Gradient and Directional Derivative Solved Problems
Learn the directional derivative formula through the same line by line method used in the workbook. Each solution begins with grad phi equal to nabla phi and then follows the unit vector and dot product method.
For \(\phi(x,y,z)\), write the gradient in the complete form
\[ \mathrm{grad}\,\phi=\nabla\phi =\mathbf{i}\frac{\partial\phi}{\partial x} +\mathbf{j}\frac{\partial\phi}{\partial y} +\mathbf{k}\frac{\partial\phi}{\partial z}. \]
If the direction vector is \(\mathbf{a}\), then
\[ \widehat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}. \]
Always convert the given direction vector into a unit vector before taking the directional derivative.
The directional derivative of \(\phi\) in the direction of \(\widehat{\mathbf{a}}\) is
\[ D_{\widehat{\mathbf{a}}}\phi =(\mathrm{grad}\,\phi)\cdot\widehat{\mathbf{a}} =(\nabla\phi)\cdot\widehat{\mathbf{a}}. \]
First find \(\mathrm{grad}\,\phi=\nabla\phi\). Then substitute the given point, find the unit direction vector, and take the dot product.
Note: The gradient points in the direction of maximum increase of \(\phi\).
Find the directional derivative of
at \((1,2,1)\) in the direction \(2\mathbf{i}+2\mathbf{j}+\mathbf{k}\).
Given
\[ \phi(x,y,z)=4x^2+3y-4z. \]Now,
\[ \mathrm{grad}\,\phi=\nabla\phi =\mathbf{i}\frac{\partial\phi}{\partial x} +\mathbf{j}\frac{\partial\phi}{\partial y} +\mathbf{k}\frac{\partial\phi}{\partial z}. \]The partial derivatives are
\[ \frac{\partial\phi}{\partial x}=8x, \qquad \frac{\partial\phi}{\partial y}=3, \qquad \frac{\partial\phi}{\partial z}=-4. \]Therefore,
\[ \mathrm{grad}\,\phi=\nabla\phi =8x\mathbf{i}+3\mathbf{j}-4\mathbf{k}. \]At \((1,2,1)\),
\[ \left.\mathrm{grad}\,\phi\right|_{(1,2,1)} =8\mathbf{i}+3\mathbf{j}-4\mathbf{k}. \]The given direction vector is
\[ \mathbf{a}=2\mathbf{i}+2\mathbf{j}+\mathbf{k}. \]Its magnitude is
\[ |\mathbf{a}|=\sqrt{2^2+2^2+1^2}=3. \]Hence, the unit vector is
\[ \widehat{\mathbf{a}} =\frac{2\mathbf{i}+2\mathbf{j}+\mathbf{k}}{3}. \]The directional derivative is
\[ D_{\widehat{\mathbf{a}}}\phi =(\mathrm{grad}\,\phi)\cdot\widehat{\mathbf{a}}. \]Substituting,
\[ D_{\widehat{\mathbf{a}}}\phi =(8\mathbf{i}+3\mathbf{j}-4\mathbf{k}) \cdot\frac{2\mathbf{i}+2\mathbf{j}+\mathbf{k}}{3}. \]Therefore,
\[ D_{\widehat{\mathbf{a}}}\phi =\frac{16+6-4}{3}=6. \]Find the directional derivative of
at \((1,2,1)\) in the direction \(2\mathbf{i}+2\mathbf{j}+\mathbf{k}\).
Given
\[ \phi(x,y,z)=4x^2+3y-4z. \]Now,
\[ \mathrm{grad}\,\phi=\nabla\phi =\mathbf{i}\frac{\partial\phi}{\partial x} +\mathbf{j}\frac{\partial\phi}{\partial y} +\mathbf{k}\frac{\partial\phi}{\partial z}. \]The partial derivatives are
\[ \frac{\partial\phi}{\partial x}=8x, \qquad \frac{\partial\phi}{\partial y}=3, \qquad \frac{\partial\phi}{\partial z}=-4. \]Therefore,
\[ \mathrm{grad}\,\phi=\nabla\phi =8x\mathbf{i}+3\mathbf{j}-4\mathbf{k}. \]At \((1,2,1)\),
\[ \left.\mathrm{grad}\,\phi\right|_{(1,2,1)} =8\mathbf{i}+3\mathbf{j}-4\mathbf{k}. \]The given direction vector is
\[ \mathbf{a}=2\mathbf{i}+2\mathbf{j}+\mathbf{k}. \]Its magnitude is
\[ |\mathbf{a}|=\sqrt{2^2+2^2+1^2}=3. \]Hence, the unit vector is
\[ \widehat{\mathbf{a}} =\frac{2\mathbf{i}+2\mathbf{j}+\mathbf{k}}{3}. \]The directional derivative is
\[ D_{\widehat{\mathbf{a}}}\phi =(\mathrm{grad}\,\phi)\cdot\widehat{\mathbf{a}}. \]Substituting,
\[ D_{\widehat{\mathbf{a}}}\phi =(8\mathbf{i}+3\mathbf{j}-4\mathbf{k}) \cdot\frac{2\mathbf{i}+2\mathbf{j}+\mathbf{k}}{3}. \]Therefore,
\[ D_{\widehat{\mathbf{a}}}\phi =\frac{16+6-4}{3}=6. \]