Finding the Directional Derivatives of f(x,y)=x^2 + sin(xy) at (1,0)

[Punkyc7]

find the directions in which the directional derivative of f(x,y)=x^2 + sin(xy) has a value of 1 at the point (1,0)

Fx=2x+ycos(xy)=2
Fy=xcos(xy)=1

So we have <2,1> and we need to find vectors that dotted with <2,1> =1
<2,1>.<x1,x2>=1

2x1+x2=1

So whn x1 is 0 we have x2 is 1

so one of the directions is <0,1>

im not sure how to find the other

 

[lanedance]

so you also know it shoudl be a unit vector
so
<2,1>.<x1,x2>=1
and
<x1,x2>.<x1,x2>=1

 

[lanedance]

note however that by geometrical reasoning the directional derivative can take the same value for at most 2 directions (barring the trivial case)...

so you could find the other (if it exists) by geometrical/symmetrical reasoning, noting the gradient is the direction of maximum rate of change...

 

[Stephen Tashi]

If "directions" means to use unit vectors, you should include
\sqrt{ x_1^2 + x_2^2} = 1
as a condition and solve two simultaneous equations for x_1 and x_2.

 

[lanedance]

however as mentioned, the vectors will be symmetric around the gradient direction <2,1>, so you coudl probably draw <2,1> and <0,1> on a graph and read off the symmetric vector to <0,1>

 

[HallsofIvy]

find the directions in which the directional derivative of f(x,y)=x^2 + sin(xy) has a value of 1 at the point (1,0)

Fx=2x+ycos(xy)=2
Fy=xcos(xy)=1

So we have <2,1> and we need to find vectors that dotted with <2,1> =1
<2,1>.<x1,x2>=1

2x1+x2=1

So whn x1 is 0 we have x2 is 1

so one of the directions is <0,1>

im not sure how to find the other

There are, of course, an infinite number of solutions. As others have said, requiring that x_1^2+ x_2^2= 1 reduces to only two solutions.

If you want a formal method, rewrite 2x_1+ x_2= 1 as 2x_2= 1- x_1 and square both sides: 4x_2^2= 1- 2x_1+ x_1^2. Now, since x_2^2= 1- x_1^2 we can write that as 4(1- x_1^2)= 4- 4x_1^2= 1- 2x_1+ x_1^2 which reduces to the quadratic equation 5x_1^2- 2x_1- 3= (x_1- 1)(5x_1+ 3)= 0.

 

[lanedance]

halls is right as always

that said I would definitely take some time to understand the goemetric argument i gave as that really helped me to understand a directional derivative is just a dot prouct of the gradient with a unit direction vector.

So for example straight away you can say:
- the gradient direction has the maximum magnitude directional derivative
- perpindicular to the gradient the directional derivative is zero

 

[Punkyc7]

Thanks hallsofivy that makes sense