INSTRUCTOR: Calculate D sub u of f of x comma y comma z and D sub u of f of 0 comma negative 2 comma 5 in the direction of the vector v equal to negative 3i plus 12j minus 4k for the function f of x comma y comma z equal to 3x squared plus xy minus 2y squared plus 4yz minus z squared plus 2xz. To begin, we want the direction of the vector v. So let's first go ahead and find its magnitudes. The magnitude of vector v is equal to the square root of parentheses negative 3 close parentheses squared plus 12 squared plus open parentheses negative 4 close parentheses squared. And it's going to be the square root of 169, which is equal to 13, and we need a unit vector in that same direction. So the vector v divided by the magnitude of vector v will give us a unit vector and that will be equal to negative 3 over 13i plus 12 over 13j minus 4 over 13k.

This is a unit vector in the same direction as v and it also gives us some other information. So we can say that cosine alpha is equal to negative 3 13ths from the i component cosine of beta is equal to 12 13ths from the j component and cosine gamma equals negative 4 13ths from the k component. Now, D sub u of f of x comma y comma z by definition is f sub x multiplied by cosine alpha plus f sub y multiply by cosine of beta plus f sub z multiply by cosine of gamma. So we're going to go ahead and take our partial. In fact, I'll just write these over here.

So F sub x is equal to 6x plus y plus 2z. fy is equal to x minus 4y plus 4z. And F sub z is equal to 4y minus 2z plus 2x. So putting this together, D sub u of f of x comma y comma z is equal to negative 3 over 13 open parentheses 6x plus y plus 2z plus 12 over 13 open parenthesis x minus 4y plus 4z close parentheses minus 4 over 13 open parentheses 4y minus 2z plus 2x close parentheses. And we could potentially combine like terms and things.

But all I really care about at the moment is finding D sub u for this function and evaluating it at the point 0 comma negative 2 comma 5. So D sub u of f of 0 comma negative 2 comma 5. Evaluating that at all the values we know, that is 384 over 13.

And I think it's worth noting here actually that there is a notational error in what I have because we're finding D sub u, that would be the direction of derivative in the direction of u, but my vector is actually vector v. So I probably should go back and fix that. I'm going to leave it up to you to understand the difference that we adapt our statements to the vectors that we actually have the names of those vectors. Might even just be easier to say vector v and rename it as you.