00:01
In this question, where are asked to find the equations of the tangent plane and the normal line at the given point? let f capital of x, y, z be equal to yz minus 9 ln of x plus z.
00:19
Then the equation of the tangent plane is the normal vector of the gradient of f at 0 .01.
00:30
Dot product x y and z minus 1 equals to 0.
00:37
This is the equation of the tangent plane.
00:40
Now we need to find the gradient of the function f.
00:44
And the gradient is a vector with the components f capital x, f capital y, and f capital z.
00:58
So the first component is the derivative of f capital with respect to x.
01:05
And the derivative of f capital with respect to x is going to be yz cancels because it doesn't depend on x.
01:14
And the derivative of ln equals to 1 over x plus z, multiplied by the derivative of x plus z with respect to x, which is 1.
01:24
So it's going to be negative 9 over x plus z.
01:28
The y coordinate is going to be, so the derivative of f with respect to y equals to z.
01:41
And ln gets cancelled because it doesn't depend on y.
01:46
Finally, the derivative of f capital with respect to z equals to y minus the derivative of ln minus 9 times 1 over x plus z multiplied by the derivative of x plus z with respect to z and that's going to be 1.
02:07
So it's y minus 9 over x plus z...