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- Feb 29, 2012

- 342

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.