Directional Derivatives

Hi, everyone. Welcome back to integralcalc.com. Today, we’re going to be talking about directional
derivatives. And in this one, we’ve been given the function f(x,y) = x^2 + 2xy + 3y^2
and we’ve been asked to find the directional derivative of this function at the point P
which is (2,1) in the direction of these vectors here. So there’s going to be three parts
to this problem. The first is going to be to convert the vector v to a unit vector.
The second will be to find the partial derivatives of f with respect to x and y. And then the
third will be to combine those two pieces of information along with the point P to find
the directional derivative. So first, let’s talk about how to find the
unit vector from these vectors v. The way we’re going to do it is with a really simple
formula and this is the same every time. We’re going to be taking the square root of the
information inside these vectors here. So basically, we’ve got the vector here, a,b.
We’re going to take the square root of a^2 + b^2. And if you have a vector with a, b
and c, you can do the same thing. You would just add c^2 under the square root sign. But
we’re going to take a^2 + b^2. So in our case, that’s the square root of 1^2 and
1^2 because both a and b are equal to 1 which simplifies to the square root of 2. So now,
we’re going to take this value, the square root of 2, and we’re going to say that the
unit vector is equal to; we will take a and b which we take as 1 and 1 and put those in
the numerators and then we’re going to take the square root of 2, the value we just found
and put those in the denominators. And so now instead of v = (1,1), we have u = 1 over
the square root of 2, 1 over the square root of 2. And this is now our unit vector. So
that’s the first piece of information. The second is we need to take the partial
derivatives of f with respect to x and y. So let’s go ahead and take the partial derivative
of f with respect to x. And remember when we do that, we’re treating x as the variable
and holding y constant. So let’s take the derivative term by term. The derivative of
x^2 with respect to x is going to be 2x. The derivative of 2xy with respect to x will simply
be 2y because remember if we were just taking the derivative of x on its own, the derivative
of x will be 1 so we’d end up with 2y(1) which is just going to be 2y. And the derivative
of 3y^2 with respect to x will just be zero so we’ll go ahead and leave it. Now, let’s
go ahead and find the partial derivative of f with respect to y. So doing that, obviously,
we’ll treat y as the variable and hold x constant. So the derivative of x squared with
respect to y is zero. Because there’s no y variable involved in this term here. The
derivative of 2xy with respect to y is just 2x because similarly to when we took the derivative
with respect to x, if we just took the derivative of y, the derivative of y with respect to
y is 1. So we would have 1(2x) and we’re just left with the 2x so the derivative of
that term is 2x. The derivative of 3y^2 is 6y. So we have these two partial derivatives.
Now, the directional derivative is going to be a combination of these two pieces of information.
The directional derivative is going to be denoted as d sub u of f(x,y) and it’s going
to be equal to, we’ll call these values in the unit vector here a,b just like we did
in the vector v, so the directional derivative is going to be a times the partial derivative
of f with respect to x plus b times the partial derivative of f with respect to y. So if we
plug in the information we already have into this formula, we’ll get the directional
derivative of f(x,y) equals a which we already know as 1 over the square root of 2 times
our partial derivative with respect to x which is just 2x + 2y. Then if we add to that b
which is also 1 over the square root of 2, and multiply by the partial derivative with
respect to y, we’ll get 1 over the square root of 2 times 2x + 6y.
So that, essentially, is our directional derivative. But remember, we care about the directional
derivative at the point P which is (2,1). So what we need to do is plug the point P
into our directional derivative and when we do that, we’ll get d sub u for the directional
derivative of f of the point (2,1). So we plug that in there and then we’re plugging
in 2 for x and 1 for y. So we’ll get 1 over the square root of 2 times 2(2) + 2(1), plugging
in 2 for x and 1 for y plus 1 over the square root of 2 times 2(2) + 6(1). And when we simplify
that d sub u f(2,1) is going to be equal to 1 over the square root of 2 times 4 + 2 so
times 6; plus 1 over he square root of 2 times 4 + 6 so times 10. So as you can see, we’re
going to get 16 over the square root of 2. Now, we could leave it at that but it’s
really not proper to leave a square root at the denominator of a fraction so what we can
do to eliminate it is multiply by the square root of 2 over the square root of 2. That’s
just like multiplying by 1 but we’ll cancel out the square root of 2 in the denominator.
So when we do that, we’ll get 16 times the square root of 2 divided by the square root
of 2 times the square root of 2 is 2 and now we can simplify 16/2. The two will cancel
and we’ll just be left with 8 in the numerator so 8 times the square root of 2. And that
is our directional derivative. So this is our final answer. This is the directional
derivative of the function f(x,y) at the point P which is the point (2,1) in the direction
of these vectors here (1,1). So that’s it. I hope that video helped you guys and I will
see you in the next one. Bye!