminding us how we found the equation of a plane
when we had a normal vector.
Remember we use the components of the normal vector
to help us determine
the equation of the plane.
And we use a similar idea when it comes to the equation of a tangent plane.
Let's take a look at how we're gonna do this.
But first
instead of defining a surface explicitly as z equals f of x, y.
We can also define a surface implicitly by
big F of x, y, z equal to zero.
And we're going to have to do this to determine the equation of the tangent plane.
So if big F is differentiable at the point
x sub zero, y sub zero, z sub zero.
Then we can determine the equation of a tangent plane
to the surface
expressed
implicitly
by the following formula.
The first thing I want you to notice about this equation for the tangent plane
is that it does involve
the first order partial derivatives of big F here.
With respect to x here with respect to y.
And here with respect to z.
And then notice were multiplying
these partial derivatives by
x minus x sub zero
y minus y sub zero and z minus z sub zero.
And the sum of these products is equal to zero.
And I will prove this formula in another video.
But the idea behind the proof is that
because
the gradient of big F is going to be normal to the tangent plane
at the given point.
It must be orthogonal to every vector in the tangent plane.
So the idea here is that if we take the dot product
of the gradient of big F
and
vector v
in the tangent plane it will always equal zero.
And this dot product
is what gives us the equation of this tangent plane.
Let's go and take a look at a couple of examples.
We want to determine the equation of the tangent plane to the surface
at the given point.
So the first thing we need to do is write an equation for big F of x, y, z.
And that's going to be equal to x cubed.
Minus three xy.
Plus y cubed.
And we have minus z.
Equal zero.
So moving this z over to the other side
and setting it equal to
big F of x, y, z.
For the next step were gonna go ahead and determine
these partial derivatives
and then we'll evaluate them at the point one, two, three.
So the partial derivative respect to x
is going to be three x squared.
This will be minus three y.
This should be zero and this should be zero.
The partial with respect to y.
Would be zero.
Negative three x.
Plus three y squared.
Should be zero.
And the partial with respect to z.
Gonna be zero, zero, zero and then negative one.
And now we need to evaluate these partials
at the point one, two, three.
When x is one we have three times one that be three.
Minus
when y is two were gonna have minus six.
This will give us negative three.
Now the partial respect to y at the point one, two, three.
Well x is ones so this will be negative three.
Y is two so we have four times three plus 12.
So this will be nine.
And the partial respect to z will remain negative one.
This is all we need in order to determine the equation of the tangent plane.
Solve this is equal to negative three. So we have negative three.
Times x minus
x sub zero.
Well x sub zero would be one.
Plus nine.
Times y
minus two.
Minus one.
Times
z minus three
must equal zero.
So again we have our
partial derivatives
here
and here.
Evaluated at the point one, two, three
were multiplying it by x
minus x sub zero.
Y minus y sub zero.
And z minus z sub zero.
Now we distribute
and that will give us the equation of the tangent plane.
It three x plus three.
Plus nine y minus 18.
Minus z plus three equals zero.
So we have negative three x.
Plus nine y.
Minus z equals
plus five we have three minus 18 plus three
that's gonna be negative 12 move to the other side that will be positive 12.
In a lot of times text books like to have a leading coefficient positive
so we can multiply through by negative one.
That would give us three x
minus nine y
plus z
equals negative 12.
Of course it's a same equation in a different form.
Let's take a look at this graphically.
So in gray we have our surface.
In red is our point of tangency.
In yellow is the tangent plane.
And I also plotted the gradient.
Which show that the gradient is normal or orthogonal to that tangent plane.
Let's take a look at one more example.
Here's our surface.
We want to determine the equation of a tangent plane at the given point.
So first
big
F of x, y, z.
Is equal to
x squared
plus y squared
to the one half
minus z.
Let's go and determine our partial derivatives.
The partial derivative with respect to x.
Is going to be
one half
times x squared plus
y squared to the negative one half
times
two x
and derivative of z would be zero.
Let's go and clean this up two simplify out.
So we have
the partial with respect to x would be
x.
All over
the square root of x squared plus y squared.
And the partial with respect to y.
It's going to be one half x squared plus y squared.
The negative one half times two y.
Again this simplifies in a similar way.
So we have y over the square root of x squared plus y squared.
And the partial with respect to z
is just gonna be negative one.
I will evaluate these at the point negative three, four, five.
So our numerator is going to be negative three and our denominator
were gonna have the square root of
negative three squared plus four squared that's gonna give us a square root of 25 which is equal to five.
Partial with respect to y at the given point.
Our denominator is gonna be the same. Our numerator now gonna be y so we have four fifth for y.
And the partial with respect to z
stays negative one.
Now we have all the information we need to determine this equation.
Were gonna have
negative three fifth
times x minus negative three that will become x plus three.
Plus
four fifth
times
y minus four
minus one times z minus five.
And again this must equal zero.
Now we probably want to clear these fractions out of here
so were gonna multiply both sides of the equation by five.
That will give us negative three times x plus three.
Plus four times y minus four.
Minus five times z minus five.
Equal zero.
Let's go and take this onto the next slide and finish it up.
So we go ahead distribute we have negative three x minus nine
plus four y
minus 16 minus five z plus 25
equal zero.
So we have negative three x plus four y.
Minus five z. Let's get our constant terms now.
We have negative
nine
minus 16 that's negative 25 plus 25.
That's actually gonna be zero.
And again if we wanted to we would rewrite this as a positive leading coefficient.
We have three x minus four y
plus five z
equal zero.
Let's go and verify our answer graphically.
In gray we have our surface.
In red is the point of tangency.
And in yellow we have our tangent plane.
So it looks good.
And once again I went ahead and
graph the
gradient vector so we could see that it is orthogonal to
that tangent plane.
I hope you found these two examples helpful.
Thank you for watching.