We're given f of x comma y equals x y,
and now to determine the
directional derivative
of the function in the
direction of the vector
with components of negative two comma six,
and at the point one comma negative three.
Let's begin by looking
at this graphically.
So here we see the graph of the surface,
and we're asked to find the value
of the directional derivative
at the point one comma negative three,
which would be this red
point in the x-y plane.
To find the corresponding
point on the surface,
we need to find the z-coordinate
by substituting one for x
and negative three for y
into our function, and that would give us
this corresponding point
here on the surface.
And again, we're asked to find the value
of the directional
derivative from this point
in the direction of the vector
with components negative two comma six,
which would be this red vector right here.
So if we're on the surface at this point,
and move in the direction
of this red vector,
to find the directional derivative
would be the slope of this blue
tangent line to the surface,
at this point in the
direction of the given vector.
And notice how if we're on the surface,
and we move in the
direction of the red vector,
we would be going uphill,
and therefore the directional derivative
is going to be positive.
So hopefully this
illustration helps explain
what the value of the directional
derivative actually means.
Now, let's go back to our work.
For a quick review, if
we're given a function,
f of x comma y, which is differentiable,
the directional derivative
of the function, f,
in the direction of the unit vector
is given by the partial derivative of f,
with respect to x, evaluated
at the point, x comma y,
times cosine theta, plus
the partial derivative,
with respect to y, evaluated
at the point, x comma y,
times sine theta, so to find the value
of our direction derivative,
we'll need to find
the partial derivatives,
then evaluate those partials
at the point, one comma negative three,
and we'll also have to
find the unit vector
in the direction of the given vector.
So we're given f of x comma y, equals x y.
Let's first find the partial
derivative, with respect to x,
so we'll differentiate with respect to x,
treating y as a constant,
so the derivative of x y,
with respect to x, would just be y,
again, because we're
treating y as a constant,
and the derivative of x,
with respect to x, is one.
Now, we need to evaluate this
at the point, one comma negative three,
which is just going to be the
value of the y-coordinate,
which is negative three.
And now, we'll find the partial derivative
with respect to y, so we
differentiate with respect to y,
treating x as a constant,
so the derivative of x y,
with respect to y would just be x.
We evaluate this at the point,
one comma negative three,
which would just be the
value of the x-coordinate,
which is one.
And now, we're given
the directional vector.
Let's call it vector v.
Has components, negative two comma six.
We need the unit vector
in the same direction,
which would be equal to
vector v, the given vector,
divided by the magnitude of vector v,
so we'd have negative two comma six,
divided by the square root
of negative two squared,
plus six squared.
So we have the given vector,
divided by, this is going to
be the square root of four,
plus 36, so the square root of 40,
which does simplify,
because 40 is equal to four,
times 10, and the square
root of four is equal to two,
so let's rewrite this as
negative two, divided by two,
square root of 10, comma
six, divided by two,
square root of 10, and
now we'll go and simplify
these components.
Remember, because this is a unit vector,
the x and y components are
going to give us the value
of cosine theta and sine theta.
So we know the unit
vector and the direction
of the given vector.
Would have an x component of,
this would be negative one,
divided by the square root of 10,
and a y component of three,
divided by the square root of 10,
which, again, is equal to
cosine theta comma sine theta.
So now, we have all
the information we need
in order to find the value
of our directional derivative
at the point one comma three
in the direction of the vector
with components, negative two comma six.
We know the value of
the partial derivative,
with respect to x, at the given
point equals negative three,
we know the value of
the partial derivative,
with respect to y, at the
point is positive one,
and we also know cosine
theta equals negative one,
divided by the square root of 10,
and sine theta is equal to three,
divided by the square root of 10.
So let's go ahead and find the value
of the directional
derivative on the next slide.
So the directional derivative of f,
at the point one comma negative three,
is going to be equal to the partial,
with respect to x at the given point,
is equal to negative
three, times cosine theta,
which equals negative one,
divided by the square root of 10,
plus the partial, with respect
to y at the given point,
has a value of one, times sine theta,
which equals three, divided
by the square root of 10.
So we have three, divided
by the square root of 10,
plus three, divided by
the square root of 10,
which equals six, divided
by the square root of 10,
so this would be the exact value
of the slope of the tangent
line at the given point
in the direction of the given vector.
Let's also give a decimal
approximation for this.
We would have six, divided
by the square root of 10,
which, to four decimal places,
would be approximately 1.8974.
And let's go back to
our graph one more time.
Again, we just found the slope
of this blue tangent line
to the surface from this
point in the direction
of this red vector.
Notice how if we're on the surface,
walking in the direction
of this red vector,
we would be walking uphill
along this blue tangent line,
and, again, that's why the value
of the directional derivative is positive.
I hope you found this helpful.
