Okay, we are going to find the equation to the tangent plane. Problem says find the equation
of the plane tangent to the given surface P equals f of (x, y) at the indicated point
P. So, the problem that they gave me is z, and we’ll do that little line to make sure
that that’s not a two, equals x squared plus y squared and the point that they gave
is (three, four, twenty five). So they’re asking for… they’re asking for the equation
of the plane intersecting the graph of this function at this point. Just like in… in
calculus, we’re only using the single variable, they are asking for the line intersecting
the graph at certain point, it’s just the plane intersecting the graph at certain point,
so, it’s just like dreaming instead of doing. So, this isn’t actually really all that
difficult. All we need to do is take the derivative of this equation with respect to both x and
y and then, kind of, plug in. So, what we’re going to do is go ahead and say partial derivative
of the function z with respect to x and then partial derivative of the function z with
respect to y, we need to find these two things, so partial derivatives. First, we’re taking
with respect to x and the derivative of this equation here… here with respect to x is
two x, the y squared goes away. With respect to y, it’s two y because the x squared goes
away. And, now that we have the partial derivatives, what we do is plug in this point to these
to get the slope in the x direction and slope in the y direction. So, this here, keep in
mind, is (x, y, z) point always. So, we’re going to plug in three for x so it’s going
to be two times three, so this will be six. And then, here, we’re going to plug in y,
y is four, so two times four is eight. So, now we… we now have the slope in the
x direction and slope in the y direction and so what we have to do is plug in. The formula
that we’re looking at, in mine it’s called the plane tangent to a surface, and what that
basically is going to look like is z… Let’s see if I can… I don’t know if I can write
this whole thing out this time… What we’re going to do is z minus the z… the z coordinate
here, twenty five, equals and then we always plug things in the same spot so the twenty
five, the z coordinate always goes here, then we have slope in the x direction which is
six times x minus the x point, which is three, plus the slope in the y direction, eight,
times y minus y point, which is four. We always plug things in this order. We always have
z here, always have the point at z here, slope in the x direction, x minus x point, slope
in the y direction, y minus y point, always, it’s how we start this equation. So, now
that we’ve plugged everything in and you notice before we get started they gave us,
you can see, they gave us z, x, and y, we solved for six and eight, slope in the x direction,
slope in the y direction, so those are the elements we’ve solved for and we write in
z, x, and y here, so, that… that’s how we, kind of, break down those elements.
So now, all we have to do is simplify and solve, really just simplify. So, we’re just
going to multiply out. You could either give your answer in this form or you could change
it to the following form, I’ll show you both. With a…with a z minus twenty five
equals, and now we’re just going to … we’re just going to multiply out here, so, six x
minus eighteen plus eight y minus thirty two. z equals, we’re going to add the twenty
five to the other side, but you can always do six x plus eight y first, the x plus the
y, and then… and then the constant. So, negative eighteen minus thirty two is negative
forty… negative fifty… plus twenty five… is still negative twenty five. Okay, so we
combine the twenty five, eighteen, and thirty two to get the twenty five and that’s it.
So, either your answer can look like this, and that’s usually perfectly acceptable,
or it can look like this, which is also great. Same thing, we just simplified it here. I,
kind of, preferred this way, unless they asked you for this specific equation. That’s it!