Temperature Gradient Questions

[Mindscrape]

Is there a direction A in which the rate of change of the temperature function $$T(x,y,x)=2xy - yz$$ at P(1,-1,1) is -3ºC/ft? Give reasons for your answer.

For this problem I found the gradient of at the point P. So
$$\nabla f = 2y|_p \mathbf{i} + 2x-z|_p \mathbf{j} - y|_p \mathbf{k}$$
which I then found was
$$\nabla f = -2 \mathbf{i} + 1 \mathbf{j} + 1 \mathbf{k}$$
Then we want to know if the gradient, in the direction of A will be 3ºC, so
$$\nabla f \cdot \frac{A}{|A|} = \frac{3º}{ft}$$
which means that we want an x, y, z, such that
$$\frac{-2x +y +z}{\sqrt{x^2 + y^2 + z^2}} = \frac{3º}{ft}$$
So I believe this is right so far, but I don't think this eqn can be solved for.

Here is the other one:
The Celsius temperature in a region in space is given by $$T(x,y,z) = 2x^2 -xyz$$. A particle is moving in this region and its position at time t is given by $$x=2t^2$$, $$y=3t$$, $$z=-t^2$$, where time is measured in seconds and distanace in meters.
a) How fast is the temp experienced by the particle changing in ºC/m when the particle is at the point P(8,6,-4)?
b) How fast is the temp experienced by the particle changing in ºC/sec at P?

So, using the chain rule
$$\frac{dT}{dt} = (4x-yz)4t \mathbf{i} - (xz)3 \mathbf{j} + (xy)2t \mathbf{k}$$
For the change per meter I will just solve for one of the t eqns, and then substitute back into the equation to find the gradient?
Then for the change per second, it will be the opposite?

 

[Mindscrape]

Nevermind, I figured them out.