MHB How to Find the Directional Derivative of a Function at a Given Point?

[Fantini]

Here's the problem:

Find the directional derivative of the function $V(x,y) = x^3 -3xy +4y^2$ at the point $P(2,1)$ in the direction of the unit vector $\vec{u}$ given by the angle $\theta = \pi /6$.

So I found the gradient of the function $$\nabla V(x,y) = (3x^2 -3y, 8y -3x),$$ and then at the point desired: $$\nabla V(2,1) = (3(2)^2 -3(1), 8(1) -3(2)) = (12-3, 8-6) = (9,2).$$ Single he gave me the angle, I can find it using the scalar product definition: $D_u V(2,1) = | \nabla V(2,1) | \cdot \cos \theta$, which means $$D_u V(2,1) = \sqrt{ 9^2 +2^2} \cdot \frac{\sqrt{3}}{2} = \sqrt{85} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{255}}{2}.$$

The problem is that the answer is $D_u V(2,1) = 1 + 4.5 \sqrt{3}$ which is simply multiplying each of the gradient's coordinate by the cosine and summing. I disagree with it entirely; if the vector wasn't given, there's no way to do the scalar product by multiplying and summing the coordinates, so we have to use the other form.

 

[Mathwebster]

Here's the problem:

Find the directional derivative of the function $V(x,y) = x^3 -3xy +4y^2$ at the point $P(2,1)$ in the direction of the unit vector $\vec{u}$ given by the angle $\theta = \pi /6$.

So I found the gradient of the function $$\nabla V(x,y) = (3x^2 -3y, 8y -3x),$$ and then at the point desired: $$\nabla V(2,1) = (3(2)^2 -3(1), 8(1) -3(2)) = (12-3, 8-6) = (9,2).$$ Single he gave me the angle, I can find it using the scalar product definition: $D_u V(2,1) = | \nabla V(2,1) | \cdot \cos \theta$, which means $$D_u V(2,1) = \sqrt{ 9^2 +2^2} \cdot \frac{\sqrt{3}}{2} = \sqrt{85} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{255}}{2}.$$

The problem is that the answer is $D_u V(2,1) = 1 + 4.5 \sqrt{3}$ which is simply multiplying each of the gradient's coordinate by the cosine and summing. I disagree with it entirely; if the vector wasn't given, there's no way to do the scalar product by multiplying and summing the coordinates, so we have to use the other form.

I believe that the unit vector you look for is

$\vec{u} = < \cos \theta, \sin \theta>$

BTW - with your definition $\theta$, $\theta$ is the angle between $\nabla V(2,1) = <9,2>$ and the unit vector $<\dfrac{\sqrt{3}}{2},\dfrac{1}{2}>$ and not $\theta = \dfrac{\pi}{6}$

 

[Reckoner]

Single he gave me the angle, I can find it using the scalar product definition: $D_u V(2,1) = | \nabla V(2,1) | \cdot \cos \theta$, which means $$D_u V(2,1) = \sqrt{ 9^2 +2^2} \cdot \frac{\sqrt{3}}{2} = \sqrt{85} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{255}}{2}.$$

What's going on here? The direction vector is \(\mathbf{u} = \cos\theta\,\mathbf{i} + \sin\theta\,\mathbf{j} = \frac{\sqrt3}2\mathbf{i} + \frac12\mathbf{j}\), so

\[D_{\mathbf u}V(x, y) = \nabla V(x,y)\cdot\mathbf{u}\]

\[=\left(3x^2-3y\right)\cdot\frac{\sqrt3}2+\left(8y-3x\right)\cdot\frac12\]

\[\Rightarrow D_{\mathbf u}V(2,1) = \frac{9\sqrt3}2 + 1.\]

 

[Fantini]

I supposed I misread the question then. What I did was assume he spoke of a general unit vector giving the angle between it and the gradient as $\theta = \pi/6.$ My bad then. Thanks for clearing it up! :D

 

[HallsofIvy]

A "general unit vector giving the angle between it and the gradient as \theta = \pi/6"?
There is only one such vector!