I forgot directional derivatives
by EpicSkills32, Oct 2, 2016, 6:37 AM
I learned this less than a year ago but now it's so cryptic for me. . .
Alright, say I have some weird 3d function, and I want to find the rate of change in the function at a certain point in the direction of another point. To do this, I'll have to use the directional derivative.
Alright we have this function:
![\[ f(x,y)=3xy^2-4x^2y \]](//latex.artofproblemsolving.com/7/d/7/7d77747d7e293284009d911ac1b8f45089bd0379.png)
What would be the rate of change at the point 
in the direction of 
(or whatever the output is there)?
We'll first need to compute the gradient.
gives us the rate of change in specific directions, sort of like multiple derivatives in 2d.
The gradient is made up of partial derivatives, or rates of changes in just one direction respectively. It is a vector.
![\[ \vec{\nabla} f(x,y) = < \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y} > \]](//latex.artofproblemsolving.com/6/5/1/6511b7016824d6b2cc29f0322ab3f75e485a0ab9.png)
![\[ \vec{\nabla} f(x,y) = < 3y^2-8xy , 6xy-4x^2 > \]](//latex.artofproblemsolving.com/0/a/2/0a26ae52ae779b3217fdddefe0555d64ddd7e910.png)
We evaluate our gradient at the point we're looking at:
![\[ \vec{\nabla} f(2,1) = < -13,-4> \]](//latex.artofproblemsolving.com/4/4/6/446d4af5cdfcbe9267d99deba21e98f4906b998b.png)
These values show what our rate of change looks like in their respective directions. To find the rate of change in the direction that we want, we'll take the dot product of this vector with a unit vector that corresponds to our (second) point.
We want to know our rate of change in the direction. We'll need a unit vector so that we're just multiplying by one (essentially).
The point corresponds to the vector 
For our unit vector, we'll take 
The unit vector pointing in the direction of our point would be:
Finally, to calculate the directional derivative, we go:
![\[ D_{<3,-4>}f(2,1) = \vec{\nabla} f(2,1) \cdot \vec{u} \]](//latex.artofproblemsolving.com/8/6/2/862d04a10211a71813c2b08bcb5b379cecd0ff4f.png)
![\[ = <-13,-4> \cdot < \dfrac{3}{5},-\dfrac{4}{5} > \]](//latex.artofproblemsolving.com/6/d/a/6da8ea1585a80a52c8cdc5a0261445befca88445.png)
![\[ = -\dfrac{39}{5} + \dfrac{16}{5} = -\dfrac{23}{5} \]](//latex.artofproblemsolving.com/a/3/b/a3bf2afac82265cdfd6a8474e2d623feeab6c13d.png)
![\[ = \boxed{-\dfrac{23}{5}} \]](//latex.artofproblemsolving.com/b/e/3/be3bb89a636add342cb8d0c752d4d66513d45de8.png)
Alright, say I have some weird 3d function, and I want to find the rate of change in the function at a certain point in the direction of another point. To do this, I'll have to use the directional derivative.
Alright we have this function:
![\[ f(x,y)=3xy^2-4x^2y \]](http://latex.artofproblemsolving.com/7/d/7/7d77747d7e293284009d911ac1b8f45089bd0379.png)
What would be the rate of change at the point 
in the direction of 
(or whatever the output is there)?We'll first need to compute the gradient.
gives us the rate of change in specific directions, sort of like multiple derivatives in 2d.The gradient is made up of partial derivatives, or rates of changes in just one direction respectively. It is a vector.
![\[ \vec{\nabla} f(x,y) = < \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y} > \]](http://latex.artofproblemsolving.com/6/5/1/6511b7016824d6b2cc29f0322ab3f75e485a0ab9.png)
![\[ \vec{\nabla} f(x,y) = < 3y^2-8xy , 6xy-4x^2 > \]](http://latex.artofproblemsolving.com/0/a/2/0a26ae52ae779b3217fdddefe0555d64ddd7e910.png)
We evaluate our gradient at the point we're looking at:![\[ \vec{\nabla} f(2,1) = < -13,-4> \]](http://latex.artofproblemsolving.com/4/4/6/446d4af5cdfcbe9267d99deba21e98f4906b998b.png)
These values show what our rate of change looks like in their respective directions. To find the rate of change in the direction that we want, we'll take the dot product of this vector with a unit vector that corresponds to our (second) point.
We want to know our rate of change in the direction
. We'll need a unit vector so that we're just multiplying by one (essentially).The point
corresponds to the vector 
For our unit vector, we'll take 
The unit vector pointing in the direction of our point would be:
Finally, to calculate the directional derivative, we go:
![\[ D_{<3,-4>}f(2,1) = \vec{\nabla} f(2,1) \cdot \vec{u} \]](http://latex.artofproblemsolving.com/8/6/2/862d04a10211a71813c2b08bcb5b379cecd0ff4f.png)
![\[ = <-13,-4> \cdot < \dfrac{3}{5},-\dfrac{4}{5} > \]](http://latex.artofproblemsolving.com/6/d/a/6da8ea1585a80a52c8cdc5a0261445befca88445.png)
![\[ = -\dfrac{39}{5} + \dfrac{16}{5} = -\dfrac{23}{5} \]](http://latex.artofproblemsolving.com/a/3/b/a3bf2afac82265cdfd6a8474e2d623feeab6c13d.png)
![\[ = \boxed{-\dfrac{23}{5}} \]](http://latex.artofproblemsolving.com/b/e/3/be3bb89a636add342cb8d0c752d4d66513d45de8.png)
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