Directional and partial derivatives help please

[iampaul]

Directional and partial derivatives help please!

I have read that the partial derivative of a function z=f(x,y) :∂z/∂x, ∂z/∂y at the point (x_o,y_o,z_o)are just the tangent lines at (x_o,y_o,z_o) along the planes y=y_o and x=x_o. Directional derivatives were explained to be derivatives at a particular direction defined by the unit vector a^→=cos∅x^λ sin∅y^λand it is given by D_af(x,y)=
∂z/∂xcos∅ + ∂z/∂ysin∅. Partial derivatives were also said to be special cases of directional derivatives when ∅=0° or 90°.

What confuses me is that if ∅=180° or 270°,the directional derivative equals the negative of the partial derivative. The angle was just increasedby 180 so the tangent line remains at the same plane and i think should it not change its value, it is still the same tangent line as before.

 

[theorem4.5.9]

You have answered your own question, so I'm not sure if anything I say will be helpful.

Think of the equation for the tangent line given by the partial derivatives. It's a line parameterized by a single variable, let's say $t$. Now when $t$ increases we need to pick which direction we want to traverse the tangent line. When we rotate by the angels you mentioned, all we are doing is choosing to traverse the tangent line in the opposite direction.

Part of the reason for this is that our math shouldn't be bogged down into which coordinate system we use. Instead of the $x$ and $y$ axes, we may want to use two different vectors to serve as our axes.

 

[Vargo]

The directional derivative tells the rate of change of z as the point in your domain moves at unit speed in the given direction. If you flip your direction of motion (in the domain), then z switches from increasing to decreasing (or vice-versa), which is why the sign of the directional derivative flips. So yes, it is the same tangent line, but you are going "down" along it instead of "up" along it.

 

[iampaul]

thanks a lot :)