Differential Operators Solved Problems
Example. 1. Find the directional derivative of f = xy + yz + zx in the direction of the vector i + 2j + 2k at the point (1, 2, 0)
Solution:
Given f = xy + yz + zx
grad f \(\nabla \mathbf{f}=\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\)
= (y + z) i + (z + x) j + (x + y) k
If e is the unit vector in the direction of the vector i + 2 j + 2k, then
⇒ \(\mathbf{e}=-\frac{\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}}{\sqrt{\left(1+2^2+2^2\right)}}=\frac{1}{3}(\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}) \)
The directional derivative = e.∇f
= \(\frac{1}{3}(i+2 j+2 k) \cdot[(y+z) \mathbf{i}+(z+x) \mathbf{j}+(x+y) \mathbf{k}]=\frac{1}{3}(4 x+3 y+3 z)=\frac{10}{3} \text { at }(1,2,0)\)
Gradient And Level Surface Solved Examples
Example. 2. Find the directional derivative of the function xy2+ yz2 +zx2 along the tangent to the curve x =t1 y = t2,z = t3 at the point (1, 1, 1).
Solution:
Given
xy2+ yz2 +zx2
Here f = xy2+ yz2 + zx2
⇒ \(\begin{equation} \nabla f=\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}=\left(y^2+2 x z\right) \mathbf{i}+\left(z^2+2 x y\right) \mathbf{j}+\left(x^2+2 y z\right) \mathbf{k} \end{equation}\)
= 3 (i + j + k) at (1,1,1)
Let r be the position vector of the curve x = t1
y = t2 z = t3 , then r = t1i + t2j + t3k
⇒ \(\begin{equation} \frac{d \mathbf{r}}{d t}=\mathbf{i}+2 t \mathbf{j}+3 t^2 \mathbf{k} \end{equation}\) = (i+2j + 3k) at (1,1,1)
dr/dt is the vector along the tangent to the curve.
The unit vector along the tangent is T
⇒ \( \begin{equation} \mathbf{T}=\frac{\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}{\sqrt{(1+4+9)}}=\frac{1}{\sqrt{14}}(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}) \end{equation}\)
The directional derivative along the tangent
⇒ \(\begin{equation} =\mathbf{T} \cdot \nabla f=\frac{1}{\sqrt{14}}(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}) \cdot 3(\mathbf{i}+\mathbf{j}+\mathbf{k})=\frac{3}{\sqrt{14}}(1+2+3)=\frac{18}{\sqrt{14}} \end{equation}\)
Example. 3. Find the directional derivative of the function f = x² – y² + 2z² at the pt P(l, 2, 3) in the direction of the line PQ where Q = (5, 0,4) 2
Solution:
Given
f = x² – y² + 2z² at the pt P(l, 2, 3) in the direction of the line PQ where Q = (5, 0,4) 2
The position vectors of P and Q with respect to the origin O are
OP =i + 2j + 3k and OQ = 5i + 4k PQ = OQ (or) OP = 4i- 2j + k
If e is the unit vector in the direction of PQ then e = \( \begin{equation}\frac{4 \mathbf{i}-2 \mathbf{j}+\mathbf{k}}{\sqrt{4^2+2^2+1^2}}=\frac{1}{\sqrt{21}}(4 \mathbf{i}-2 \mathbf{j}+\mathbf{k})\end{equation}\)
Again grad f = \(\begin{equation} \mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}=2 x \mathbf{i}-2 y \mathbf{j}+4 z \mathbf{k} \end{equation} \)
The directional derivative = e.∇f
⇒ \(\begin{equation} \frac{1}{\sqrt{21}}(4 \mathbf{i}-2 \mathbf{j}+\mathbf{k}) \cdot(2 x \mathbf{i}-2 y \mathbf{j}+4 z \mathbf{k})=\frac{1}{\sqrt{21}}(8 x+4 y+4 z) \end{equation}\)
⇒ \(\begin{equation} =\frac{1}{\sqrt{21}}(8+8+12)=\frac{28}{\sqrt{21}} \text { at }(1,2,3) \end{equation}\)
Applications Of Level Surfaces In Calculus Problems
Example. 4. Find the greatest value of the directional derivative of the function f = x2y z3 at (2,1,- 1)
Solution: grad f =2xyz3i + x2z3f + 3x2yz2k =- 4i- 4j + 12k at (2,1,-1)
Greatest value of directional derivative of f =|∇f|=\(\sqrt{16+16+144}\) \(=4 \sqrt{11}\)
Example. 5. Find the directional, derivative of Φ = x2yz + 4xz2 in the direction of vector 2i−j− 2k at (1,-2,-1)
Solution: Let Φ = x2yz + 4xz2 be the given function
Differentiating partially, we get \(\begin{equation} \frac{\partial \phi}{\partial x}=2 x y z+4 z^2, \frac{\partial \phi}{\partial y}=x^2 z, \frac{\partial \phi}{\partial z}=x^2 y+8 x z \end{equation}\)
grad Φ= ∇Φ\(=\frac{\partial \phi}{\partial x} \mathbf{i}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial z} \mathbf{k}\) (2xyz+4z2)i+(x2z)j+x2y+8xz)k
Given vector is 2i-j-k Let e is the unit vector its direction.
⇒ \(\begin{equation}\mathbf{e}=\frac{2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}}{\sqrt{4+1+4}}=\frac{2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}}{3}\end{equation}\)
The direction derivative of Φ in the direction of 2i- j- 2k
⇒ \(\begin{equation} =\nabla \phi \cdot \mathbf{e}=2\left(2 x y z+4 z^2\right)-\left(x^2 z\right)-2\left(x^2 y+8 x z\right) \end{equation}\)
Its value at (1,-2,-1) = 8 + 1 + 20 = 29
Differential Operators Example 6
Example. 6. Find the angle of intersection at (4,- 3,2) of spheres x2+ y2 + z2 = 29 and x2 + y2 + z2 + 4x- 6y- 8z- 47 = 0.
Solution:
Given
(4,- 3,2) of spheres x2+ y2 + z2 = 29 and x2 + y2 + z2 + 4x- 6y- 8z- 47 = 0
Let f= x2 + y2 + z2– 29
g = x2 + y2 +z2 + 4x- 6y- 8z- 47
Then grad f = 2xi + 2yj + 2zk and grad g = (2x + 4) i + (2y- 6) j + (2z- 8) k
The angle between two surfaces at a point is the angle between the normal to the surfaces at that point.
Let n1= grad f at (4,- 3, 2) = 8i- + 4k
and n2 =grad g at (4, − 3, 2) = 12i − 12j − 4k
The vectors n1 and n2 are along the normals to the two surfaces at (4,- 3, 2). If 0 is the angle between these vectors, then.
n1.n2 =|n1| |n2| cos0 ⇒ 96 + 72−16 \(=\sqrt{(64+36+16)} \sqrt{(144+144+16)} \cos \theta\)
∴ cos θ\(=\frac{152}{\sqrt{(116)(304)}}\) cos-1
⇒ θ=cos-1 \(\sqrt{(19 / 29)}\)
Differential Operators In Level Surface Examples
Example: 7 Show that (1)(a.∇)Φ = a.∇Φ (2) (a.∇) r=a
Solution: (1) Let a= a1i+a2j+a3k, Then
a. ∇=(a1i+a2j+a3k ).\(\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)\)
⇒ \(=a_1 \frac{\partial}{\partial x}+a_2 \frac{\partial}{\partial y}+a_3 \frac{\partial}{\partial z}\)
⇒ \(=\sum a_1 \frac{\partial}{\partial x}\)
∴ (a.∇) Φ \(=\left(\Sigma a_1 \frac{\partial}{\partial x}\right)(\phi)\)\(=\Sigma a_1 \frac{\partial \phi}{\partial x}\)
Again a.∇ =(a1i+a2j+a3k). \(\left(\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right)\)\(=a_1 \frac{\partial \phi}{\partial x}+a_2 \frac{\partial \phi}{\partial y}+a_3 \frac{\partial \phi}{\partial z}\)
Hence (a.∇ ) Φ = a.∇Φ
(2) r=xi+yj+zk
∴\(\frac{\partial \mathbf{r}}{\partial x}=\mathbf{i}, \frac{\partial \mathbf{r}}{\partial y}=\mathbf{j}, \frac{\partial \mathbf{r}}{\partial z}=\mathbf{k}\)
∴ (a.∇) r \(=\left(\Sigma a_1 \frac{\partial}{\partial x}\right) \mathbf{r}\)
∴ \(=\Sigma a_1 \frac{\partial \mathbf{r}}{\partial x}\)= a1i+a2j+a3k =a .