The gradient of a scalar point function \( \phi (x,y,z) \), denoted by \( \text{grad}\,\phi \) or \( \nabla \phi \), is defined by
\[ \nabla \phi = \hat{i}\frac{\partial \phi}{\partial x} + \hat{j}\frac{\partial \phi}{\partial y} + \hat{k}\frac{\partial \phi}{\partial z} = \sum \hat{i}\frac{\partial \phi}{\partial x} \]
Note:The gradient of a scalar point function is a vector point function.
Properties of a Gradient
Property 1: \( \nabla \phi \) is a vector normal to the level surface \( \phi (x,y,z) = c \) and has a magnitude equal to the rate of \( \phi \) along this normal. Thus \( \left| \nabla \phi \right| \) gives the greatest rate of change of \( \phi (x,y,z) = c \).
Property 2: The angle between any two surfaces \( \phi_{1}(x,y,z) = c_{1} \) and \( \phi_{2}(x,y,z) = c_{2} \) is the angle between their corresponding normals, given by \( \nabla \phi_{1} \) and \( \nabla \phi_{2} \) respectively. Thus, if \( \theta \) is the angle between two surfaces \( \phi_{1}(x,y,z) = c_{1} \) and \( \phi_{2}(x,y,z) = c_{2} \), then
\[ \cos \theta = \frac{\nabla \phi_{1} \cdot \nabla \phi_{2}}{\left| \nabla \phi_{1} \right| \left| \nabla \phi_{2} \right|} \]
Property 3: Two surfaces \( \phi_{1}(x,y,z) = c_{1} \) and \( \phi_{2}(x,y,z) = c_{2} \) cut orthogonally \( \Leftrightarrow \nabla \phi_{1} \cdot \nabla \phi_{2} = 0 \).
Example 1: Find \( \text{grad}\,\phi \) for \( \phi = xy^{2}z^{3} \) at \( (1,1,1) \).
Solution: The gradient of a scalar function \( \phi \) is given by:
\[ \nabla\phi = \frac{\partial\phi}{\partial x}\hat{i} + \frac{\partial\phi}{\partial y}\hat{j} + \frac{\partial\phi}{\partial z}\hat{k} \]
Given \( \phi = xy^{2}z^{3} \), we compute the partial derivatives:
\[ \frac{\partial\phi}{\partial x} = y^{2}z^{3} \]
\[ \frac{\partial\phi}{\partial y} = 2xyz^{3} \]
\[ \frac{\partial\phi}{\partial z} = 3xy^{2}z^{2} \]
Therefore, the gradient is:
\[ \nabla\phi = y^{2}z^{3}\hat{i} + 2xyz^{3}\hat{j} + 3xy^{2}z^{2}\hat{k} \]
At the point \( (1,1,1) \):
\[ \left.\frac{\partial\phi}{\partial x}\right|_{(1,1,1)} = (1)^{2}(1)^{3} = 1 \]
\[ \left.\frac{\partial\phi}{\partial y}\right|_{(1,1,1)} = 2(1)(1)(1)^{3} = 2 \]
\[ \left.\frac{\partial\phi}{\partial z}\right|_{(1,1,1)} = 3(1)(1)^{2}(1)^{2} = 3 \]
Thus, the gradient at \( (1,1,1) \) is:
\[ \nabla\phi = \hat{i} + 2\hat{j} + 3\hat{k} \]
Example 2:Determine the constants \( a \) and \( b \) so that surfaces \( 5x^{2}-2yz-9x=0 \) and \( ax^{2}y+bz^{2}=4 \) cut orthogonally at the point \( (1,-1,2) \).
Solution: Let \( \phi_{1}(x,y,z) = 5x^{2}-2yz-9x \) and \( \phi_{2}(x,y,z) = ax^{2}y+bz^{2}-4 \).
\[ \therefore \nabla \phi_{1} = (10x-9)\hat{i} - 2z\hat{j} - 2y\hat{k} \quad \text{and} \quad \nabla \phi_{2} = 2axy\hat{i} + ax^{2}\hat{j} + 2bz\hat{k} \]
At \( (1,-1,2) \):
\[ \nabla \phi_{1} = \hat{i} - 4\hat{j} + 2\hat{k} \quad \text{and} \quad \nabla \phi_{2} = -2a\hat{i} + a\hat{j} + 4b\hat{k} \]
Two surfaces cut orthogonally if \( \nabla \phi_{1} \cdot \nabla \phi_{2} = 0 \):
\[ \nabla \phi_{1} \cdot \nabla \phi_{2} = 0 \Rightarrow (\hat{i} - 4\hat{j} + 2\hat{k}) \cdot (-2a\hat{i} + a\hat{j} + 4b\hat{k}) = 0 \]
\[ \Rightarrow -2a -4a + 8b = 0 \quad \Rightarrow \quad 8b - 6a = 0 \quad \text{(1)} \]
Since the point \( (1,-1,2) \) lies on both surfaces:
\[ \phi_{2}(1,-1,2) = 0 \Rightarrow -a + 4b = 4 \quad \text{(2)} \]
Solving equations (1) and (2), we get:
From (1): \( b = \frac{3a}{4} \)
Substitute into (2): \( -a + 4\left(\frac{3a}{4}\right) = 4 \)
\( -a + 3a = 4 \) ⇒ \( 2a = 4 \) ⇒ \( a = 2 \)
Then \( b = \frac{3 \times 2}{4} = \frac{3}{2} \)
Vector Calculus Exercises
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Find unit normal to the surface \( x^{2}y + 2xz = 4 \) at the point \( (2,-2,3) \).
Answer: \( \frac{2\hat{i} + 2\hat{j} + 4\hat{k}}{\sqrt{24}} \) or \( \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k} \) -
Find the angle between the tangent planes to the surfaces \( x\log z = y^{2}-1 \) and \( x^{2}y = 2-z \) at the point \( (1,1,1) \).
Answer: \( \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \) or approximately \( 65.9^\circ \) -
Calculate the angle between the normals to the surface \( xy = z^{2} \) at the points \( (4,1,2) \) and \( (3,3,-3) \).
Answer: \( \cos^{-1}\left(\frac{1}{\sqrt{42}}\right) \) or approximately \( 81.1^\circ \) -
Find the constants \( a \) and \( b \) so that the surface \( ax^{2} - byz = (a+2)x \) will be orthogonal to the surface \( 4x^{2}y + z^{3} = 4 \) at the point \( (1,-1,2) \).
Answer: \( a = 1 \), \( b = \frac{3}{2} \) -
If \( u = x + y + z \), \( v = x^{2} + y^{2} + z^{2} \), \( w = yz + zx + xy \), prove that \( \text{grad}\, u \), \( \text{grad}\, v \) and \( \text{grad}\, w \) are coplanar vectors.
Answer: The scalar triple product \( [\nabla u\ \nabla v\ \nabla w] = 0 \), proving they are coplanar -
Find \( \nabla \phi \), if \( \phi = \log\left(x^{2} + y^{2} + z^{2}\right) \).
Answer: \( \nabla \phi = \frac{2x}{x^{2}+y^{2}+z^{2}}\hat{i} + \frac{2y}{x^{2}+y^{2}+z^{2}}\hat{j} + \frac{2z}{x^{2}+y^{2}+z^{2}}\hat{k} \)