00:01
To find the equation of the tangent plane to this surface, let's first write the equation for this surface in the form f of x y z is equal to zero.
00:14
Okay so i'm just going to take this term and subtract it on both sides so we have f of x y z is equal to e to the z y power plus x z squared minus 6 x y to the fourth times z cubed.
00:32
Okay which is equal to zero.
00:35
Then the equation of the tangent plane to the surface at this point has the form partial derivative of f with respect to x evaluated at the point negative one zero one times x minus the x coordinate of the point which is negative one plus the partial derivative of f with respect to y evaluated at the point multiplied by y minus the y coordinate of the point which is zero plus the partial derivative with respect to z evaluated at the point multiplied by z minus the z coordinate of the point which is one.
01:14
Okay and this is equal to zero.
01:18
So let's solve this.
01:20
We have zero is equal to the partial derivative of f with respect to x is z squared minus 6 y to the fourth z cubed.
01:32
Okay so we need to evaluate this at the point negative one zero one.
01:39
Okay and then we're multiplying by x plus one.
01:43
Next we have the partial derivative with respect to y.
01:46
This is equal to z times e to the z y power minus six times four which is 24 x y cubed z cubed.
01:57
Okay again evaluated at the point multiplied by y and then we have the partial derivative with respect to z which is y times e to the z y power plus 2 x z minus three times six which is 18 x y to the fourth z squared evaluated at the point multiplied by z minus one.
02:32
Okay so we have zero is equal to let's plug in the values of the point into this partial derivative...