Hi everyone! Welcome back to integralcalc.com. We’re gonna be doing another linear approximation
problem today. This one is asking us to find the equation of the tangent line and it was
submitted by one of our viewers. The function that's given or the equation
I should say that's given is e to the x y plus x squared minus y squared equals
five and they ask us to find the equation of the tangent plane at the point
two zero. 5={e^{xy}}+{x^2}-{y^2}~~;~~P(2, 0)
Okay so when we're dealing with linear approximation or asked to find the tangent line or the tangent
plane, when we have two variables as we do here with x and y, normally if we're finding
the tangent line and there’s one variable, we just take the derivative and then use
some work. In this case, we have two variables. We have to take partial derivatives. So we'll
first need to take partial derivatives with respect to x and y. Actually before I do
that, let me go ahead and write out the linear approximation equation, the equation that
we're gonna use or the formula that we're gonna use to build our equation of the tangent
line. It looks like this. z minus z one equals a times x minus x one plus b times y minus y one.
z-{z_1} = a(x-{x_1})+b(y-{y_1}) When you’re dealing with two variables,
as we are here, x and y, this is always the equation that we wanna use to find the
equation of the tangent plane or the tangent line. So each of these components is represented
by something we’re gonna solve for. You’re gonna end up leaving z, x and y intact
and we're gonna be solving for z one, x one, and y one, as well as a and b and I'll
let you know what each of those represent as we go through.
So like I said, first thing we do partial derivatives with respect to x and y. So
first, the partial derivative with respect to x. We go term by term, so partial derivative
with respect to x of e to the x y. remember when we're taking partial derivatives, we're
holding the opposite value of the constant. So in this case, since we're taking with respect
to x, we're treating y as a constant. So I like to think of it as the number two
or the number three, a constant. So if this is, if y is three, then this term here instead
of e to the x y ends up looking like e to the three x, right? Because if we replace
that with three, it will become the coefficient on this x term, then we have e to the
three x. Well, the derivative of e to the three x, right is three e to the three x,
well in this case, what does three represent? It represents y so it's not three e to
the three x, it's y e to the y x, in that case, we would reorganize those variables.
You always wanna have your variables in alphabetic order so instead of three e to the three x,
it would be y e to the xy. So y e to the xy is the derivative of this first term,
the partial derivative of this first term with respect to x. Here, at the second term
we know, that's much simpler. It's just two x and then this term here, the y squared is actually
gonna be zero because if we plug in three for y, again we're treating it as a constant
like five here. If we plug in three, right? We’ll get nine which is still constant and
the derivative of any constant is just zero. So we end up with zero for this term. And
then the other thing I did wanna do before we've moved on. This equation here, when you're
doing linear approximation, what I should have done before I went through or these steps
was actually subtracted, if you can imagine, if you take this equation and subtract five
from both sides, this will become zero equals and this whole thing instead of equals five,
the five was over here. We subtracted it from both sides and we end up with the zero on
the other side and negative five on this side. We wanna do that because it's important that
we atleast acknowledge that the five is in here and that I need to make sure that I'm
dealing with the five when I'm taking partial derivatives. Of course in this case, it's
gonna go away because it's a constant and the derivative of any constant is zero, so
it doesn't factor into this partial derivative but you should always first subtract that
constant from both sides and get this equals to zero. So let's go ahead and continue.
5={e^{xy}}+{x^2}-{y^2} 5-5={e^{xy}}+{x^2}-{y^2}-5
0={e^{xy}}+{x^2}-{y^2}-5 {partial {f}}/{partial {x}} = ye^{xy}+2x
This is the entire partial derivative with respect to x. So now let's go ahead and take
the partial derivative with respect to y, same as before except now we're holding x constant.
So in this case if we hold x constant, we treat x as let's call it the number four.
So pretend x is four. e to the four y, we're taking the derivative with respect to y when y is
our variable. So the derivative f e to the four y would be four e to the four y,
so in this case, four represents x, so we're looking at x e to the xy for our derivative.
Then we move on to x squared, the derivative of x squared with respect to y is zero.
With respect to x of course it's two x, but wth respect to y, we're holding x constant,
so this is treated like four, four squared is sixteen and sixteen is constant so the
derivative is zero. Of course then this y squred, y is actually the variable we're dealing with here,
so this becaomes minus two y and of course the five is still a constant so the derivative
there is zero. 0={e^{xy}}+{x^2}-{y^2}-5
{partial {f}}/{partial {y}} = xe^{xy}-2y So these are the partial derivatives with
respect to x and y and now what we need to do is three things. We need to plug in
our point two zero to three differrent things. The original equation and both partial derivatives.
So let's go ahead and plug it in to our original equation first. So two is representing x here
obviously and zero, y. So let's go ahead and say zero equals e x so that's two times y,
which is zero plus x squared, so two squared minus y squared, y is zero and then minus
five. Okay so two times zero is zero, so we end up with e to the zero plus four minus
zero minus five. e to the zero, in fact anything raised to zero power is one, so we
end up with one plus four minus five which is zero. Okay, so when we plug in the point
to the original equation, when we get our answer here, what we were solving for is z one,
right her. So z one, let's just go ahead and right is above here is gonna be zero.
0={e^{xy}}+{x^2}-{y^2}-5~~;~~P(2, 0) P(2_x, 0_y)
{f}(x, y)={e^{xy}}+{x^2}-{y^2}-5 {f}(2, 0)={e^{(2)(0)}}+{(2)^2}-{(0)^2}-5
{f}(2, 0)=e^0+4-0-5 Rule:~(anything)^0~=~1
{f}(2, 0)=1+4-0-5 {f}(2, 0)=0
z_1=0 Zero and two here for y and x are what
we plug in for x one and y one, so we plug in two for x one and we plug in zero
for y one. You can imagine, we're looking at points in a three dimensional coordinate
space and we have a point in that space, x, y, z, and that point is two zero zero. So when we
plugged in this point here, we got an answer for z one, we solved for that, that third
coordinate of the point so we plugged in xy and then whatever we solve for, which is z one.
So we plug in those things. Remember I said we're leaving alone x, y and z so the
only things only left for us to solve for are a and b. In this case, the way we find a and b,
in fact in every approximation case, is we solve, we plug in the point two zero to each
of our partial derivatives here. So when I plug in two zero to the partial derivative
with respect to x, I get, let's see plug in zero for y so e to the two times zero
plus two times two. Okay, so this whole thing is multiplied by zero, so it becomes zero
and then we have plus four so that's four and then we plug it in for x, it's called
the slope in the x direction, but really you just need to remember right before plugging
in the point to the partial derivative with respect to x, naturally it's gonna be a,
because a is grouped here with the x's so we plug in four for a.
P(2, 0) P(2_x, 0_y)
slope~in the~x~direction {partial {f}}/{partial {x}} = ye^{xy}+2x
{partial {f}}/{partial {x}}~(2,0)= (0)e^{(2)(0)}+2(2) {partial {f}}/{partial {x}}~(2,0)= 0+4
{partial {f}}/{partial {x}}~(2,0)= 4 And then when we plug in the point two zero
to the partial derivative with respect to y, it will be for b, because it's grouped here
with the y's. Let’s go ahead and do that, we have two e to the two times zero minus
two times zero. Okay so this is gonna be e to the zero which is one, so we have two times
one minus and then this two times zero here is just zero. I'm sorry i'm working backwards,
but so it's two, two times one minus zero. So two is the slope in the y direction or b,
so we've got two. P(2, 0)
P(2_x, 0_y) slope~in the~y~direction
{partial {f}}/{partial {y}} = xe^{xy}-2y {partial {f}}/{partial {y}}~(2,0)= (2)e^{(2)(0)}-2(0)
Rule:~(anything)^0~=~1 {partial {f}}/{partial {y}}~(2,0)= (2)(1)-0
{partial {f}}/{partial {y}}~(2,0)= 2 So now that we solved for all three of these
things, all we need to do is simplify this linear approximation equation. So let's go
ahead and write it out. We’ve got z minus zero, I don't have to write zero, it just
goes away. z equals four times x minus two plus two times y. I don't need to write
minus zero. I can just leave that. So now, you can leave it like this. It's acceptable
but it's a little cleaner to multiply things out and reorganize so we'll end up with z equals
four x minus eight plus two y and then you always wanna do your variables in alphabetical
order, put your constants in the end. Your cleanest answer would be z equals four x plus
two y minus eight, just moving that eight there to the end, with the variables up front.
z-{z_1} = a(x-{x_1})+b(y-{y_1}) z-(0) = (4)[x-(2)]+(2)[y-(0)]
z= (4)[x-(2)]+(2)(y) ~or~
z= 4x-8+2y z= 4x+2y-8
So that's linear approximation. Remember, subtract five from both sides to get this
set equal to zero, take partial derivatives with respect to x and y, then go ahead
and plug in the point that they give you to the original function to find z one to
the partial derivative with respect to x to find a, and to the partial derivative with
respect to y to find b, use your point here zero and two or two and zero sorry to
reperesent x one and y one and then go ahead, plug everything in, simplify and
solve. So your final answer is z equals four x plus two y minus eight.
final~answer:~ z= 4x+2y-8 See you guys next time!