Question

Find the tangent plane to F=(1+y^2e^x+z, -yz^2, yzsinx) at (x,y,z)=(0,1,2)
I'm given the div.F=-3.(not sure if this is part of this question, tad messy)
MANY MEDALS AWAIT!!! :)

Answer 1

The gradient of F
\( \bigtriangledown F= <\partial F/\partial x , \partial F/\partial y,\partial F/\partial z >\)
is perpendicular to F. Find the gradient of F, and evaluate it at (0,1,2)
this will provide a vector normal to F at that plane.
the equation for a plane is

\[ N \cdot P = k\]
where N is the normal, P is <x,y,z> and k is a constant
to find k, find the dot product of N and (0,1,2)

Answer 2

to get ▽F=<∂F/∂x,∂F/∂y,∂F/∂z> do i simply derivate with respect to the variable defined in each component. i.e) ∂F/∂x= y^2e^x. When do we use the chain rule then??

Answer 3

yes, that is correct. When you take a partial derivative, you treat all other variables in the expression as constants.
see http://www.khanacademy.org/math/calculus/partial_derivatives_topic/partial_derivatives/v/partial-derivatives
for a video on this

and here is one on the gradient
http://www.khanacademy.org/math/calculus/partial_derivatives_topic/gradient/v/gradient-1