0:00
Hi there.
00:01
In this problem, we're asked to find the directional derivative of this function g at the point 0 .00, the origin, in the direction of this vector v.
00:11
So let's start by noticing that this directional vector v is not a unit vector.
00:15
It's a little too long.
00:17
So let's start by finding a unit vector in the same direction.
00:20
So we'll call that u.
00:21
And the way we'll get it is just by taking v.
00:24
And we'll scale it down by dividing by the length of v.
00:27
So the length of v here, what it is is the square root of the sum of the squares of the components of v.
00:40
In other words, we just need two squared plus one squared plus negative two squared.
00:52
That number times b.
00:56
Well, two squared is four, one squared is one, four plus one is five, plus another four gives us nine.
01:04
Over the square of 9 is the same as one -third.
01:08
That means that this vector here had a length of 3.
01:14
And that means that the vector will actually use for a direction that'll be a unit vector.
01:19
We just divide the components of v by 3.
01:21
So we get 2 thirds, 1 3rd, negative 2 thirds.
01:29
Okay, so now we are ready to find our directional derivative.
01:36
Remember what it's defined as the gradient of g dotted with that unit vector u.
01:42
The gradient of g is defined as the partial derivative of each variable in each component.
01:50
So really we need now next to find the partial derivatives.
01:54
So let's begin by finding the partial of g with respect to x.
01:58
So we're holding y and z constant.
02:04
So if we're looking, this 3 is a constant that stays there.
02:08
The cosine y, z is a constant, and the derivative of e to the x is itself.
02:12
So in this case, the partial derivative of g with respect to x is exactly g.
02:23
Next, partial of g with respect to y.
02:26
So this time, the 3, the e to the x, those are all treated as constants.
02:32
The derivative of cosine y, z with respect to y now.
02:36
So the derivative of cosine is negative sign.
02:41
We keep that interfunction yz.
02:43
And now we need to multiply by the chain rule.
02:46
We need to multiply by the partial derivative of yz.
02:49
With respect to y, which is z.
02:58
Okay.
02:59
Now finally, a partial of g with respect to z...