Hi, everyone.
Welcome back to integralcalc.com.
Today we’re going to be talking about the gradient vector.
And in this problem, we’ve been asked to find the maximum directional derivative which
is also called the gradient vector and the direction on which it occurs.
So we’ll be finding direction first and then finding the maximum directional derivative.
So in order to find the direction, we’re going to need to find the gradient vector
at the point P. and the way that we do that is by taking the partial derivatives of f
with respect to x and y and then plugging in the point P.
We denote the gradient vector as an inverted triangle and it’s also called the maximum
directional derivative.
So the gradient vector is going to be this upside down triangle here.
And that’s going to be the partial derivatives with respect to x and y.
So let’s first go ahead and find both of those partial derivatives.
When we take the partial derivative of f with respect to x, we’re treating x as the variable
and holding y constant.
So we’ll take the derivative term by term.
The derivative of 2x^2 will be 4x.
The derivative of 3xy will be 3y.
And the reason remember is because when we’re taking the derivative with respect to x, if
we just took the derivative of x, we would get 1 so we’d end up with 1 times 3y which
of course just gives us 3y.
So the derivative of that term with respect to x is 3y.
The derivative of 4x^2 is zero because there is no x variable involved in that term.
if we take the partial derivative of f with respect to y term by term, the derivative
of 2x^2 will be zero because there’s no y variable involved in that term.
The derivative of 3xy will be 3x.
And the derivative of 4y^2 will be 8y.
So these are our two partial derivatives and we can now plug them in to the gradient vector
to give us the maximum directional derivative.
So plugging them in here and we’ll get 4x + 3y and 3x + 8y.
In order to find the direction of this, we need to plug in the point (1,1).
So that’ll give us f of (1,1).
Plugging that in 1 for x and 1 for y, we’ll get 4(1) + 3(1), 3(1) + 8(1) and when we simplify
this, obviously we get (7,11).
Now this is the direction in which the maximum directional derivative or the gradient vector
occurs.
So remember we were asked to find the direction and the maximum directional derivative we
have found the direction with this.
Now finding the maximum derivative or the gradient vector is as simple as taking our
answer here (7,11) and doing the following.
First, let’s assume that this vector is a,b and we’ve got (7,11).
And if we had three variables, if we had (7,11,8), it would be a, b and c.
But right now we have two variables and we would plug them in here in the square root
at 7^2 + 11^2 and again, if we had a third vector and we called that c, we could add
that c^2 term under the square root as well.
But in this case, we just have the two.
So 7^2 + 11^2 is going to be equal to the square root of 49 + 121 which is going to
be equal to the square root of 170.
And this is going to be the maximum directional derivative.
So again, the square root of 170 is the maximum directional derivative and it occurs in the
direction, (7,11).
So that’s it.
I hope this video helped you guys and I will see you in the next one.
Bye!