00:01
So in this question, we're asked to find an equation of the tangent plane to a surface at a given point.
00:07
Now, you should always try and take pictures in numeraid, because sometimes it's very difficult to see what happens when you cut and paste.
00:15
I believe the surface that you have is f of x, y equals x squared times y plus 1.
00:26
So what do we do to start? well, the equation of a tangent plane is given by z minus the z coordinate equals the partial derivative of f with respect to x at the point x naught, y naught given times the quantity of x minus x naught plus the partial derivative of f with respect to y at the point given times the quantity of y minus y naught.
00:52
So first of all, let's figure out our z naught.
00:56
I'll figure out f of 2 comma negative 1.
00:59
I'm getting 2 squared, that's 4, times negative 1, negative 4, plus 3 gives us negative 3.
01:10
Now i'm going to need my partial derivatives.
01:12
My partial derivative of f with respect to x is just 2xy.
01:18
And if i evaluate that partial derivative of f with respect to x at the point 2, negative 1, i'm getting 2 times 2 times negative 1, that's negative 4.
01:30
While if i find my partial derivative of f with respect to y, what is that? well, the derivative of that first term is just x squared.
01:39
Of course, the plus 1 differentiates out to 0.
01:42
So f sub y at the point 2 comma negative 1, that's just 2 squared, which is 4...