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Suppose that $f: \mathbb{R}^2 \to \mathbb{R}$ is differentiable at $p$. Also suppose that $D_uf(p)=1$ and $D_vf(p)=1$ where $u=\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $v=\left( \frac{1}{2}, \frac{-\sqrt{3}}{2} \right)$. Compute $\nabla f(p)$.

I know this relationship between the gradient and directional derivative:

Let $u$ be a unit vector and $u=<u_1,u_2>$.
$D_uf(p)=\frac{\partial f}{\partial x_1}(p)\cdot u_1+\frac{\partial f}{\partial x_2}(p)\cdot u_2=<\frac{\partial f}{\partial x_1}(p)\frac{\partial f}{\partial x_2}(p)>\cdot u=\nabla f(p)$

emka
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  • Is $D_uf(p)$ the directional derivative of $f$ at $p$ in the direction of $u$? I am not familiar with your notation. (and similar for $D_vf(p)$) –  Mar 31 '13 at 02:06
  • That is correct. – emka Mar 31 '13 at 02:09

2 Answers2

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This sounds like homework, so I'll give a hint.

Start by letting $a$ and $b$ be the two components of $\nabla f(p)$, and then see what the information that you have says about $a$ and $b$.

Jim Belk
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If $f$ is differentiable at $p$, then we have the derivative matrix $\begin{bmatrix} \textbf{D} f(p) \end{bmatrix} = \begin{bmatrix} a & b \end{bmatrix}$. We are given

$$ \begin{bmatrix} a & b \end{bmatrix} \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} = 1 \ \text{and} \ \begin{bmatrix} a & b \end{bmatrix} \begin{pmatrix} 1/2 \\ -\sqrt{3}/2 \end{pmatrix} = 1 \ . $$

This gives a system of linear equations, which you can solve for $a$ and $b$ to find $\begin{bmatrix} \textbf{D} f(p) \end{bmatrix}$. The gradient $\vec{\nabla}f(p)$ is the transpose of this matrix.

  • I tried to solve this using octave, and I got ugly numbers for a and b. IS there a simpler way to do this that I am missing. – emka Mar 31 '13 at 02:28
  • I would just augment the system into a $2\times 3$ matrix and row reduce by hand to keep the radicals (a little less ugly). I don't have a fancy CAS... –  Mar 31 '13 at 02:31
  • I guess I just assumed this should come out nicely since this was an example problem in my notes (prefaced with the phrase "details left to student"). – emka Mar 31 '13 at 02:33