Directional Derivatives
It simply means (a) find derivative (b) give it direction.
Here derivative is found out by gradient .
And direction is the unit vector.
So to find Directional Derivative of a function we find derivative i.e. gradient \(\longrightarrow\) then find dot product with the unit vector of the vector that will be given in the question.
Greatest Rate of Increase
The greatest rate of increase in a function is always in the Normal direction of the function i.e. we have to find gradient of the function.
Sample Problem 1 :
Find Directional Derivative of \(\phi (x,y,z) = x^2yz + 4xz^2 \) at the point (1,0,3) in the direction of vector \(2\hat{i} - \hat{j} - 2\hat{k} \). What is the greatest rate of increase in \(\phi\) at the point (1,-2,1) ?
Solution :
Given function is \(\phi (x,y,z) = x^2yz + 4xz^2 \)
Derivative of \(\phi = \) grad(\(\phi\)) =
\[\left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)\left(x^2yz + 4xz^2 \right) \]
\[= 2\left(xyz + 2z^2 \right)\hat{i} + \left(x^2z \right)\hat{j} + \left(x^2y + 8xz \right)\hat{k} \]
\[ \text{Let } \vec{A} = 2\left(xyz + 2z^2 \right)\hat{i} + \left(x^2z \right)\hat{j} + \left(x^2y + 8xz \right)\hat{k} \]
At point (1,0,3)
\[ \vec{A} = 2(0 + 18)\hat{i} + 3\hat{j} + (0 + 24)\hat{k} \]
\[ \vec{A} = 36\hat{i} + 3\hat{j} + 24\hat{k} \]
Let the vector given in the question be \(\vec{B}\)
\[\vec{B} = 2\hat{i} - \hat{j} - 2\hat{k} \]
\[\text{Direction of }\vec{B} = \hat{B} = \frac{2\hat{i} - \hat{j} - 2\hat{k}}{\sqrt{4 + 1 + 4}} \]
\[= \frac{2\hat{i} - \hat{j} - 2\hat{k}}{3} \]
Now Directional Derivative of \(\vec{A}\) in the direction of \(\vec{B}\) = \(\vec{A}.\vec{B}\)
\[= (36\hat{i} + 3\hat{j} + 24\hat{k} )\left(\frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} \right) \]
\[ = 24 - 1 - 6 = 7 \]
Greatest rate of increase at point (1,-2,1)
\[ = 2[(1)(-2)(1) + 2]\hat{i} + [1]\hat{j} + [-2 + 8]\hat{k} \]
\[= 0\hat{i} + \hat{j} + 6\hat{k} \]
Sample Problem 2 :
Find Directional Derivative of \(\phi (x,y,z) = xy^2 + yz^3 \) at point (2,-1,1) in the direction of normal to the surface \(x\log{z} - y^2 + 4 = 0 \) at (2,-1,1).
Solution
\(\phi(x,y,z) = xy^2 + yz^3 \)
Derivative of \(\phi\) = gradient(\(\phi\)) =
\[\left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)\left(xy^2 + yz^3 \right) \]
\[= (y^2)\hat{i} + (2xy + z^3)\hat{j} + (3yz^2)\hat{k} \]
At point (2,-1,1)
\[\vec{A} = (-1)^2\hat{i} + [2(2)(-1) + 1]\hat{j} + 3(-1)(1)\hat{k} \]
\[= -\hat{i} - 3\hat{j} - 3\hat{k} \]
\[\text{Let } f = x\log{z} - y^2 + 4 \]
\[\nabla{f} = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)(x\log{z} - y^2 + 4 ) \]
\[= (logz)\hat{i} + (-2y)\hat{j} + \left(\frac{x}{z} \right)\hat{k} \]
At point (2,-1,1)
\[\vec{B} = 0\hat{i} + 2\hat{j} + 2\hat{k} \]
Direction of normal vector
\[\hat{B} = \frac{\vec{B}}{|\vec{B}|} \]
\[ = \frac{0\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{2^2 + 2^2}} \]
\[ \Rightarrow \hat{B} = 0\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k} \]
Directional Derivative
\[\vec{A}.\hat{B} = (\hat{i} - 3\hat{j} - 3\hat{k} ).\left(0\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k} \right) \]
\[= 0 - \frac{3}{\sqrt{2}} - \frac{3}{\sqrt{2}} = \boxed{-3\sqrt{2}} \]
Sample Problem 3 :
If \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) show that \(\nabla r^n = nr^{n-2}\vec{r} \)
Solution :
\[\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \] \[r = |\vec{r}| = \sqrt{x^2 + y^2 + z^2} \] Now L.H.S. = \(\nabla r^n\) = \[\left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right)r^n \] \[= \left(nr^{n-1}\frac{\partial r}{\partial x} \right)\hat{i} + \left(nr^{n-1}\frac{\partial r}{\partial y} \right)\hat{j} + \left(nr^{n-1}\frac{\partial r}{\partial z} \right)\hat{k} \hspace{10pt} \longrightarrow (1) \] \[\text{Now, as } r = \sqrt{x^2 + y^2 + z^2} \] \[\frac{\partial r}{\partial x} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}}\times 2x \] \[\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} = \frac{x}{r} \] Similarly, \[\frac{\partial r}{\partial y} = \frac{y}{r} \] \[\frac{\partial r}{\partial z} = \frac{z}{r} \] Putting values in equation (1) \[ \left(nr^{n-1}\frac{x}{r} \right)\hat{i} + \left(nr^{n-1}\frac{y}{r} \right)\hat{j} + \left(nr^{n-1}\frac{z}{r} \right)\hat{k} \] \[ = nr^{n-2}(x\hat{i} + y\hat{j} + z\hat{k}) \] \[ = nr^{n-2}\vec{r} \] = R.H.S., Hence Proved. :)
Angle between surfaces :
Angle between surfaces will be equal to the angle between the normals of these surfaces, so find normals and then find the angle between the normals.
Sample Problem 4 :
Find the angle between surfaces \(x^2 + y^2 + z^2 = 9 \) and \(z = x^2 + y^2 - 3 \) at the point (2,-1,2).
Solution :
Equation of first surface
\[x^2 + y^2 + z^2 = 9 \]
\[\Rightarrow x^2 + y^2 + z^2 - 9 = 0 \]
\[\text{Let } f = x^2 + y^2 + z^2 - 9 \hspace{10pt} \longrightarrow (1) \]
Equation of second surface
\[z = x^2 + y^2 - 3 \]
\[\Rightarrow x^2 + y^2 - z - 3 = 0 \]
\[\text{Let } \phi = x^2 + y^2 - z - 3 \hspace{10pt} \longrightarrow (2) \]
Now, Normal vector to the first surface,
\[\vec{N}_1 = \nabla f \]
\[= \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)(x^2 + y^2 + z^2 - 9) \]
\[\Rightarrow \vec{N}_1 = 2x\hat{i} + 2y\hat{j} + 2z\hat{k} \]
Normal vector to the second surface,
\[\vec{N}_2 = \nabla \phi \]
\[= \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)(x^2 + y^2 - z - 3) \]
\[\Rightarrow \vec{N}_2 = 2x\hat{i} + 2y\hat{j} - \hat{k} \]
At point (2,-1,2)
\[\vec{N}_1 = 4\hat{i} - 2\hat{j} + 4\hat{k} \]
\[\vec{N}_2 = 4\hat{i} - 2\hat{j} - \hat{k} \]
Now to find angle between normals,
\[ \vec{N}_1.\vec{N}_2 = |\vec{N}_1||\vec{N}_2|\cos{\theta} \]
\[ \theta = \cos^{-1}\frac{\vec{N}_1.\vec{N}_2}{|\vec{N}_1||\vec{N}_2|} \]
\[ = \cos^{-1}{\left(\frac{16 + 4 - 4}{\sqrt{4^2 + (-2)^2 + (4)^2}\sqrt{4^2 + (-2)^2 + (-1)^2}} \right) }\]
\[ \theta = \cos^{-1}\left(\frac{8}{3\sqrt{21}} \right) \]
Here we were given two surfaces at one point so we had two normals. You can also be given only one surface and two points so that you have to find two normal vectors at those two different points.
Sample Problem 6 :
Find constants \(a\) and \(b\) so that the surface \(ax^2 - byz = (a + 2)x \) is orthogonal to the surface \(4x^2y + z^3 = 4 \) at point (1,-1,2).
Solution :
Othogonal means perpendicular.
Equation of first surface is
\[ ax^2 - byz = (a + 2)x \]
\[ ax^2 - (a + 2)x - byz = 0 \]
\[ \text{Let } f = ax^2 - (a + 2)x - byz \hspace{10pt} \longrightarrow (1) \]
Equation of second surface
\[ 4x^2y + z^3 = 4 \]
\[ 4x^2y + z^3 - 4 = 0 \]
\[ \text{Let } \phi = 4x^2y + z^3 - 4 \hspace{10pt} \longrightarrow (2) \]
Normal vector of the first surface
\[\vec{N}_1 = \nabla f \]
\[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)[ax^2 - (a + 2)x - byz] \]
\[\Rightarrow \vec{N}_1 = (2ax - a - 2)\hat{i} - bz\hat{j} - by\hat{k} \]
Normal vector to the second surface
\[ \vec{N}_2 = \nabla f \]
\[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)(4x^2y + z^3 - 4) \]
\[\Rightarrow \vec{N}_2 = 8xy\hat{i} + 4x^2\hat{j} + 3z^2\hat{k} \]
At point (1,-1,2)
\[ \vec{N}_1 = (a-2)\hat{i} - 2b\hat{j} + b\hat{k} \]
\[ \vec{N}_2 = -8\hat{i} + 4\hat{j} + 12\hat{k} \]
Since both surfaces are Othogonal
\[ \vec{N}_1.\vec{N}_2 = 0 \]
\[ \Rightarrow [(a-2)\hat{i} - 2b\hat{j} + b\hat{k}].[-8\hat{i} + 4\hat{j} + 12\hat{k}] \]
\[ \Rightarrow (a-2)(-8) + (-2b)(4) + (b)(12) = 0 \]
\[ \Rightarrow 2a - b = 4 \hspace{10pt} \longrightarrow (3) \]
Here as we have two variables \(a\) and \(b\), so we need two equations. Second we will form by using given point. Since the given point lies on both the surfaces so it will satisfy equations of both the surfaces.
Putting point in equation (1)
\[ a(1)^2 - b(-1)(2) = (a+2)1 \]
\[ \Rightarrow \boxed{b = 1} \]
Putting \(b\) in euqation (3)
\[ 2a - 1 = 4 \]
\[ \Rightarrow \boxed{a= \frac{5}{2}} \]
