We're given f of x comma y
equals natural log of the
quantity x squared plus y squared
and asked to determine the
directional derivative of f
in the direction of the
vector with components,
negative two comma two,
and at the point, one comma three.
Again, let's first look
at this graphically.
Again, our goal is to find
the value of the directional derivative
at the point three comma one,
which would be the point
here in the xy plane.
The second point up here on the surface
would be the corresponding point.
We're to find the
coordinate of this point,
we'd have to substitute the x
and y values from this point
into our function f of x comma y
and then from this point, we want to find
the slope of the tangent line
in the direction of this red vector,
which is the given vector with components,
negative two comma two.
So, again, the slope of
this blue tangent line
from this point in the
direction of this red vector
would be the value of the
directional derivative
that we're looking for.
Let's go ahead and rotate this,
so we can see this from different angles.
Also notice how I included the plane
that includes the directional vector
to add better perspective.
Going back to our work,
we know from our previous examples
this is our we're going to find
the value of the directional derivative.
So, we'll have to find the
partial derivatives of f
and then evaluate those
partial derivatives
at the point one comma three
and then define cosine
theta and sine theta,
we'll fine the unit vector
in the direction of
our directional vector,
with components negative two comma two.
So, starting with the given function,
we'll first find the partial derivatives.
Let's first find the partial derivative
with respect to x.
So, we'll differentiate the given function
with respect to x,
treating y as a constant.
Remember the derivative of natural log e,
with respect to x,
would be equal to one
over u times u prime,
or one over u times dudx.
So the derivative of natural log
of the quantity of x
squared plus y squared
with respect to x
would be one over the quantity
x squared plus y squared
times the derivative of
x squared plus y squared
with respect to x, which
would just be two x.
Remember the derivative
of y squared would be zero
with respect to x because
we're treating y as a constant.
So, this would just be two x divided by
the quantity x squared plus y squared,
and now we need to evaluate this
at the point three comma one.
So, we would have, two times x,
which is three,
and notice how the
denominator is going to be
x squared plus y squared,
which would be three
squared plus one squared
which equals 10.
So, the partial derivative
with respect to x
at the point three comma one
would be equal to 6/10,
which simplifies to 3/5.
And now we'll find the partial derivative
with respect to y.
So, now we're treating x as a constant
and differentiating with respect to y.
So, derivative of natural
log of the quantity
x squared plus y squared
with respect to y would be
one over the quantity x
squared plus y squared
times the derivative of
x squared plus y squared
with respect to y, which would be two y.
Again, the derivative of x
squared with respect to y
would be zero because you're
treating x as a constant.
So, here we have two y divided by
the quantity x squared plus y squared,
which we now need to evaluate
again at the point three comma one.
So, here you we'd have
two times y which is one
and all divided by, again,
three squared plus one squared.
So here we have 2/10.
So the partial derivative
with respect to y
at the point three comma
one is equal to 2/10
which equals 1/5.
So we need this value and this value
to find the directional derivative
at the given point in the given direction,
but we also need to find
cosine theta and sine theta
by determining the unit vector
in the direction of the vector
with components negative two comma two.
And let's do this on the next slide.
The unit vector is going to be equal to,
if we let this vector be vector v,
we'd have vector v divided
by the magnitude of vector v.
So, vector v has components
negative two comma two.
We'll divide by the magnitude
which would be equal to
the square root of negative
two squared plus two squared.
So we have the given vector divided by,
this is going to be equal
to the square root of
four plus four, the square root of eight,
because eight is equal to four times two.
The square root of eight
simplifies to two square root two.
So, let's write this as negative two
divided by two square root two comma
two divided by two square root two.
So, the unit vector in the same direction
as the given directional vector
would have an x component of negative one
divided by square root two
and a y component on one
divided by square root two.
So now we know that cosine theta
is equal to negative one
divided by square root two
and we know sine theta
is equal to one divided
by square root two.
Now we have all the information we need
in order to find the value of
the directional derivative.
The value of the directional derivative
at the point three comma one
in the direction of the vector
with components negative two comma two
is going to be equal to
the partial derivative
with respect to f
at the point which is equal to 3/5
time cosine theta which equals
negative one divided by square root two
plus the partial derivative
of f with respect to y
evaluated at the point which is 1/5
times sine theta which equals
one divided by square root two.
So, here we have negative three
divided by five square root two
plus one divided by five square root two.
So, notice how we have the exact value
of negative two divided
by five square root two.
Let's also get our decimal approximation.
So we have negative two divided by,
and then in parenthesis, we
have five square root two,
right arrow, close parenthesis and enter.
So, notice how the directional derivative
is approximately equal to negative 0.2828.
Now let's go back to our graph
and take a look at this again.
Notice how, for on this
point on the surface,
and we move in the
direction of this red vector
notice how the line is slanted
down in this direction,
and that's the reason why
the directional derivative,
in this case, is negative.
I hope you found this example helpful.
