Assignment 02 Solution:¶
Q-1. If \(\varphi(x,y,z) = 3x^2y - y^3z^2\), find \(\nabla\varphi\) at the point \((1, -2, -1)\).
Solution
\(\dfrac{\partial\varphi}{\partial x} = 6xy, \qquad \dfrac{\partial\varphi}{\partial y} = 3x^2 - 3y^2z^2, \qquad \dfrac{\partial\varphi}{\partial z} = -2y^3z\)
At \((1,\ -2,\ -1)\):
\(\dfrac{\partial\varphi}{\partial x} = 6(1)(-2) = -12\)
\(\dfrac{\partial\varphi}{\partial y} = 3(1)^2 - 3(-2)^2(-1)^2 = 3 - 12 = -9\)
\(\dfrac{\partial\varphi}{\partial z} = -2(-2)^3(-1) = -2(-8)(-1) = -16\)
Q-4. Show that \(\nabla r^n = nr^{n-2}\vec{r}\), where \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).
Solution
Here \(r = |\vec{r}| = \sqrt{x^2+y^2+z^2}\), so:
\(\dfrac{\partial r^n}{\partial x} = nr^{n-1}\cdot\dfrac{\partial r}{\partial x} = nr^{n-1}\cdot\dfrac{x}{r} = nxr^{n-2}\)
\(\Rightarrow \dfrac{\partial r^n}{\partial y} = nyr^{n-2}, \qquad \dfrac{\partial r^n}{\partial z} = nzr^{n-2}\)
\(\Rightarrow \nabla r^n = nxr^{n-2}\hat{i} + nyr^{n-2}\hat{j} + nzr^{n-2}\hat{k}\)
\(\Rightarrow \nabla r^n = nr^{n-2}(x\hat{i}+y\hat{j}+z\hat{k})\)
Q-5. Show that \(\nabla\varphi\) is a vector perpendicular to the surface \(\varphi(x,y,z) = c\) where \(c\) is a constant.
Solution
Let \(\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}\) be any arbitrary curve lying on the surface \(\varphi(x,y,z) = c\).
Since every point of the curve lies on the surface:
Differentiating both sides with respect to \(t\):
\(\Rightarrow \nabla\varphi \cdot \dfrac{d\vec{r}}{dt} = 0\)
Since \(\dfrac{d\vec{r}}{dt}\) is tangent to the curve lying on the surface, and this holds for any curve on the surface, \(\nabla\varphi\) is perpendicular to every tangent vector on the surface.
Q-7. Find the equation of the tangent plane to the surface \(2xz^2 - 3xy - 4x = 7\) at the point \((1,\ -1,\ 2)\).
Solution
Let \(F(x,y,z) = 2xz^2 - 3xy - 4x - 7 = 0\)
The normal to the surface is \(\nabla F\):
\(\dfrac{\partial F}{\partial x} = 2z^2 - 3y - 4, \qquad \dfrac{\partial F}{\partial y} = -3x, \qquad \dfrac{\partial F}{\partial z} = 4xz\)
At \((1,\ -1,\ 2)\):
\(\dfrac{\partial F}{\partial x} = 2(4) - 3(-1) - 4 = 8+3-4 = 7\)
\(\dfrac{\partial F}{\partial y} = -3(1) = -3\)
\(\dfrac{\partial F}{\partial z} = 4(1)(2) = 8\)
The tangent plane equation is:
\(7(x-1) - 3(y+1) + 8(z-2) = 0\)
\(\Rightarrow 7x - 7 - 3y - 3 + 8z - 16 = 0\)
Q-10. Find the directional derivative of \(\varphi = x^2yz + 4xz^2\) at \((1,\ -2,\ -1)\) in the direction \(2\hat{i} - \hat{j} - 2\hat{k}\).
Solution
\(\dfrac{\partial\varphi}{\partial x} = 2xyz + 4z^2, \qquad \dfrac{\partial\varphi}{\partial y} = x^2z, \qquad \dfrac{\partial\varphi}{\partial z} = x^2y + 8xz\)
At \((1,\ -2,\ -1)\):
\(\dfrac{\partial\varphi}{\partial x} = 2(1)(-2)(-1) + 4(1) = 4+4 = 8\)
\(\dfrac{\partial\varphi}{\partial y} = (1)^2(-1) = -1\)
\(\dfrac{\partial\varphi}{\partial z} = (1)^2(-2) + 8(1)(-1) = -2-8 = -10\)
\(\Rightarrow \nabla\varphi = 8\hat{i} - \hat{j} - 10\hat{k}\)
Unit vector in the direction \(\vec{a} = 2\hat{i} - \hat{j} - 2\hat{k}\):
\(|\vec{a}| = \sqrt{4+1+4} = \sqrt{9} = 3\)
\(\Rightarrow \hat{a} = \dfrac{2\hat{i} - \hat{j} - 2\hat{k}}{3}\)
Directional derivative:
\(D_{\hat{a}}\varphi = \nabla\varphi \cdot \hat{a} = (8\hat{i} - \hat{j} - 10\hat{k}) \cdot \dfrac{(2\hat{i}-\hat{j}-2\hat{k})}{3}\)
\(\Rightarrow = \dfrac{(8)(2)+(-1)(-1)+(-10)(-2)}{3}\)
\(\Rightarrow = \dfrac{16+1+20}{3}\)
Q-11. (a) In what direction from the point \((2,1,-1)\) is the directional derivative of \(\varphi = x^2y^2z^3\) a maximum? (b) What is the magnitude of this maximum?
Solution
The directional derivative is maximum in the direction of \(\nabla\varphi\), and the maximum value equals \(|\nabla\varphi|\).
At \((2,\ 1,\ -1)\):
\(\dfrac{\partial\varphi}{\partial x} = 2(2)(1)^2(-1)^3 = -4\)
\(\dfrac{\partial\varphi}{\partial y} = 2(2)^2(1)(-1)^3 = -8\)
\(\dfrac{\partial\varphi}{\partial z} = 3(2)^2(1)^2(-1)^2 = 12\)
\(\Rightarrow \nabla\varphi = -4\hat{i} - 8\hat{j} + 12\hat{k}\)
(a) The directional derivative is maximum in the direction of:
(b) The magnitude of this maximum:
\(|\nabla\varphi| = \sqrt{(-4)^2+(-8)^2+(12)^2} = \sqrt{16+64+144} = \sqrt{224}\)
Q-12. Find the angle between the surfaces \(x^2+y^2+z^2=9\) and \(z=x^2+y^2-3\) at the point \((2,-1,2)\).
Solution
Let \(F_1 = x^2+y^2+z^2-9 = 0\) and \(F_2 = x^2+y^2-z-3 = 0\)
\(\Rightarrow\) at \((2,-1,2)\): \(\nabla F_1 = 4\hat{i}-2\hat{j}+4\hat{k}\)
\(\Rightarrow\) at \((2,-1,2)\): \(\nabla F_2 = 4\hat{i}-2\hat{j}-\hat{k}\)
The angle between the surfaces is the angle between their normals:
\(\nabla F_1 \cdot \nabla F_2 = (4)(4)+(-2)(-2)+(4)(-1) = 16+4-4 = 16\)
\(|\nabla F_1| = \sqrt{16+4+16} = \sqrt{36} = 6\)
\(|\nabla F_2| = \sqrt{16+4+1} = \sqrt{21}\)
\(\Rightarrow \cos\theta = \dfrac{16}{6\sqrt{21}} = \dfrac{8}{3\sqrt{21}}\)
Q-17. Prove that \(\nabla^2\left(\dfrac{1}{r}\right) = 0\), where \(\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}\).
Solution
Using the result \(\nabla r^n = nr^{n-2}\vec{r}\) (Q-4) with \(n = -1\):
\(\Rightarrow \nabla^2\left(\dfrac{1}{r}\right) = \nabla\cdot\nabla\left(\dfrac{1}{r}\right) = -\nabla\cdot\left(\dfrac{\vec{r}}{r^3}\right)\)
Now computing \(\nabla\cdot\left(\dfrac{\vec{r}}{r^3}\right) = \dfrac{\partial}{\partial x}\left(\dfrac{x}{r^3}\right) + \dfrac{\partial}{\partial y}\left(\dfrac{y}{r^3}\right) + \dfrac{\partial}{\partial z}\left(\dfrac{z}{r^3}\right)\)
\(\dfrac{\partial}{\partial x}\left(\dfrac{x}{r^3}\right) = \dfrac{1}{r^3} + x\cdot\left(\dfrac{-3x}{r^5}\right) = \dfrac{1}{r^3} - \dfrac{3x^2}{r^5}\)
Similarly: \(\dfrac{\partial}{\partial y}\left(\dfrac{y}{r^3}\right) = \dfrac{1}{r^3} - \dfrac{3y^2}{r^5}\) and \(\dfrac{\partial}{\partial z}\left(\dfrac{z}{r^3}\right) = \dfrac{1}{r^3} - \dfrac{3z^2}{r^5}\)
\(\Rightarrow \nabla\cdot\left(\dfrac{\vec{r}}{r^3}\right) = \dfrac{3}{r^3} - \dfrac{3(x^2+y^2+z^2)}{r^5} = \dfrac{3}{r^3} - \dfrac{3r^2}{r^5} = \dfrac{3}{r^3} - \dfrac{3}{r^3} = 0\)
Q-22. Determine the constant \(c\) such that the vector \(\vec{V} = (x+y)\hat{i} + (bx+cy-z)\hat{j} + (x+2y+2z)\hat{k}\) is solenoidal.
Solution
For a vector to be solenoidal, \(\nabla\cdot\vec{V} = 0\):
\(\Rightarrow 1 + c + 2 = 0\)
\(\Rightarrow c + 3 = 0\)
Q-23. If \(\vec{A} = x^2y\hat{i} - 2xyz\hat{j} + 2yz^2\hat{k}\), find \(\nabla\times\vec{A}\) (or curl \(\vec{A}\)) at the point \((1,1,1)\).
Solution
\(\Rightarrow \hat{i}\left(\dfrac{\partial(2yz^2)}{\partial y} - \dfrac{\partial(-2xyz)}{\partial z}\right) - \hat{j}\left(\dfrac{\partial(2yz^2)}{\partial x} - \dfrac{\partial(x^2y)}{\partial z}\right) + \hat{k}\left(\dfrac{\partial(-2xyz)}{\partial x} - \dfrac{\partial(x^2y)}{\partial y}\right)\)
\(\Rightarrow \hat{i}(2z^2 - (-2xy)) - \hat{j}(0 - 0) + \hat{k}(-2yz - x^2)\)
\(\Rightarrow (2z^2+2xy)\hat{i} + 0\hat{j} + (-2yz-x^2)\hat{k}\)
At \((1,\ 1,\ 1)\):
\(\Rightarrow (2+2)\hat{i} + (-2-1)\hat{k}\)
Q-12. If \(\phi(x,y,z)=3x^{2}y-y^{3}z^{2}\), find \(\nabla\phi\) (or \(\text{grad }\phi\)) at the point \((1, -2, -1)\).
Solution: The gradient of a scalar function \(\phi(x, y, z)\) in three-dimensional Cartesian space is defined by the vector differential operator \(\nabla\) (del):
Given the scalar field function:
We compute the first-order partial derivatives with respect to each variable independently.
Differentiating \(\phi\) partially with respect to \(x\) (treating \(y\) and \(z\) as constants):
\(\implies \frac{\partial\phi}{\partial x} = \frac{\partial}{\partial x}(3x^{2}y - y^{3}z^{2})\)
\(\implies \frac{\partial\phi}{\partial x} = 6xy - 0\)
\(\implies \frac{\partial\phi}{\partial x} = 6xy\)
Differentiating \(\phi\) partially with respect to \(y\) (treating \(x\) and \(z\) as constants):
\(\implies \frac{\partial\phi}{\partial y} = \frac{\partial}{\partial y}(3x^{2}y - y^{3}z^{2})\)
\(\implies \frac{\partial\phi}{\partial y} = 3x^{2}(1) - 3y^{2}z^{2}\)
\(\implies \frac{\partial\phi}{\partial y} = 3x^{2} - 3y^{2}z^{2}\)
Differentiating \(\phi\) partially with respect to \(z\) (treating \(x\) and \(y\) as constants):
\(\implies \frac{\partial\phi}{\partial z} = \frac{\partial}{\partial z}(3x^{2}y - y^{3}z^{2})\)
\(\implies \frac{\partial\phi}{\partial z} = 0 - y^{3}(2z)\)
\(\implies \frac{\partial\phi}{\partial z} = -2y^{3}z\)
Substituting these partial derivatives back into the vector definition statement for \(\nabla\phi\):
Now, we evaluate the exact components of this vector field at the specified point \((x, y, z) = (1, -2, -1)\):
Evaluating the \(\hat{i}\) component:
\(\implies 6xy = 6(1)(-2)\)
\(\implies 6xy = -12\)
Evaluating the \(\hat{j}\) component:
\(\implies 3x^{2} - 3y^{2}z^{2} = 3(1)^{2} - 3(-2)^{2}(-1)^{2}\)
\(\implies 3x^{2} - 3y^{2}z^{2} = 3(1) - 3(4)(1)\)
\(\implies 3x^{2} - 3y^{2}z^{2} = 3 - 12\)
\(\implies 3x^{2} - 3y^{2}z^{2} = -9\)
Evaluating the \(\hat{k}\) component:
\(\implies -2y^{3}z = -2(-2)^{3}(-1)\)
\(\implies -2y^{3}z = -2(-8)(-1)\)
\(\implies -2y^{3}z = -16\)
Combining these numerical scalar components back into our vector format:
The value of \(\text{grad }\phi\) at the point \((1, -2, -1)\) is:
Q-17. Prove that \(\nabla^{2}\left(\frac{1}{r}\right)=0\) where \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\).
Solution: Let the position vector be defined as:
$\(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\)$
The magnitude \(r\) satisfies the relation:
Differentiating partially with respect to \(x\) on both sides:
\(\implies 2r\frac{\partial r}{\partial x} = 2x\)
\(\implies \frac{\partial r}{\partial x} = \frac{x}{r}\)
By symmetry, the partial derivatives with respect to \(y\) and \(z\) are:
The Laplacian operator \(\nabla^2\) acting on a scalar function is defined as:
First, we find the first-order partial derivative with respect to \(x\) using the chain rule:
\(\implies \frac{\partial}{\partial x}\left(\frac{1}{r}\right) = -\frac{1}{r^2} \cdot \frac{\partial r}{\partial x}\)
\(\implies \frac{\partial}{\partial x}\left(\frac{1}{r}\right) = -\frac{1}{r^2} \cdot \left(\frac{x}{r}\right) = -x r^{-3}\)
Now, we find the second-order partial derivative with respect to \(x\) using the product rule:
\(\implies \frac{\partial^2}{\partial x^2}\left(\frac{1}{r}\right) = \frac{\partial}{\partial x}\left(-x r^{-3}\right)\)
\(\implies \frac{\partial^2}{\partial x^2}\left(\frac{1}{r}\right) = (-1)r^{-3} + (-x)\left(-3r^{-4}\frac{\partial r}{\partial x}\right)\)
\(\implies \frac{\partial^2}{\partial x^2}\left(\frac{1}{r}\right) = -\frac{1}{r^3} + 3xr^{-4}\left(\frac{x}{r}\right)\)
\(\implies \frac{\partial^2}{\partial x^2}\left(\frac{1}{r}\right) = -\frac{1}{r^3} + \frac{3x^2}{r^5}\)
By symmetry, the second-order partial derivatives with respect to \(y\) and \(z\) are:
Substituting these three expressions back into the definition for \(\nabla^2\left(\frac{1}{r}\right)\):
\(\implies \nabla^2\left(\frac{1}{r}\right) = \left(-\frac{1}{r^3} + \frac{3x^2}{r^5}\right) + \left(-\frac{1}{r^3} + \frac{3y^2}{r^5}\right) + \left(-\frac{1}{r^3} + \frac{3z^2}{r^5}\right)\)
\(\implies \nabla^2\left(\frac{1}{r}\right) = -\frac{3}{r^3} + \frac{3(x^2 + y^2 + z^2)}{r^5}\)
Substituting \(x^2 + y^2 + z^2 = r^2\) into the numerator:
\(\implies \nabla^2\left(\frac{1}{r}\right) = -\frac{3}{r^3} + \frac{3(r^2)}{r^5}\)
\(\implies \nabla^2\left(\frac{1}{r}\right) = -\frac{3}{r^3} + \frac{3}{r^3}\)
\(\implies \nabla^2\left(\frac{1}{r}\right) = 0\)
Hence Proved.
Q-24. If \(\vec{A} = x^2y\hat{i} - 2xz\hat{j} + 2yz\hat{k}\), find \(\nabla\times\vec{A}\), find curl curl \(\vec{A}\).
Solution
\(\hat{i}:\quad \dfrac{\partial(2yz)}{\partial y} - \dfrac{\partial(-2xz)}{\partial z} = 2z+2x\)
\(\hat{j}:\quad -\left[\dfrac{\partial(2yz)}{\partial x} - \dfrac{\partial(x^2y)}{\partial z}\right] = -[0-0] = 0\)
\(\hat{k}:\quad \dfrac{\partial(-2xz)}{\partial x} - \dfrac{\partial(x^2y)}{\partial y} = -2z-x^2\)
Now curl curl \(\vec{A}\):
\(\hat{i}:\quad \dfrac{\partial(-(x^2+2z))}{\partial y} - \dfrac{\partial(0)}{\partial z} = 0\)
\(\hat{j}:\quad -\left[\dfrac{\partial(-(x^2+2z))}{\partial x} - \dfrac{\partial(2x+2z)}{\partial z}\right] = -[-2x-2] = 2(x+1)\)
\(\hat{k}:\quad \dfrac{\partial(0)}{\partial x} - \dfrac{\partial(2x+2z)}{\partial y} = 0\)
Q-30. If \(\vec{v} = \vec{\omega}\times\vec{r}\), prove \(\vec{\omega} = \dfrac{1}{2}\) curl \(\vec{v}\) where \(\vec{\omega}\) is a constant vector.
Solution
Let \(\vec{\omega} = \omega_1\hat{i}+\omega_2\hat{j}+\omega_3\hat{k}\) and \(\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}\)
\(\Rightarrow \vec{v} = (\omega_2z-\omega_3y)\hat{i} + (\omega_3x-\omega_1z)\hat{j} + (\omega_1y-\omega_2x)\hat{k}\)
curl \(\vec{v} = \nabla\times\vec{v}\):
\(\hat{i}:\quad \dfrac{\partial(\omega_1y-\omega_2x)}{\partial y} - \dfrac{\partial(\omega_3x-\omega_1z)}{\partial z} = \omega_1-(-\omega_1) = 2\omega_1\)
\(\hat{j}:\quad -\left[\dfrac{\partial(\omega_1y-\omega_2x)}{\partial x} - \dfrac{\partial(\omega_2z-\omega_3y)}{\partial z}\right] = -[-\omega_2-\omega_2] = 2\omega_2\)
\(\hat{k}:\quad \dfrac{\partial(\omega_3x-\omega_1z)}{\partial x} - \dfrac{\partial(\omega_2z-\omega_3y)}{\partial y} = \omega_3-(-\omega_3) = 2\omega_3\)
\(\Rightarrow\) curl \(\vec{v} = 2\omega_1\hat{i}+2\omega_2\hat{j}+2\omega_3\hat{k} = 2\vec{\omega}\)
Q-32. Find \(a, b, c\) so that \(\vec{V} = (x+2y+az)\hat{i}+(bx-3y-z)\hat{j}+(4x+cy+2z)\hat{k}\) is irrotational.
Solution
For \(\vec{V}\) to be irrotational, curl \(\vec{V} = \vec{0}\):
\(\hat{i}:\quad \dfrac{\partial(4x+cy+2z)}{\partial y} - \dfrac{\partial(bx-3y-z)}{\partial z} = c+1 = 0\)
\(\Rightarrow \boxed{c = -1}\)
\(\hat{j}:\quad -\left[\dfrac{\partial(4x+cy+2z)}{\partial x} - \dfrac{\partial(x+2y+az)}{\partial z}\right] = -(4-a) = 0\)
\(\Rightarrow \boxed{a = 4}\)
\(\hat{k}:\quad \dfrac{\partial(bx-3y-z)}{\partial x} - \dfrac{\partial(x+2y+az)}{\partial y} = b-2 = 0\)
\(\Rightarrow \boxed{b = 2}\)
Q-42. If \(\Phi = 2xz^4 - x^2y\), find \(\nabla\Phi\) and \(|\nabla\Phi|\) at \((2,-2,-1)\).
Solution
At \((2,-2,-1)\):
\(\Rightarrow [2(1)-2(2)(-2)]\hat{i}+[-(4)]\hat{j}+[8(2)(-1)]\hat{k}\)
\(\Rightarrow [2+8]\hat{i}-4\hat{j}-16\hat{k}\)
\(|\nabla\Phi| = \sqrt{(10)^2+(-4)^2+(-16)^2} = \sqrt{100+16+256} = \sqrt{372}\)
Q-43. If \(\vec{A} = 2x^2y\hat{i}-3yz^2\hat{j}+xyz^2\hat{k}\) and \(\Phi = 2z-x^2y\), find (a) \(\vec{A}\cdot\nabla\Phi\) and (b) \(\nabla\cdot(\Phi\vec{A})\) at \((1,0,-2)\).
Solution
At \((1,0,-2)\): \(\nabla\Phi = 0\hat{i}-\hat{j}+2\hat{k}\)
\(\vec{A}\) at \((1,0,-2)\): \(\vec{A} = 0\hat{i}+0\hat{j}+0\hat{k} = \vec{0}\)
(a)
(b) Using the identity \(\nabla\cdot(\Phi\vec{A}) = \Phi(\nabla\cdot\vec{A})+\vec{A}\cdot\nabla\Phi\):
\(\nabla\cdot\vec{A} = 4xy-3z^2+2xyz\)
At \((1,0,-2)\): \(\nabla\cdot\vec{A} = 0-12+0 = -12\)
\(\Phi\) at \((1,0,-2)\): \(\Phi = 2(-2)-(1)^2(0) = -4\)
\(\Rightarrow \nabla\cdot(\Phi\vec{A}) = (-4)(-12)+0\)
Q-44:Now I have everything. Let me write the solution for Q-44.Q-44. If \(F = x^2z + e^{y/x}\) and \(G = 2z^2y - xy^2\), find (a) \(\nabla(F+G)\) and (b) \(\nabla(FG)\) at the point \((1, 0, -2)\).
Solution:
We use the result \(\nabla(F+G) = \nabla F + \nabla G\) and \(\nabla(FG) = F\nabla G + G\nabla F\).
First, we compute \(\nabla F\) and \(\nabla G\) individually.
\(\dfrac{\partial F}{\partial x} = 2xz - \dfrac{y}{x^2}e^{y/x}, \quad \dfrac{\partial F}{\partial y} = \dfrac{1}{x}e^{y/x}, \quad \dfrac{\partial F}{\partial z} = x^2\)
At \((1, 0, -2)\):
\(\dfrac{\partial F}{\partial x} = 2(1)(-2) - 0 = -4\)
\(\Rightarrow \dfrac{\partial F}{\partial y} = \dfrac{1}{1}e^{0} = 1\)
\(\Rightarrow \dfrac{\partial F}{\partial z} = (1)^2 = 1\)
\(\Rightarrow \nabla F = -4\hat{i} + \hat{j} + \hat{k}\)
Now for \(G = 2z^2y - xy^2\):
\(\dfrac{\partial G}{\partial x} = -y^2, \quad \dfrac{\partial G}{\partial y} = 2z^2 - 2xy, \quad \dfrac{\partial G}{\partial z} = 4yz\)
At \((1, 0, -2)\):
\(\dfrac{\partial G}{\partial x} = -(0)^2 = 0\)
\(\Rightarrow \dfrac{\partial G}{\partial y} = 2(-2)^2 - 2(1)(0) = 8\)
\(\Rightarrow \dfrac{\partial G}{\partial z} = 4(0)(-2) = 0\)
\(\Rightarrow \nabla G = 0\hat{i} + 8\hat{j} + 0\hat{k} = 8\hat{j}\)
Also, the values of \(F\) and \(G\) at \((1, 0, -2)\):
\(F = (1)^2(-2) + e^{0/1} = -2 + 1 = -1\)
\(\Rightarrow G = 2(-2)^2(0) - (1)(0)^2 = 0\)
(a)
\(\nabla(F+G) = \nabla F + \nabla G\)
\(\Rightarrow = (-4\hat{i} + \hat{j} + \hat{k}) + (8\hat{j})\)
\(\Rightarrow \boxed{\nabla(F+G) = -4\hat{i} + 9\hat{j} + \hat{k}}\)
(b)
\(\nabla(FG) = F\nabla G + G\nabla F\)
\(\Rightarrow = (-1)(8\hat{j}) + (0)(-4\hat{i} + \hat{j} + \hat{k})\)
\(\Rightarrow = -8\hat{j} + \mathbf{0}\)
\(\Rightarrow \boxed{\nabla(FG) = -8\hat{j}}\)
Q-45. Find \(\nabla|\vec{r}|^3\).
We have \(|\vec{r}| = r = \sqrt{x^2 + y^2 + z^2}\), so \(|\vec{r}|^3 = r^3\).
Using the result \(\nabla r^n = nr^{n-2}\vec{r}\) with \(n = 3\):
\(\nabla r^3 = 3r^{3-2}\vec{r}\)
\(\Rightarrow = 3r\vec{r}\)
\(\Rightarrow \boxed{\nabla|\vec{r}|^3 = 3r\vec{r}}\)
Q-62. Find the directional derivative of \(\phi = 4xz^3 - 3x^2y^2z\) at \((2, -1, 2)\) in the direction \(2\hat{i} - 3\hat{j} + 6\hat{k}\).
\(\nabla\phi = \left(\frac{\partial\phi}{\partial x}\right)\hat{i} + \left(\frac{\partial\phi}{\partial y}\right)\hat{j} + \left(\frac{\partial\phi}{\partial z}\right)\hat{k}\)
\(\frac{\partial\phi}{\partial x} = 4z^3 - 6xy^2z, \quad \frac{\partial\phi}{\partial y} = -6x^2yz, \quad \frac{\partial\phi}{\partial z} = 12xz^2 - 3x^2y^2\)
At \((2, -1, 2)\):
\(\frac{\partial\phi}{\partial x} = 4(2)^3 - 6(2)(-1)^2(2) = 32 - 24 = 8\)
\(\Rightarrow \frac{\partial\phi}{\partial y} = -6(2)^2(-1)(2) = 48\)
\(\Rightarrow \frac{\partial\phi}{\partial z} = 12(2)(2)^2 - 3(2)^2(-1)^2 = 96 - 12 = 84\)
\(\Rightarrow \nabla\phi = 8\hat{i} + 48\hat{j} + 84\hat{k}\)
The unit vector in the direction of \(2\hat{i} - 3\hat{j} + 6\hat{k}\):
\(|\vec{a}| = \sqrt{(2)^2 + (-3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\)
\(\Rightarrow \hat{a} = \frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{7}\)
The directional derivative is:
\(\nabla\phi \cdot \hat{a} = (8\hat{i} + 48\hat{j} + 84\hat{k}) \cdot \frac{1}{7}(2\hat{i} - 3\hat{j} + 6\hat{k})\)
\(\Rightarrow = \frac{1}{7}[(8)(2) + (48)(-3) + (84)(6)]\)
\(\Rightarrow = \frac{1}{7}[16 - 144 + 504]\)
\(\Rightarrow = \frac{376}{7}\)
\(\Rightarrow \boxed{\nabla\phi \cdot \hat{a} = \frac{376}{7}}\)
Q-63. Find the directional derivative of \(P = 4e^{2x-y+z}\) at the point \((1, 1, -1)\) in a direction toward the point \((-3, 5, 6)\).
\(\nabla P = \frac{\partial P}{\partial x}\hat{i} + \frac{\partial P}{\partial y}\hat{j} + \frac{\partial P}{\partial z}\hat{k}\)
\(\frac{\partial P}{\partial x} = 8e^{2x-y+z}, \quad \frac{\partial P}{\partial y} = -4e^{2x-y+z}, \quad \frac{\partial P}{\partial z} = 4e^{2x-y+z}\)
At \((1, 1, -1)\), the exponent is \(2(1) - 1 + (-1) = 0\), so \(e^0 = 1\):
\(\frac{\partial P}{\partial x} = 8\)
\(\Rightarrow \frac{\partial P}{\partial y} = -4\)
\(\Rightarrow \frac{\partial P}{\partial z} = 4\)
\(\Rightarrow \nabla P = 8\hat{i} - 4\hat{j} + 4\hat{k}\)
The direction from \((1, 1, -1)\) toward \((-3, 5, 6)\) is:
\(\vec{d} = (-3-1)\hat{i} + (5-1)\hat{j} + (6-(-1))\hat{k} = -4\hat{i} + 4\hat{j} + 7\hat{k}\)
\(|\vec{d}| = \sqrt{(-4)^2 + (4)^2 + (7)^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9\)
\(\Rightarrow \hat{d} = \frac{-4\hat{i} + 4\hat{j} + 7\hat{k}}{9}\)
The directional derivative is:
\(\nabla P \cdot \hat{d} = (8\hat{i} - 4\hat{j} + 4\hat{k}) \cdot \frac{1}{9}(-4\hat{i} + 4\hat{j} + 7\hat{k})\)
\(\Rightarrow = \frac{1}{9}[(8)(-4) + (-4)(4) + (4)(7)]\)
\(\Rightarrow = \frac{1}{9}[-32 - 16 + 28]\)
\(\Rightarrow = \frac{-20}{9}\)
\(\Rightarrow \boxed{\nabla P \cdot \hat{d} = -\frac{20}{9}}\)
Since this is negative, \(P\) is decreasing in this direction.
Q-65. Find the values of the constants \(a, b, c\) so that the directional derivative of \(\phi = axy^2 + byz + cz^2x^3\) at \((1, 2, -1)\) has a maximum of magnitude \(64\) in a direction parallel to the \(z\)-axis.
\(\nabla\phi = \frac{\partial\phi}{\partial x}\hat{i} + \frac{\partial\phi}{\partial y}\hat{j} + \frac{\partial\phi}{\partial z}\hat{k}\)
\(\frac{\partial\phi}{\partial x} = ay^2 + 3cz^2x^2, \quad \frac{\partial\phi}{\partial y} = 2axy + bz, \quad \frac{\partial\phi}{\partial z} = by + 2czx^3\)
At \((1, 2, -1)\):
\(\frac{\partial\phi}{\partial x} = a(2)^2 + 3c(-1)^2(1)^2 = 4a + 3c\)
\(\Rightarrow \frac{\partial\phi}{\partial y} = 2a(1)(2) + b(-1) = 4a - b\)
\(\Rightarrow \frac{\partial\phi}{\partial z} = b(2) + 2c(-1)(1)^3 = 2b - 2c\)
\(\Rightarrow \nabla\phi = (4a+3c)\hat{i} + (4a-b)\hat{j} + (2b-2c)\hat{k}\)
The maximum directional derivative occurs in the direction of \(\nabla\phi\) and its magnitude is \(|\nabla\phi|\).
Since the maximum is in a direction parallel to the \(z\)-axis, \(\nabla\phi\) must be parallel to \(\hat{k}\), meaning the \(\hat{i}\) and \(\hat{j}\) components must be zero:
\(4a + 3c = 0 \quad \cdots (1)\)
\(\Rightarrow 4a - b = 0 \quad \cdots (2)\)
The magnitude of the maximum directional derivative equals \(64\):
\(|2b - 2c| = 64\)
\(\Rightarrow 2b - 2c = 64\)
\(\Rightarrow b - c = 32 \quad \cdots (3)\)
From \((1)\): \(c = -\dfrac{4a}{3}\)
Substituting into \((2)\): \(b = 4a\)
Substituting into \((3)\):
\(4a - \left(-\frac{4a}{3}\right) = 32\)
\(\Rightarrow 4a + \frac{4a}{3} = 32\)
\(\Rightarrow \frac{12a + 4a}{3} = 32\)
\(\Rightarrow \frac{16a}{3} = 32\)
\(\Rightarrow a = 6\)
\(\Rightarrow b = 4(6) = 24\)
\(\Rightarrow c = -\frac{4(6)}{3} = -8\)
\(\Rightarrow \boxed{a = 6, \quad b = 24, \quad c = -8}\)
Q-66. Find the acute angle between the surfaces \(xy^2z = 3x + z^2\) and \(3x^2 - y^2 + 2z = 1\) at the point \((1, -2, 1)\).
The angle between two surfaces at a point is the angle between their normals at that point.
Let \(\phi_1 = xy^2z - 3x - z^2\) and \(\phi_2 = 3x^2 - y^2 + 2z\).
\(\nabla\phi_1 = \frac{\partial\phi_1}{\partial x}\hat{i} + \frac{\partial\phi_1}{\partial y}\hat{j} + \frac{\partial\phi_1}{\partial z}\hat{k}\)
\(\frac{\partial\phi_1}{\partial x} = y^2z - 3, \quad \frac{\partial\phi_1}{\partial y} = 2xyz, \quad \frac{\partial\phi_1}{\partial z} = xy^2 - 2z\)
At \((1, -2, 1)\):
\(\frac{\partial\phi_1}{\partial x} = (-2)^2(1) - 3 = 4 - 3 = 1\)
\(\Rightarrow \frac{\partial\phi_1}{\partial y} = 2(1)(-2)(1) = -4\)
\(\Rightarrow \frac{\partial\phi_1}{\partial z} = (1)(-2)^2 - 2(1) = 4 - 2 = 2\)
\(\Rightarrow \nabla\phi_1 = \hat{i} - 4\hat{j} + 2\hat{k}\)
\(\nabla\phi_2 = \frac{\partial\phi_2}{\partial x}\hat{i} + \frac{\partial\phi_2}{\partial y}\hat{j} + \frac{\partial\phi_2}{\partial z}\hat{k}\)
\(\frac{\partial\phi_2}{\partial x} = 6x, \quad \frac{\partial\phi_2}{\partial y} = -2y, \quad \frac{\partial\phi_2}{\partial z} = 2\)
At \((1, -2, 1)\):
\(\frac{\partial\phi_2}{\partial x} = 6(1) = 6\)
\(\Rightarrow \frac{\partial\phi_2}{\partial y} = -2(-2) = 4\)
\(\Rightarrow \frac{\partial\phi_2}{\partial z} = 2\)
\(\Rightarrow \nabla\phi_2 = 6\hat{i} + 4\hat{j} + 2\hat{k}\)
Now applying:
\(\nabla\phi_1 \cdot \nabla\phi_2 = |\nabla\phi_1||\nabla\phi_2|\cos\theta\)
\(\nabla\phi_1 \cdot \nabla\phi_2 = (1)(6) + (-4)(4) + (2)(2) = 6 - 16 + 4 = -6\)
\(|\nabla\phi_1| = \sqrt{(1)^2 + (-4)^2 + (2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21}\)
\(\Rightarrow |\nabla\phi_2| = \sqrt{(6)^2 + (4)^2 + (2)^2} = \sqrt{36 + 16 + 4} = \sqrt{56}\)
\(\cos\theta = \frac{-6}{\sqrt{21} \cdot \sqrt{56}} = \frac{-6}{\sqrt{1176}} = \frac{-6}{14\sqrt{6}} = \frac{-3}{7\sqrt{6}}\)
Since we need the acute angle, we take the absolute value:
\(\cos\theta = \frac{3}{7\sqrt{6}}\)
\(\Rightarrow \boxed{\theta = \arccos\left(\frac{3}{7\sqrt{6}}\right) \approx 79°55'}\)
Q-70. If \(\vec{A} = 3xyz^2\,\hat{i} + 2xy^3\,\hat{j} - xyz\,\hat{k}\) and \(\Phi = 3x^2 - yz\), find (a) \(\nabla \cdot \vec{A}\), (b) \(\vec{A} \cdot \nabla\Phi\), (c) \(\nabla \cdot (\Phi \vec{A})\), (d) \(\nabla(\nabla \cdot \Phi)\) at the point \((1,-1,1)\).
(a)
\(\Rightarrow 3yz^2 + 6xy^2 - xy\)
\(\Rightarrow 3(-1)(1)^2 + 6(1)(-1)^2 - (1)(-1) \quad \text{at } (1,-1,1)\)
\(\Rightarrow -3 + 6 + 1 = \mathbf{4}\)
(b)
\(\Rightarrow \nabla\Phi \text{ at } (1,-1,1) = 6\hat{i} - \hat{j} + \hat{k}\)
\(\Rightarrow 18x^2yz^2 - 2xy^3z + xy^2z\)
\(\Rightarrow 18(1)^2(-1)(1)^2 - 2(1)(-1)^3(1) + (1)(-1)^2(1) \quad \text{at } (1,-1,1)\)
\(\Rightarrow -18 + 2 + 1 = \mathbf{-15}\)
(c)
Using the identity \(\nabla \cdot (\Phi\vec{A}) = (\nabla\Phi) \cdot \vec{A} + \Phi(\nabla \cdot \vec{A})\)
At \((1,-1,1)\): \(\Phi = 3(1)^2 - (-1)(1) = 3 + 1 = 4\)
\(\Rightarrow \nabla\Phi \cdot \vec{A}\) at \((1,-1,1)\):
\(\Rightarrow (\nabla\Phi)\cdot\vec{A} = (6)(-3) + (-1)(-2) + (1)(1) = -18 + 2 + 1 = -15\)
\(\Rightarrow \nabla \cdot (\Phi\vec{A}) = -15 + (4)(4) = -15 + 16 = \mathbf{1}\)
(d)
\(\nabla \cdot \Phi\) has no meaning since \(\Phi\) is a scalar, so \(\nabla(\nabla \cdot \Phi)\) is interpreted as \(\nabla^2\Phi\) with gradient applied â actually \(\nabla(\nabla \cdot \Phi)\) is undefined; the intended expression from the book context is \(\nabla^2\Phi\):
\(\Rightarrow 6 + 0 + 0 = \mathbf{6}\)
Q-71. Evaluate \(\nabla \cdot (2x^2z\,\hat{i} - xy^2z\,\hat{j} + 3yz^2\,\hat{k})\).
\(\Rightarrow 4xz - 2xyz + 6yz\)
Q-72. If \(\Phi = 3x^2z - y^2z^3 + 4x^3y + 2x - 3y - 5\), find \(\nabla^2\Phi\).
\(\Rightarrow (6z + 24xy) + (-2z^3) + (-6y^2z)\)
\(\Rightarrow \mathbf{6z + 24xy - 2z^3 - 6y^2z}\)
Q-75. If \(\vec{F} = (3x^2y - z)\,\hat{i} + (xz^3 + y^4)\,\hat{j} - 2x^3z^2\,\hat{k}\), find \(\nabla(\nabla \cdot \vec{F})\) at the point \((2,-1,0)\).
First compute \(\nabla \cdot \vec{F}\):
\(\Rightarrow 6xy + 4y^3 - 4x^3z\)
Now compute \(\nabla(\nabla \cdot \vec{F})\):
At \((2,-1,0)\):
\(\Rightarrow (6(-1) - 12(4)(0))\,\hat{i} + (6(2) + 12(1))\,\hat{j} + (-4(8))\,\hat{k}\)
\(\Rightarrow -6\,\hat{i} + 24\,\hat{j} - 32\,\hat{k}\)
\(\Rightarrow \mathbf{-6\hat{i} + 24\hat{j} - 32\hat{k}}\)
Q-76. If \(\vec{\omega}\) is a constant vector and \(\vec{V} = \vec{\omega} \times \vec{r}\), prove that \(\text{div}\,\vec{V} = 0\).
Let \(\vec{\omega} = \omega_1\hat{i} + \omega_2\hat{j} + \omega_3\hat{k}\) where \(\omega_1, \omega_2, \omega_3\) are constants, and \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).
\(\Rightarrow (\omega_2 z - \omega_3 y)\,\hat{i} - (\omega_1 z - \omega_3 x)\,\hat{j} + (\omega_1 y - \omega_2 x)\,\hat{k}\)
\(\Rightarrow 0 + 0 + 0 = 0\)
Since each partial derivative is zero (as \(\omega_1, \omega_2, \omega_3\) are constants and each component of \(\vec{V}\) is independent of the corresponding variable), we conclude:
Question Q-84: Prove that the vector \(\vec{A} = 3y^4z^2\hat{i} + 4x^3z^2\hat{j} - 3x^2y^2\hat{k}\) is solenoidal.
Solution:
A vector is solenoidal if \(\nabla \cdot \vec{A} = 0\).
\(\Rightarrow 0 + 0 + 0 = 0\)
Since \(\nabla \cdot \vec{A} = 0\), \(\vec{A}\) is solenoidal. \(\blacksquare\)
Question Q-85: Show that \(\vec{A} = (2x^2+8xy^2)\hat{i}+(3x^3y-3xy)\hat{j}-(4y^2z^2+2x^3z)\hat{k}\) is not solenoidal.
Solution:
\(\Rightarrow (4x+8y^2) + (3x^3-3x) + (-8y^2z-2x^3)\)
\(\Rightarrow 4x + 8y^2 + 3x^3 - 3x - 8y^2z - 2x^3\)
\(\Rightarrow x + x^3 + 8y^2 - 8y^2z \neq 0\)
Since \(\nabla \cdot \vec{A} \neq 0\), \(\vec{A}\) is not solenoidal. \(\blacksquare\)
Question Q-102: Show that \(\vec{A} = (6xy+z^3)\hat{i}+(3x^2-z)\hat{j}+(3xz^2-y)\hat{k}\) is irrotational and find \(\Phi\) such that \(\vec{A} = \nabla\Phi\).
Solution:
\(\hat{i}:\ \dfrac{\partial(3xz^2-y)}{\partial y} - \dfrac{\partial(3x^2-z)}{\partial z} = -1-(-1) = 0\)
\(\hat{j}:\ -\left[\dfrac{\partial(3xz^2-y)}{\partial x} - \dfrac{\partial(6xy+z^3)}{\partial z}\right] = -[3z^2-3z^2] = 0\)
\(\hat{k}:\ \dfrac{\partial(3x^2-z)}{\partial x} - \dfrac{\partial(6xy+z^3)}{\partial y} = 6x-6x = 0\)
\(\Rightarrow \nabla \times \vec{A} = \vec{0}\), so \(\vec{A}\) is irrotational. \(\blacksquare\)
Since \(\nabla\Phi = \vec{A}\), we have \(\dfrac{\partial\Phi}{\partial x} = 6xy+z^3\)
\(\Rightarrow \Phi = 3x^2y+xz^3+f(y,z) \quad \cdots (1)\)
\(\dfrac{\partial\Phi}{\partial y} = 3x^2+\dfrac{\partial f}{\partial y} = 3x^2-z\)
\(\Rightarrow \dfrac{\partial f}{\partial y} = -z\)
\(\Rightarrow f = -yz+g(z) \quad \cdots (2)\)
Substituting \((2)\) into \((1)\):
\(\Rightarrow \Phi = 3x^2y+xz^3-yz+g(z)\)
\(\dfrac{\partial\Phi}{\partial z} = 3xz^2-y+g'(z) = 3xz^2-y\)
\(\Rightarrow g'(z) = 0 \Rightarrow g(z) = C\)
Chapter 5¶
Question Q-1 : If \(\vec{R}(t) = (t-t^2)\hat{i}+2t^3\hat{j}-3\hat{k}\), find (a) \(\dfrac{d\vec{R}}{dt}\), (b) \(\displaystyle\int_0^1\vec{R}\,dt\).
Solution:
(a)
\(\dfrac{d\vec{R}}{dt} = \dfrac{d}{dt}(t-t^2)\hat{i}+\dfrac{d}{dt}(2t^3)\hat{j}+\dfrac{d}{dt}(-3)\hat{k}\)
\(\Rightarrow \dfrac{d\vec{R}}{dt} = (1-2t)\hat{i}+6t^2\hat{j}\)
(b)
\(\displaystyle\int_0^1\vec{R}\,dt = \int_0^1(t-t^2)\,dt\;\hat{i}+\int_0^1 2t^3\,dt\;\hat{j}+\int_0^1(-3)\,dt\;\hat{k}\)
\(\Rightarrow \left[\dfrac{t^2}{2}-\dfrac{t^3}{3}\right]_0^1\hat{i}+\left[\dfrac{t^4}{2}\right]_0^1\hat{j}+\left[-3t\right]_0^1\hat{k}\)
\(\Rightarrow \left(\dfrac{1}{2}-\dfrac{1}{3}\right)\hat{i}+\dfrac{1}{2}\hat{j}-3\hat{k}\)
Question Q-4 (Chapter 5): If \(\vec{A} = (3x^2+6y)\hat{i}-14yz\hat{j}+20xz^2\hat{k}\), evaluate \(\displaystyle\int_C\vec{A}\cdot d\vec{r}\) from \((0,0,0)\) to \((1,1,1)\) along (a) \(x=t,\ y=t^2,\ z=t^3\), (b) straight lines from \((0,0,0)\to(1,0,0)\to(1,1,0)\to(1,1,1)\), (c) straight line joining \((0,0,0)\) and \((1,1,1)\).
Solution:
(a) \(x=t,\ y=t^2,\ z=t^3\), \(t: 0\to1\), so \(dx=dt,\ dy=2t\,dt,\ dz=3t^2\,dt\)
\(\vec{A} = (3t^2+6t^2)\hat{i}-14t^5\hat{j}+20t^7\hat{k} = 9t^2\hat{i}-14t^5\hat{j}+20t^7\hat{k}\)
\(\vec{A}\cdot d\vec{r} = 9t^2\,dt+(-14t^5)(2t\,dt)+(20t^7)(3t^2\,dt)\)
\(\Rightarrow (9t^2-28t^6+60t^9)\,dt\)
\(\Rightarrow \displaystyle\int_0^1 = \left[3t^3-4t^7+6t^{10}\right]_0^1 = 3-4+6\)
\(\boxed{= 5}\)
(b) Segment \((0,0,0)\to(1,0,0)\): \(y=0,\ z=0,\ dy=dz=0\)
\(\displaystyle\int = \int_0^1 3x^2\,dx = [x^3]_0^1 = 1\)
Segment \((1,0,0)\to(1,1,0)\): \(x=1,\ z=0,\ dx=dz=0\)
\(\displaystyle\int = \int_0^1(-14y\cdot0)\,dy = 0\)
Segment \((1,1,0)\to(1,1,1)\): \(x=1,\ y=1,\ dx=dy=0\)
\(\displaystyle\int = \int_0^1 20z^2\,dz = \left[\dfrac{20z^3}{3}\right]_0^1 = \dfrac{20}{3}\)
\(\Rightarrow\) Total \(= 1+0+\dfrac{20}{3}\)
\(\boxed{= \dfrac{23}{3}}\)
(c) Straight line: \(x=t,\ y=t,\ z=t\), \(t: 0\to1\), \(dx=dy=dz=dt\)
\(\vec{A}\cdot d\vec{r} = (3t^2+6t)\,dt+(-14t^2)\,dt+(20t^3)\,dt\)
\(\Rightarrow (20t^3-11t^2+6t)\,dt\)
\(\Rightarrow \displaystyle\int_0^1 = \left[5t^4-\dfrac{11t^3}{3}+3t^2\right]_0^1 = 5-\dfrac{11}{3}+3 = 8-\dfrac{11}{3}\)
\(\boxed{= \dfrac{13}{3}}\)
Question Q-6 (Chapter 5): Find the total work done in moving a particle in a force field given by \(\vec{F} = 3xy\hat{i} - 5z\hat{j} + 10x\hat{k}\) along the curve \(x = t^2+1,\ y = 2t^2,\ z = t^3\) from \(t=0\) to \(t=2\).
Solution:
Work done \(= \displaystyle\int_C \vec{F}\cdot d\vec{r} = \int_C F_x\,dx + F_y\,dy + F_z\,dz\)
From the parametric equations:
\(dx = 2t\,dt, \qquad dy = 4t\,dt, \qquad dz = 3t^2\,dt\)
Substituting \(x,\ y,\ z\):
\(F_x = 3xy = 3(t^2+1)(2t^2) = 6t^2(t^2+1)\)
\(F_y = -5z = -5t^3\)
\(F_z = 10x = 10(t^2+1)\)
\(\vec{F}\cdot d\vec{r} = 6t^2(t^2+1)\cdot2t\,dt + (-5t^3)\cdot4t\,dt + 10(t^2+1)\cdot3t^2\,dt\)
\(\Rightarrow [12t^3(t^2+1) - 20t^4 + 30t^2(t^2+1)]\,dt\)
\(\Rightarrow [12t^5+12t^3-20t^4+30t^4+30t^2]\,dt\)
\(\Rightarrow [12t^5+10t^4+12t^3+30t^2]\,dt\)
\(\displaystyle\int_0^2\vec{F}\cdot d\vec{r} = \int_0^2(12t^5+10t^4+12t^3+30t^2)\,dt\)
\(\Rightarrow \left[2t^6+2t^5+3t^4+10t^3\right]_0^2\)
\(\Rightarrow 2(64)+2(32)+3(16)+10(8)\)
\(\Rightarrow 128+64+48+80\)
\(\boxed{= 320}\)
Question Q-8 (Chapter 5): If \(\vec{F} = 3xy\hat{i} - y^2\hat{j}\), evaluate \(\displaystyle\int_C \vec{F}\cdot d\vec{r}\) where \(C\) is the curve in the \(xy\) plane \(y = 2x^2\), from \((0,0)\) to \((1,2)\).
Solution:
\(\vec{F}\cdot d\vec{r} = F_x\,dx + F_y\,dy = 3xy\,dx - y^2\,dy\)
Since \(y = 2x^2\), we have \(dy = 4x\,dx\), and \(x\) goes from \(0\) to \(1\).
Substituting \(y = 2x^2\) and \(dy = 4x\,dx\):
\(\vec{F}\cdot d\vec{r} = 3x(2x^2)\,dx - (2x^2)^2\cdot4x\,dx\)
\(\Rightarrow 6x^3\,dx - 4x\cdot4x^4\,dx\)
\(\Rightarrow 6x^3\,dx - 16x^5\,dx\)
\(\Rightarrow (6x^3 - 16x^5)\,dx\)
\(\displaystyle\int_C \vec{F}\cdot d\vec{r} = \int_0^1(6x^3-16x^5)\,dx\)
\(\Rightarrow \left[\dfrac{6x^4}{4} - \dfrac{16x^6}{6}\right]_0^1\)
\(\Rightarrow \left[\dfrac{3x^4}{2} - \dfrac{8x^6}{3}\right]_0^1\)
\(\Rightarrow \dfrac{3}{2} - \dfrac{8}{3}\)
\(\Rightarrow \dfrac{9}{6} - \dfrac{16}{6}\)
\(\boxed{= -\dfrac{7}{6}}\)
Question Q-9 (Chapter 5): Find the work done in moving a particle once around a circle \(C\) in the \(xy\) plane, if the circle has center at the origin and radius 3 and if the force field is \(\vec{F} = (2x-y+z)\hat{i}+(x+y-z^2)\hat{j}+(3x-2y+4z)\hat{k}\).
Solution:
Work done \(= \displaystyle\oint_C \vec{F}\cdot d\vec{r} = \oint_C F_x\,dx + F_y\,dy + F_z\,dz\)
Since \(C\) is in the \(xy\) plane, \(z = 0\) and \(dz = 0\), so:
\(\vec{F}\cdot d\vec{r} = (2x-y)\,dx + (x+y)\,dy\)
Parametrize the circle of radius 3:
\(x = 3\cos t, \qquad y = 3\sin t, \qquad t: 0 \to 2\pi\)
\(\Rightarrow dx = -3\sin t\,dt, \qquad dy = 3\cos t\,dt\)
Substituting:
\(\Rightarrow (6\cos t - 3\sin t)(-3\sin t\,dt) + (3\cos t + 3\sin t)(3\cos t\,dt)\)
\(\Rightarrow [-18\sin t\cos t + 9\sin^2 t + 9\cos^2 t + 9\sin t\cos t]\,dt\)
\(\Rightarrow [-9\sin t\cos t + 9]\,dt\)
\(\Rightarrow \left[-\dfrac{9}{2}\sin 2t + 9\right]dt\)
\(\displaystyle\oint_C \vec{F}\cdot d\vec{r} = \int_0^{2\pi}\left[-\dfrac{9}{2}\sin 2t + 9\right]dt\)
\(\Rightarrow \left[\dfrac{9}{4}\cos 2t + 9t\right]_0^{2\pi}\)
\(\Rightarrow \left(\dfrac{9}{4}\cos 4\pi + 18\pi\right) - \left(\dfrac{9}{4}\cos 0\right)\)
\(\Rightarrow \left(\dfrac{9}{4} + 18\pi\right) - \dfrac{9}{4}\)
\(\boxed{= 18\pi}\)
Question Q-9.1 (Chapter 5): If \(\vec{F}\) is conservative then show that \(\dfrac{\partial F_z}{\partial y} = \dfrac{\partial F_y}{\partial z}\), \(\dfrac{\partial F_x}{\partial z} = \dfrac{\partial F_z}{\partial x}\) and \(\dfrac{\partial F_y}{\partial x} = \dfrac{\partial F_x}{\partial y}\).
Solution:
Since \(\vec{F}\) is conservative, there exists a scalar potential \(\varphi\) such that \(\vec{F} = \nabla\varphi\), which means:
By Clairaut's theorem, mixed partial derivatives are equal, so:
\(\dfrac{\partial F_z}{\partial y} = \dfrac{\partial}{\partial y}\left(\dfrac{\partial\varphi}{\partial z}\right) = \dfrac{\partial^2\varphi}{\partial y\,\partial z} = \dfrac{\partial^2\varphi}{\partial z\,\partial y} = \dfrac{\partial}{\partial z}\left(\dfrac{\partial\varphi}{\partial y}\right) = \dfrac{\partial F_y}{\partial z}\)
\(\dfrac{\partial F_x}{\partial z} = \dfrac{\partial^2\varphi}{\partial z\,\partial x} = \dfrac{\partial^2\varphi}{\partial x\,\partial z} = \dfrac{\partial F_z}{\partial x}\)
\(\dfrac{\partial F_y}{\partial x} = \dfrac{\partial^2\varphi}{\partial x\,\partial y} = \dfrac{\partial^2\varphi}{\partial y\,\partial x} = \dfrac{\partial F_x}{\partial y}\)
This is equivalent to saying \(\nabla\times\vec{F} = \vec{0}\), i.e. curl \(\vec{F} = 0\). \(\blacksquare\)
Question Q-12 (Chapter 5): (a) Show that \(\vec{F} = (2xy+z^3)\hat{i}+x^2\hat{j}+3xz^2\hat{k}\) is a conservative force field. (b) Find the scalar potential. (c) Find the work done in moving an object in this field from \((1,-2,1)\) to \((3,1,4)\).
Solution:
(a)
\(\hat{i}:\ \dfrac{\partial(3xz^2)}{\partial y} - \dfrac{\partial(x^2)}{\partial z} = 0-0 = 0\)
\(\hat{j}:\ -\left[\dfrac{\partial(3xz^2)}{\partial x} - \dfrac{\partial(2xy+z^3)}{\partial z}\right] = -[3z^2-3z^2] = 0\)
\(\hat{k}:\ \dfrac{\partial(x^2)}{\partial x} - \dfrac{\partial(2xy+z^3)}{\partial y} = 2x-2x = 0\)
\(\Rightarrow \nabla\times\vec{F} = \vec{0}\), so \(\vec{F}\) is conservative. \(\blacksquare\)
(b)
Since \(\vec{F} = \nabla\varphi\):
\(\dfrac{\partial\varphi}{\partial x} = 2xy+z^3\)
\(\Rightarrow \varphi = x^2y+xz^3+f(y,z) \quad\cdots(1)\)
\(\dfrac{\partial\varphi}{\partial y} = x^2+\dfrac{\partial f}{\partial y} = x^2\)
\(\Rightarrow \dfrac{\partial f}{\partial y} = 0 \Rightarrow f = g(z)\)
\(\dfrac{\partial\varphi}{\partial z} = 3xz^2+g'(z) = 3xz^2\)
\(\Rightarrow g'(z) = 0 \Rightarrow g(z) = C\)
(c)
Work done \(= \varphi(3,1,4)-\varphi(1,-2,1)\)
\(\varphi(3,1,4) = (3)^2(1)+(3)(4)^3 = 9+192 = 201\)
\(\varphi(1,-2,1) = (1)^2(-2)+(1)(1)^3 = -2+1 = -1\)
\(W = 201-(-1)\)
Question Q-16 (Chapter 5): If \(\varphi = 2xyz^2\), \(x=t^2,\ y=2t,\ z=t^3\) from \(t=0\) to \(t=1\), evaluate (a) \(\displaystyle\int_C\varphi\,d\vec{r}\) and (b) \(\displaystyle\int_C\vec{F}\cdot d\vec{r}\) where \(\vec{F} = \nabla\varphi\).
Solution:
\(d\vec{r} = dx\,\hat{i}+dy\,\hat{j}+dz\,\hat{k} = 2t\,dt\,\hat{i}+2\,dt\,\hat{j}+3t^2\,dt\,\hat{k}\)
\(\varphi\) along \(C\): \(\varphi = 2(t^2)(2t)(t^6) = 4t^9\)
(a)
\(\displaystyle\int_C\varphi\,d\vec{r} = \int_0^1 4t^9(2t\,\hat{i}+2\,\hat{j}+3t^2\,\hat{k})\,dt\)
\(\Rightarrow \int_0^1(8t^{10}\,\hat{i}+8t^9\,\hat{j}+12t^{11}\,\hat{k})\,dt\)
\(\Rightarrow \left[\dfrac{8t^{11}}{11}\hat{i}+\dfrac{8t^{10}}{10}\hat{j}+\dfrac{12t^{12}}{12}\hat{k}\right]_0^1\)
(b)
\(\vec{F} = \nabla\varphi = 2yz^2\,\hat{i}+2xz^2\,\hat{j}+4xyz\,\hat{k}\)
Along \(C\): \(F_x = 2(2t)(t^6) = 4t^7,\quad F_y = 2(t^2)(t^6) = 2t^8,\quad F_z = 4(t^2)(2t)(t^3) = 8t^6\)
\(\vec{F}\cdot d\vec{r} = 4t^7(2t)\,dt+2t^8(2)\,dt+8t^6(3t^2)\,dt\)
\(\Rightarrow (8t^8+4t^8+24t^8)\,dt = 36t^8\,dt\)
\(\displaystyle\int_C\vec{F}\cdot d\vec{r} = \int_0^1 36t^8\,dt = \left[4t^9\right]_0^1\)
Question Q-29: Evaluate \(\displaystyle\int_0^{\pi/2}(3\sin u - 2\cos u)\,du\)
Solution:
\(\Rightarrow (-3\cos\frac{\pi}{2} - 2\sin\frac{\pi}{2}) - (-3\cos 0 - 2\sin 0)\)
\(\Rightarrow (0 - 2) - (-3 - 0)\)
\(\Rightarrow -2 + 3\)
Question Q-30: If \(\vec{A}(t) = (t-t^2)\hat{i} + \hat{k}\) and \(\vec{B}(t) = 2t^2\hat{i} + 6t\hat{k}\), evaluate (a) \(\displaystyle\int_0^1\vec{A}\cdot\vec{B}\,dt\) (b) \(\displaystyle\int_0^1\vec{A}\times\vec{B}\,dt\)
Solution:
(a)
\(\vec{A}\cdot\vec{B} = (t-t^2)(2t^2) + (0)(0) + (1)(6t)\)
\(\Rightarrow 2t^3 - 2t^4 + 6t\)
\(\displaystyle\int_0^1\vec{A}\cdot\vec{B}\,dt = \int_0^1(2t^3-2t^4+6t)\,dt\)
\(\Rightarrow \left[\dfrac{t^4}{2} - \dfrac{2t^5}{5} + 3t^2\right]_0^1\)
\(\Rightarrow \dfrac{1}{2} - \dfrac{2}{5} + 3 = \dfrac{5}{10} - \dfrac{4}{10} + \dfrac{30}{10}\)
(b)
\(\hat{i}:\ (0)(6t)-(1)(0) = 0\)
\(\hat{j}:\ -[(t-t^2)(6t)-(1)(2t^2)] = -[6t^2-6t^3-2t^2] = -(4t^2-6t^3) = 6t^3-4t^2\)
\(\hat{k}:\ (t-t^2)(0)-(0)(2t^2) = 0\)
\(\Rightarrow \vec{A}\times\vec{B} = (6t^3-4t^2)\hat{j}\)
\(\displaystyle\int_0^1\vec{A}\times\vec{B}\,dt = \int_0^1(6t^3-4t^2)\,dt\;\hat{j}\)
\(\Rightarrow \left[\dfrac{3t^4}{2} - \dfrac{4t^3}{3}\right]_0^1\hat{j} = \left(\dfrac{3}{2}-\dfrac{4}{3}\right)\hat{j} = \left(\dfrac{9}{6}-\dfrac{8}{6}\right)\hat{j}\)
Question Q-37: If \(\vec{A} = (2y+3)\hat{i} + xz\hat{j} + (yz-x)\hat{k}\), evaluate \(\displaystyle\int_C\vec{A}\cdot d\vec{r}\) along (a) \(x=2t^2,\ y=t,\ z=t^3\) from \(t=0\) to \(t=1\), (b) the straight lines from \((0,0,0)\) to \((0,0,1)\), then to \((0,1,1)\), then to \((2,1,1)\), (c) the straight line joining \((0,0,0)\) and \((2,1,1)\).
Solution:
\(\vec{A}\cdot d\vec{r} = (2y+3)\,dx + xz\,dy + (yz-x)\,dz\)
(a) \(x=2t^2,\ y=t,\ z=t^3,\quad dx=4t\,dt,\ dy=dt,\ dz=3t^2\,dt\)
\(\Rightarrow (2t+3)(4t)\,dt + (2t^2)(t^3)\,dt + (t\cdot t^3-2t^2)(3t^2)\,dt\)
\(\Rightarrow (8t^2+12t)\,dt + 2t^5\,dt + (t^4-2t^2)(3t^2)\,dt\)
\(\Rightarrow (8t^2+12t+2t^5+3t^6-6t^4)\,dt\)
\(\displaystyle\int_0^1 = \left[\dfrac{3t^7}{7}+\dfrac{t^6}{3}-\dfrac{6t^5}{5}+\dfrac{8t^3}{3}+6t^2\right]_0^1\)
\(\Rightarrow \dfrac{3}{7}+\dfrac{1}{3}-\dfrac{6}{5}+\dfrac{8}{3}+6 = \dfrac{45+35-126+280+630}{105}\)
(b) Segment 1: \((0,0,0)\to(0,0,1)\): \(x=0,\ y=0,\ dx=dy=0\)
\(\Rightarrow (yz-x)\,dz = (0-0)\,dz = 0 \qquad \displaystyle\int = 0\)
Segment 2: \((0,0,1)\to(0,1,1)\): \(x=0,\ z=1,\ dx=dz=0\)
\(\Rightarrow xz\,dy = 0\cdot1\,dy = 0 \qquad \displaystyle\int = 0\)
Segment 3: \((0,1,1)\to(2,1,1)\): \(y=1,\ z=1,\ dy=dz=0\)
\(\Rightarrow (2y+3)\,dx = (2+3)\,dx = 5\,dx\)
\(\displaystyle\int_0^2 5\,dx = 5[x]_0^2 = 10\)
\(\Rightarrow\) Total \(= 0+0+10\)
(c) Straight line: \(x=2t,\ y=t,\ z=t,\ t:0\to1,\quad dx=2\,dt,\ dy=dt,\ dz=dt\)
\(\Rightarrow (2t+3)(2)\,dt + (2t)(t)\,dt + (t^2-2t)\,dt\)
\(\Rightarrow (4t+6+2t^2+t^2-2t)\,dt\)
\(\Rightarrow (3t^2+2t+6)\,dt\)
\(\displaystyle\int_0^1(3t^2+2t+6)\,dt = \Big[t^3+t^2+6t\Big]_0^1 = 1+1+6\)
Question Q-38: If \(\vec{F} = (5xy-6x^2)\hat{i}+(2y-4x)\hat{j}\), evaluate \(\displaystyle\int_C\vec{F}\cdot d\vec{r}\) along the curve \(C\) in the \(xy\) plane, \(y = x^3\) from the point \((1,1)\) to \((2,8)\).
Solution:
\(\vec{F}\cdot d\vec{r} = (5xy-6x^2)\,dx + (2y-4x)\,dy\)
Since \(y = x^3\), we have \(dy = 3x^2\,dx\), and \(x: 1\to2\)
Substituting \(y = x^3\) and \(dy = 3x^2\,dx\):
\(\Rightarrow (5x\cdot x^3-6x^2)\,dx + (2x^3-4x)(3x^2)\,dx\)
\(\Rightarrow (5x^4-6x^2)\,dx + (6x^5-12x^3)\,dx\)
\(\Rightarrow (6x^5+5x^4-12x^3-6x^2)\,dx\)
\(\displaystyle\int_C\vec{F}\cdot d\vec{r} = \int_1^2(6x^5+5x^4-12x^3-6x^2)\,dx\)
\(\Rightarrow \Big[x^6+x^5-3x^4-2x^3\Big]_1^2\)
\(\Rightarrow (64+32-48-16)-(1+1-3-2)\)
\(\Rightarrow 32-(-3)\)
Question Q-55: If \(\varphi = 2xyz^2\), evaluate \(\displaystyle\int_C\varphi\,d\vec{r}\) where \(C\) is: (a) \(x=t,\ y=t^2,\ z=t^3\) from \(t=0\) to \(t=1\), (b) consists of the straight lines from \((0,0,0)\) to \((1,0,0)\), then to \((1,1,0)\), and then to \((1,1,1)\).
Solution:
\(\displaystyle\int_C\varphi\,d\vec{r} = \hat{i}\int_C\varphi\,dx + \hat{j}\int_C\varphi\,dy + \hat{k}\int_C\varphi\,dz\)
(a) \(dx=dt,\ dy=2t\,dt,\ dz=3t^2\,dt\), and \(\varphi = 2(t)(t^2)(t^6) = 2t^9\)
\(\hat{i}:\quad \displaystyle\int_0^1 2t^9\,dt = \left[\dfrac{t^{10}}{5}\right]_0^1 = \dfrac{1}{5}\)
\(\hat{j}:\quad \displaystyle\int_0^1 2t^9\cdot2t\,dt = \int_0^1 4t^{10}\,dt = \left[\dfrac{4t^{11}}{11}\right]_0^1 = \dfrac{4}{11}\)
\(\hat{k}:\quad \displaystyle\int_0^1 2t^9\cdot3t^2\,dt = \int_0^1 6t^{11}\,dt = \left[\dfrac{t^{12}}{2}\right]_0^1 = \dfrac{1}{2}\)
(b) Segment 1: \((0,0,0)\to(1,0,0)\): \(y=0,\ z=0 \Rightarrow \varphi=0\), all integrals \(= 0\)
Segment 2: \((1,0,0)\to(1,1,0)\): \(x=1,\ z=0 \Rightarrow \varphi=0\), all integrals \(= 0\)
Segment 3: \((1,1,0)\to(1,1,1)\): \(x=1,\ y=1,\ dx=dy=0,\ \varphi = 2z^2\)
\(\hat{i}: 0, \qquad \hat{j}: 0\)
\(\hat{k}:\quad \displaystyle\int_0^1 2z^2\,dz = \left[\dfrac{2z^3}{3}\right]_0^1 = \dfrac{2}{3}\)
\(\Rightarrow\) Total \(= 0+0+\dfrac{2}{3}\hat{k}\)
Question Q-56: If \(\vec{F} = 2y\hat{i}-z\hat{j}+x\hat{k}\), evaluate \(\displaystyle\int_C\vec{F}\cdot d\vec{r}\) along the curve \(x=\cos t,\ y=\sin t,\ z=2\cos t\) from \(t=0\) to \(t=\pi/2\).
Solution:
\(dx = -\sin t\,dt, \qquad dy = \cos t\,dt, \qquad dz = -2\sin t\,dt\)
\(\vec{F}\cdot d\vec{r} = 2y\,dx - z\,dy + x\,dz\)
\(\Rightarrow 2\sin t(-\sin t)\,dt - 2\cos t(\cos t)\,dt + \cos t(-2\sin t)\,dt\)
\(\Rightarrow -2\sin^2 t\,dt - 2\cos^2 t\,dt - 2\sin t\cos t\,dt\)
\(\Rightarrow -2(\sin^2 t+\cos^2 t)\,dt - \sin 2t\,dt\)
\(\Rightarrow (-2-\sin 2t)\,dt\)
\(\displaystyle\int_C\vec{F}\cdot d\vec{r} = \int_0^{\pi/2}(-2-\sin 2t)\,dt\)
\(\Rightarrow \left[-2t+\dfrac{\cos 2t}{2}\right]_0^{\pi/2}\)
\(\Rightarrow \left(-\pi+\dfrac{\cos\pi}{2}\right)-\left(0+\dfrac{\cos 0}{2}\right)\)
\(\Rightarrow \left(-\pi-\dfrac{1}{2}\right)-\dfrac{1}{2}\)
Q-56. If \(\vec{F} = 2y\hat{i} - z\hat{j} + x\hat{k}\), evaluate \(\int_C \vec{F} \times d\vec{r}\) along the curve \(x = \cos t,\ y = \sin t,\ z = 2\cos t\) from \(t = 0\) to \(t = \pi/2\).
We have \(d\vec{r} = dx\,\hat{i} + dy\,\hat{j} + dz\,\hat{k}\)
\(x = \cos t \Rightarrow dx = -\sin t\, dt\)
\(\Rightarrow y = \sin t \Rightarrow dy = \cos t\, dt\)
\(\Rightarrow z = 2\cos t \Rightarrow dz = -2\sin t\, dt\)
\(\Rightarrow d\vec{r} = (-\sin t\,\hat{i} + \cos t\,\hat{j} - 2\sin t\,\hat{k})\,dt\)
Along the curve, \(\vec{F}\) becomes:
\(\vec{F} = 2\sin t\,\hat{i} - 2\cos t\,\hat{j} + \cos t\,\hat{k}\)
Now we compute \(\vec{F} \times d\vec{r}\):
\(\vec{F} \times d\vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2\sin t & -2\cos t & \cos t \\ -\sin t & \cos t & -2\sin t \end{vmatrix} dt\)
\(\hat{i}\) component: \([(-2\cos t)(-2\sin t) - (\cos t)(\cos t)] = 4\sin t\cos t - \cos^2 t\)
\(\Rightarrow \hat{j}\) component: \(-[(2\sin t)(-2\sin t) - (\cos t)(-\sin t)] = -[-4\sin^2 t + \sin t\cos t] = 4\sin^2 t - \sin t\cos t\)
\(\Rightarrow \hat{k}\) component: \([(2\sin t)(\cos t) - (-2\cos t)(-\sin t)] = 2\sin t\cos t - 2\sin t\cos t = 0\)
\(\Rightarrow \vec{F} \times d\vec{r} = [(4\sin t\cos t - \cos^2 t)\,\hat{i} + (4\sin^2 t - \sin t\cos t)\,\hat{j} + 0\,\hat{k}]\,dt\)
Now integrating from \(t = 0\) to \(t = \pi/2\):
\(\int_C \vec{F} \times d\vec{r} = \hat{i}\int_0^{\pi/2}(4\sin t\cos t - \cos^2 t)\,dt + \hat{j}\int_0^{\pi/2}(4\sin^2 t - \sin t\cos t)\,dt\)
For the \(\hat{i}\) component:
\(\int_0^{\pi/2}4\sin t\cos t\,dt = 2\int_0^{\pi/2}\sin 2t\,dt = 2\left[-\frac{\cos 2t}{2}\right]_0^{\pi/2} = [-\cos 2t]_0^{\pi/2}\)
\(\Rightarrow = [-\cos\pi + \cos 0] = [1 + 1] = 2\)
\(\int_0^{\pi/2}\cos^2 t\,dt = \int_0^{\pi/2}\frac{1+\cos 2t}{2}\,dt = \left[\frac{t}{2} + \frac{\sin 2t}{4}\right]_0^{\pi/2} = \frac{\pi}{4}\)
\(\Rightarrow \hat{i}\) component \(= 2 - \dfrac{\pi}{4}\)
For the \(\hat{j}\) component:
\(\int_0^{\pi/2}4\sin^2 t\,dt = 4\int_0^{\pi/2}\frac{1-\cos 2t}{2}\,dt = 2\left[t - \frac{\sin 2t}{2}\right]_0^{\pi/2} = 2\cdot\frac{\pi}{2} = \pi\)
\(\int_0^{\pi/2}\sin t\cos t\,dt = \frac{1}{2}\int_0^{\pi/2}\sin 2t\,dt = \frac{1}{2}\left[-\frac{\cos 2t}{2}\right]_0^{\pi/2} = \frac{1}{2}[1+1]\cdot\frac{1}{2} = \frac{1}{2}\)
\(\Rightarrow \hat{j}\) component \(= \pi - \dfrac{1}{2}\)
Therefore:
Q-1. State the Divergence theorem of Gauss, Stoke's theorem, Green's theorem in the plane.
Divergence Theorem of Gauss:
If \(V\) is the volume bounded by a closed surface \(S\) and \(\vec{A}\) is a vector function of position having continuous derivatives, then:
\(\iint\limits_S \vec{A} \cdot d\vec{S} = \iiint\limits_V \nabla \cdot \vec{A}\ dV\)
or \(= \iiint\limits_V \text{div}\ \vec{A}\ dV\)
where \(\hat{n}\) is the outward drawn unit normal to \(S\).
Stoke's Theorem:
If \(S\) is an open two-sided surface bounded by a closed non-self-intersecting curve \(C\), and if \(\vec{A}\) has continuous partial derivatives, then:
\(\oint_C \vec{A} \cdot d\vec{r} = \iint\limits_S (\nabla \times \vec{A}) \cdot d\vec{S}\)
\(\Rightarrow = \iint\limits_S (\nabla \times \vec{A}) \cdot \hat{n}\ dS\)
where \(\hat{n}\) is the positive unit normal to \(S\) and \(C\) is traversed in the positive direction.
Green's Theorem in the Plane:
If \(R\) is a closed region in the \(xy\)-plane bounded by a simple closed curve \(C\), and if \(M\) and \(N\) are continuous functions of \(x\) and \(y\) having continuous partial derivatives in \(R\), then:
\(\oint_C (M\,dx + N\,dy) = \iint\limits_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)dx\,dy\)
where \(C\) is traversed in the positive (counterclockwise) direction.
Q-2. Verify Green's theorem in the plane for \(\displaystyle\oint_C (xy + y^2)\,dx + x^2\,dy\) where \(C\) is the closed curve of the region bounded by \(y = x\) and \(y = x^2\).
Green's theorem states:
\(\oint_C (M\,dx + N\,dy) = \iint\limits_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)dx\,dy\)
Here \(M = xy + y^2\), \(N = x^2\).
\(\frac{\partial N}{\partial x} = 2x, \quad \frac{\partial M}{\partial y} = x + 2y\)
\(\Rightarrow \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 2x - x - 2y = x - 2y\)
The curves \(y = x\) and \(y = x^2\) intersect where \(x = x^2\), i.e. \(x = 0\) and \(x = 1\).
In the region \(R\), for \(0 \leq x \leq 1\): \(x^2 \leq y \leq x\).
Right side (double integral):
\(\iint\limits_R (x - 2y)\,dx\,dy = \int_0^1\int_{x^2}^{x}(x-2y)\,dy\,dx\)
\(\Rightarrow = \int_0^1 \left[xy - y^2\right]_{x^2}^{x}dx\)
\(\Rightarrow = \int_0^1 \left[(x\cdot x - x^2) - (x\cdot x^2 - x^4)\right]dx\)
\(\Rightarrow = \int_0^1 \left[0 - x^3 + x^4\right]dx\)
\(\Rightarrow = \int_0^1 (x^4 - x^3)\,dx\)
\(\Rightarrow = \left[\frac{x^5}{5} - \frac{x^4}{4}\right]_0^1\)
\(\Rightarrow = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20}\)
Left side (line integral):
The boundary \(C\) consists of two paths:
\(C_1\): along \(y = x^2\) from \((0,0)\) to \((1,1)\)
\(C_2\): along \(y = x\) from \((1,1)\) to \((0,0)\)
Along \(C_1\): \(y = x^2\), \(dy = 2x\,dx\), \(x\) from \(0\) to \(1\):
\(\int_{C_1} = \int_0^1 [(x\cdot x^2 + x^4)\,dx + x^2\cdot 2x\,dx]\)
\(\Rightarrow = \int_0^1 (x^3 + x^4 + 2x^3)\,dx\)
\(\Rightarrow = \int_0^1 (3x^3 + x^4)\,dx\)
\(\Rightarrow = \left[\frac{3x^4}{4} + \frac{x^5}{5}\right]_0^1 = \frac{3}{4} + \frac{1}{5} = \frac{19}{20}\)
Along \(C_2\): \(y = x\), \(dy = dx\), \(x\) from \(1\) to \(0\):
\(\int_{C_2} = \int_1^0 [(x\cdot x + x^2)\,dx + x^2\,dx]\)
\(\Rightarrow = \int_1^0 (x^2 + x^2 + x^2)\,dx\)
\(\Rightarrow = \int_1^0 3x^2\,dx\)
\(\Rightarrow = \left[x^3\right]_1^0 = 0 - 1 = -1\)
\(\oint_C = \int_{C_1} + \int_{C_2} = \frac{19}{20} + (-1) = \frac{19}{20} - \frac{20}{20} = -\frac{1}{20}\)
Since both sides equal \(-\dfrac{1}{20}\), Green's theorem is verified. \(\blacksquare\)
Q-9. Evaluate \(\displaystyle\oint_C (y - \sin x)\,dx + \cos x\,dy\), where \(C\) is the triangle with vertices \(O(0,0)\), \(A(\pi/2, 0)\), \(B(\pi/2, 1)\).
(a) Directly:
\(C\) consists of three paths \(C_1\), \(C_2\), \(C_3\).
Along \(C_1\): \(OA\), \(y = 0\), \(dy = 0\), \(x\) from \(0\) to \(\pi/2\)
\(\int_{C_1} = \int_0^{\pi/2}(0 - \sin x)\,dx\)
\(\Rightarrow = \Big[\cos x\Big]_0^{\pi/2}\)
\(\Rightarrow = \cos\frac{\pi}{2} - \cos 0\)
\(\Rightarrow = 0 - 1 = -1\)
Along \(C_2\): \(AB\), \(x = \dfrac{\pi}{2}\), \(dx = 0\), \(y\) from \(0\) to \(1\)
\(\int_{C_2} = \int_0^1 \cos\frac{\pi}{2}\,dy\)
\(\Rightarrow = \int_0^1 0\,dy = 0\)
Along \(C_3\): \(BO\), from \((\pi/2,\, 1)\) to \((0,\, 0)\), the line \(OB\) gives \(y = \dfrac{2x}{\pi}\), so \(dy = \dfrac{2}{\pi}\,dx\), \(x\) from \(\pi/2\) to \(0\)
\(\int_{C_3} = \int_{\pi/2}^{0}\left[\left(\frac{2x}{\pi} - \sin x\right)dx + \cos x \cdot \frac{2}{\pi}\,dx\right]\)
\(\Rightarrow = \int_{\pi/2}^{0}\left(\frac{2x}{\pi} - \sin x + \frac{2\cos x}{\pi}\right)dx\)
\(\Rightarrow = \left[\frac{x^2}{\pi} + \cos x + \frac{2\sin x}{\pi}\right]_{\pi/2}^{0}\)
\(\Rightarrow = \left[0 + 1 + 0\right] - \left[\frac{\pi}{4} + 0 + \frac{2}{\pi}\right]\)
\(\Rightarrow = 1 - \frac{\pi}{4} - \frac{2}{\pi}\)
Adding all three paths:
\(\oint_C = -1 + 0 + 1 - \frac{\pi}{4} - \frac{2}{\pi}\)
\(\Rightarrow \boxed{\oint_C = -\dfrac{\pi}{4} - \dfrac{2}{\pi}}\)
(b) By Green's Theorem:
\(\oint_C (M\,dx + N\,dy) = \iint\limits_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)dx\,dy\)
Here \(M = y - \sin x\), \(N = \cos x\)
\(\frac{\partial N}{\partial x} = -\sin x\)
\(\Rightarrow \frac{\partial M}{\partial y} = 1\)
\(\Rightarrow \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -\sin x - 1\)
The hypotenuse \(OB\) has equation \(y = \dfrac{2x}{\pi}\). For the region \(R\): \(x\) from \(0\) to \(\dfrac{\pi}{2}\), \(y\) from \(0\) to \(\dfrac{2x}{\pi}\)
\(\iint\limits_R (-\sin x - 1)\,dx\,dy = \int_0^{\pi/2}\int_0^{2x/\pi}(-\sin x - 1)\,dy\,dx\)
\(\Rightarrow = \int_0^{\pi/2}(-\sin x - 1)\Big[y\Big]_0^{2x/\pi}\,dx\)
\(\Rightarrow = \int_0^{\pi/2}(-\sin x - 1)\cdot\frac{2x}{\pi}\,dx\)
\(\Rightarrow = \frac{2}{\pi}\int_0^{\pi/2}(-x\sin x - x)\,dx\)
\(\Rightarrow = \frac{-2}{\pi}\left[\int_0^{\pi/2}x\sin x\,dx + \int_0^{\pi/2}x\,dx\right]\)
For \(\displaystyle\int_0^{\pi/2} x\sin x\,dx\), using integration by parts, \(u = x\), \(dv = \sin x\,dx\)
\(\Rightarrow du = dx,\quad v = -\cos x\)
\(\Rightarrow = \Big[-x\cos x\Big]_0^{\pi/2} + \int_0^{\pi/2}\cos x\,dx\)
\(\Rightarrow = \left[-\frac{\pi}{2}\cdot 0 + 0\right] + \Big[\sin x\Big]_0^{\pi/2}\)
\(\Rightarrow = 0 + 1 = 1\)
For \(\displaystyle\int_0^{\pi/2} x\,dx = \left[\frac{x^2}{2}\right]_0^{\pi/2} = \frac{\pi^2}{8}\)
Substituting back:
\(= \frac{-2}{\pi}\left[1 + \frac{\pi^2}{8}\right]\)
\(\Rightarrow = \frac{-2}{\pi} - \frac{\pi}{4}\)
\(\Rightarrow \boxed{\oint_C = -\dfrac{2}{\pi} - \dfrac{\pi}{4}}\)
Both results are equal, hence verified. \(\blacksquare\)