The directional derivative of a scalar point function \( \phi (x,y,z) \) in the direction of a non-unit vector \( \vec{a} \) is:
\[ \nabla \phi \cdot \frac{\vec{a}}{|\vec{a}|} \]
This represents the rate of change of \( \phi \) in the direction of \( \vec{a} \).
The directional derivative of a scalar point function \( \phi (x,y,z) \) is maximum in the direction of \( \nabla \phi \).
The maximum directional derivative of \( \phi \) is \( |\nabla \phi| \).
- The directional derivative is maximal in the direction of the gradient vector.
- It is zero in directions perpendicular to the gradient.
- The directional derivative in direction \( \vec{u} \) can be expressed as: \[ D_{\vec{u}}\phi = \nabla \phi \cdot \vec{u} \] where \( \vec{u} \) is a unit vector.
Example 1: Find the directional derivative of \( \phi(x,y,z) = x^{2}yz + 4xz^{2} \) at the point \( (1,-2,-1) \) in the direction of the normal to the surface \( f(x,y,z) = x\log z - y^{2} \) at \( (-1,2,1) \).
Solution:
\( \phi(x,y,z) = x^{2}yz + 4xz^{2} \) at \( (1,-2,-1) \)
\( f(x,y,z) = x\log z - y^{2} \) at \( (-1,2,1) \)
\[ \nabla \phi = \frac{\partial \phi}{\partial x}\hat{i} + \frac{\partial \phi}{\partial y}\hat{j} + \frac{\partial \phi}{\partial z}\hat{k} = (2xyz + 4z^{2})\hat{i} + (x^{2}z)\hat{j} + (x^{2}y + 8xz)\hat{k} \]
At \( (1,-2,-1) \):
\[ \nabla \phi_{(1,-2,-1)} = [2(1)(-2)(-1) + 4(-1)^2]\hat{i} + [1^2(-1)]\hat{j} + [1^2(-2) + 8(1)(-1)]\hat{k} \]
\[ = (4 + 4)\hat{i} - \hat{j} + (-2 - 8)\hat{k} = 8\hat{i} - \hat{j} - 10\hat{k} \]
\[ \nabla f = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k} = \log z\hat{i} - 2y\hat{j} + \frac{x}{z}\hat{k} \]
At \( (-1,2,1) \):
\[ \nabla f_{(-1,2,1)} = \log(1)\hat{i} - 2(2)\hat{j} + \frac{-1}{1}\hat{k} = 0\hat{i} - 4\hat{j} - \hat{k} \]
Let \( \vec{a} = \nabla f_{(-1,2,1)} = -4\hat{j} - \hat{k} \)
Magnitude: \( |\vec{a}| = \sqrt{(-4)^2 + (-1)^2} = \sqrt{17} \)
\[ D_{\vec{u}}\phi = \nabla \phi \cdot \frac{\vec{a}}{|\vec{a}|} = (8\hat{i} - \hat{j} - 10\hat{k}) \cdot \frac{(-4\hat{j} - \hat{k})}{\sqrt{17}} \]
\[ = \frac{(8)(0) + (-1)(-4) + (-10)(-1)}{\sqrt{17}} = \frac{0 + 4 + 10}{\sqrt{17}} = \frac{14}{\sqrt{17}} \]
Example 2: The temperature of points in space is given by \( T(x,y,z) = x^{2} + y^{2} - z \). A mosquito located at \( (1,1,2) \) desires to fly in such a direction that it will get warm as soon as possible. In what direction should it move?
Solution: The temperature increases most rapidly in the direction of the gradient vector \( \nabla T \).
\[ \nabla T = \frac{\partial T}{\partial x}\hat{i} + \frac{\partial T}{\partial y}\hat{j} + \frac{\partial T}{\partial z}\hat{k} = 2x\hat{i} + 2y\hat{j} - \hat{k} \]
\[ \nabla T_{(1,1,2)} = 2(1)\hat{i} + 2(1)\hat{j} - \hat{k} = 2\hat{i} + 2\hat{j} - \hat{k} \]
\[ \text{Unit vector} = \frac{\nabla T}{|\nabla T|} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{9}} = \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \]
Conclusion: The mosquito should fly in the direction of the gradient vector to get warm as quickly as possible.
Model problems on Directional Derivatives
- Find the directional derivative of \(\phi = x^{2}y^{2}z^{2}\) in the direction of the vector \(\hat{i}+2\hat{j}+2\hat{k}\) at \((2,1,-1)\)
Answer: \(\frac{4}{3}\) - Find the directional derivative of \(f(x,y,z)=4e^{2x-y+z}\) at the point \(P(1,1,-1)\) in the direction towards \(PQ\) where \(Q(-3,5,6)\)
Answer: \(\frac{-20}{9}\) - Find the directional derivative of \(\phi =5x^{2}y-5y^{2}z+2.5z^{2}x\) at the point \((1,1,1)\) in the direction of the line \(\frac{x-1}{2}=\frac{y-3}{-2}=z\).
Answer: \(\frac{35}{3}\) - Find the directional derivative of \(\phi =xy^{2}+yz^{3}\) at the point \((2,-1,1)\) in the direction of the normal to the surface \(x\log z-y^{2}=-4\) at \((-1,-2,1)\)
Answer: \(\frac{-9}{\sqrt{17}}\) - Find the greatest rate of increase of \(u=xyz^{2}\) at the point \((1,0,3)\)
Answer: \(9\) - Find the values of constants \(a\), \(b\), \(c\) so that the maximum value of the directional derivative of \(\phi =axy^{2}+byz+cx^{3}z^{2}\) at \((1,2,-1)\) has a maximum magnitude 64 in the direction parallel to the \(z\)-axis.
Answer: \(a=6\), \(b=24\), \(c=-8\) - Find the directional derivative of \(\phi =x^{4}+y^{4}+z^{4}\) at the point \((1,-2,1)\) in the direction \(AB\) where \(B\) is \((2,6,-1)\).
Answer: \(\frac{-260}{\sqrt{69}}\) - In what direction from \((3,1,-2)\) is the directional derivative of \(\phi =x^{2}y^{2}z^{4}\) maximum? Find also the magnitude of this maximum.
Answer: \(96\sqrt{19}\)