Directional Derivatives Help

[Joe20]

Find the directions in which the directional derivative of f(x,y) = x^2+ xy^3 at the point (2,1) has the value of 2.

What I have done so far which I am not sure how to continue:

partial derivative of fx = 2x + y^3 and fy = 3xy^2

gradient vector, <fx,fy> at (2,1) = <5,6>

Let u = <a,b>

Directional derivative at (2,1) = gradient vector at (2,1) . u => <5,6> . <a, b> = 5a+6b = 2

Hope someone can advise. Thanks.

 

[Opalg]

Find the directions in which the directional derivative of f(x,y) = x^2+ xy^3 at the point (2,1) has the value of 2.

What I have done so far which I am not sure how to continue:

partial derivative of fx = 2x + y^3 and fy = 3xy^2

gradient vector, <fx,fy> at (2,1) = <5,6>

Let u = <a,b>

Directional derivative at (2,1) = gradient vector at (2,1) . u => <5,6> . <a, b> = 5a+6b = 2

Hope someone can advise. Thanks.

Instead of taking $\mathbf{u} = \langle a,b\rangle$, it might be better to take $\mathbf{u} = \langle \cos\theta,\sin\theta\rangle$, so that $\mathbf{u}$ is a unit vector making an angle $\theta$ with the $x$-axis. The equation then becomes $5\cos\theta + 6\sin\theta = 2.$ Do you know how to solve that sort of trigonometric equation?

 

[Joe20]

Instead of taking $\mathbf{u} = \langle a,b\rangle$, it might be better to take $\mathbf{u} = \langle \cos\theta,\sin\theta\rangle$, so that $\mathbf{u}$ is a unit vector making an angle $\theta$ with the $x$-axis. The equation then becomes $5\cos\theta + 6\sin\theta = 2.$ Do you know how to solve that sort of trigonometric equation?

May I ask why are we using cosine teta and sine teta? Yes , using R-formulae to solve for the teta.

 

[HOI]

May I ask why are we using cosine teta and sine teta? Yes , using R-formulae to solve for the teta.

Opalg told you that: u is a unit vector making an angle θ with the x-axis. The problem asked for the direction and giving that angle is a good way to designate the x- direction.

The only difficulty with your method is that there are an infinite number of vectors pointing in a given direction, all with different lengths. When you took such a vector to be <a, b> you arrived at the equation 5a+ 6b= 2, a single equation in two variables. Given different values for a, b, satisfying that equation you get different vectors, of different lengths, all pointing in the same direction. If you want a single vector as answer, you have to impose some other condition. If you require that the vector be of unit length, that a^2+ b^2= 1, then you will have two equations in a and b and will get opalg's answer. But you could also simply require that a, the x-component of the vector be equal to -2 and have the equation 5(-2)+ 6b= -10+ 6b= 2 and have 6b= 10= 2= 12 so b= 2. The vector <-2, 2> is a vector in the correct direction.