The base vectors in two dimensional Cartesian coordinates are the unit vector i in the positive direction of the x-axis and the unit vector j in the y direction (see figure bottom left). In three dimensions we also require k, the unit vector in the z direction.
The position vector of a point P (x, y) in two dimensions is xi + yj. We will often denote this important vector by r (see figure bottom right). In three dimensions the position vector is r = xi + yj + zk.
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The vector differential operator ∇, called 'del' or 'nabla', is defined in three dimensions to be:
∇ = ∂/∂xi + ∂/∂yj + ∂/∂zk
- note that these are partial derivatives!
This vector operator may be applied to (differentiable) scalar functions (scalar fields) and the result is a special case of a vector field, called a gradient vector field.
Here are two warming up exercises on partial differentiation:
Quiz 1: Select the following partial derivative: ∂/∂z(xyzx):
(a) x2yzx−1 Correct - well done!
(b) 0 Incorrect - please try again!
(c) xy logx(z) Incorrect - please try again!
(d) yzx−1 Incorrect - please try again!
Solution:The partial derivative of xyzx with respect to the variable z is:
∂/∂z (xyzx) = xy × ∂/∂z (zx) = xy × x × zx−1 = x2yzx−1
Quiz 2: Choose the following partial derivative ∂/∂x(x cos (y) + y):
(a)cos (y) Correct - well done!
(b)cos (y) − x sin (y) + 1 Incorrect - please try again!
(c) cos (y) + x sin (y) + 1 Incorrect - please try again!
(d)−sin (y)
Solution: Consider the function ƒ (x,y) = x cos (y) + y. Its derivative with respect to the variable x is:
| ∂/∂x ƒ (x,y) | = | ∂/∂x cos (y) + y |
| = | ∂/∂x (x) × cos (y) + ∂/∂x y | |
| = | 1×cos (y) + 0=cos (y) |
The gradient of a function, ƒ (x,y), in two dimensions is defined as:
grad ƒ (x,y) = ∇ƒ (x,y) = ∂ƒ/∂x i + ∂ƒ/∂y j
The gradient of a function is a vector field. It is obtained by applying the vector operator ∇ to the scalar function ƒ (x,y). Such a vector field is called a gradient (or conservative) vector field.
Example 1: The gradient of the function ƒ (x,y) = x + y2 is given by:
| ∇ƒ (x,y) | = | ∂ƒ/∂x i + ∂ƒ/∂y j |
| = | ∂/∂x (x + y2)i + ∂/∂y (x + y2)j | |
| = | (1 + 0)i + (0 + 2y)j | |
| = | i + 2yj |
Quiz 3: Choose the gradient of ƒ (x,y) = x2y3:
(a) 2xi + 3y2j Incorrect - please try again!
(b) x2i + y3j Incorrect - please try again!
(c) 2xy3i + 3x2y2j Correct - well done!
(d) y3i + x2j
Solution: The gradient of the function ƒ (x,y) =x2y3 is given by:
| ∇ƒ (x,y) | = | ∂ƒ/∂x i + ∂ƒ/∂y j |
| = | ∂/∂x (x2y3)i + ∂/∂y (x2y3)j | |
| = | ∂/∂x (x2) × y3i + x2 × ∂/∂y (y3)j | |
| = | 2x2−1 × y3i + 3x2 × y3−1j | |
| = | 2xy3i + 3x2y2j |
The definition of the gradient may be extended to functions defined in three dimensions ƒ (x,y,z):
∇ ƒ (x,y,z) = ∂ƒ/∂x i + ∂ƒ/∂y j + ∂ƒ/∂z k
Exercise 1: Calculate the gradient of the following functions:
(a)ƒ (x,y) = x + 3y2
Solution: The function ƒ (x,y) = x + 3y2 has gradient:
| ∇ƒ (x,y) | = | ∂ƒ/∂x i + ∂ƒ/∂y j |
| = | ∂/∂x (x + 3y2)i + ∂/∂y (x + 3y2)j | |
| = | (1 + 0)i + (0 + 3 × 2y2−1)j | |
| = | i + 6yj |
(b)ƒ (x,y) = √x2 + y2
Solution: The gradient of the function ƒ (x,y) = √x2 + y2 = (x2 + y2)½ is given by:
| ∇ƒ (x,y) | = | ∂ƒ/∂x i + ∂ƒ/∂y j = ∂/∂x (x2 + y2)½i + ∂/∂y (x2 + y2)½j |
| = | 1/2 (x2 + y2)½−1 × ∂/∂x (x2)i + 1/2 (x2 + y2)½−1 × ∂/∂y (y2)j | |
| = | 1/2 (x2 + y2)−½ × 2x2−1i + 1/2 (x2 + y2)−½ × 2y2−1j | |
| = | (x2 + y2)−½ xi + (x2 + y2)−½ yj | |
| = | x/ √x2 + y2i + y/ √x2 + y2j |
(c) ƒ (x,y,z) = 3x2√y + cos (3z)
Solution: The gradient of the function ƒ (x,y,z) = 3x2√y + cos (3z) = 3x2y½ + cos (3z) is given by:
| ∇ƒ (x,y,z) | = | ∂ƒ/∂xi + ∂ƒ/∂yj + ∂ƒ/∂z k |
| = | 3y½∂/∂x (x2)i + 3x2∂/∂y (y½)j + ∂/∂y (cos (3z))k | |
| = | 3y½ × 2x2−1i + 3x2 × 1/2y½−1 j − 3 sin (3z)k | |
| = | 6y½xi + 3/2x2y−½ j − 3 sin (3z)k | |
| = | 6x√yi + 3/2 x2/ √yj − 3 sin (3z)k |
(d) ƒ (x,y,z) =1/ √x2 + y2 + z2
Solution: The partial derivative of the function ƒ (x,y,z) = 1/ √x2 + y2 + z2 = (x2 + y2 + z2)½ with respect to the variable x is:
∂ƒ/∂x = − (x2 + y2 + z2)½−1 × ∂(x2)/∂x = − x/ (x2 + y2 + z2)3 ⁄ 2
Similarly, the derivatives ∂ƒ/∂y and ∂ƒ/∂z are:
∂ƒ/∂y = y/ (x2 + y2 + z2)3 ⁄ 2 and z/ (x2 + y2 + z2)3 ⁄ 2 respectively.
Therefore the gradient is:
∇ƒ (x,y,z) = − xi + yj + zk / (x2 + y2 + z2)3 ⁄ 2
(e) ƒ (x,y) 4y/ (x2 + 1)
Solution: The gradient of the function ƒ (x,y) 4y/ (x2 + 1) = 4y(x2 + 1)−1 is:
| ∇ƒ (x,y) | = | 4y × ∂/∂x (x2 + 1)−1i + (x2 + 1)−1 × ∂/∂y 4yj |
| = | 4y × (−1)(x2 + 1)−1−1 ∂/∂x (x2 + 1)2i + 4(x2 + 1)−1j | |
| = | −4y(x2 + 1)−2 × 2xi + 4(x2 + 1)−1j | |
| = | − 8xy/ (x2 + 1)2i + 4/ (x2 + 1)j |
(f) ƒ (x,y,z) = sin (x) ey ln(z)
Solution: The partial derivatives of the function ƒ (x,y,z) = sin (x) ey ln(z) are:
| ∂ƒ/∂x | = | ∂/∂x (sin (x)) ey ln(z) = cos (x) ey ln(z) |
| ∂ƒ/∂y | = | sin (x)∂/∂y (ey) ln(z) = sin (x) ey ln(z) |
| ∂ƒ/∂z | = | sin (x) ey ∂/∂z (ln(z)) = sin (x) ey 1/z |
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Therefore the gradient is:
∇ƒ (x,y,z) = cos (x) ey ln(z)i + sin (x) ey ln(z)j + sin (x) ey 1/z k
To interpret the gradient of a scalar field
∇ƒ (x,y,z) = ∂ƒ/∂x i + ∂ƒ/∂y j + ∂ƒ/∂z k
we note that its component in the i direction is the partial derivative of ƒ with respect to x. This is the rate of change of ƒ in the x direction since y and z are kept constant. In general, the component of ∇ƒ in any direction is the rate of change of ƒ in that direction.
Example 2: Consider the scalar field ƒ (x,y) = 3x + 3 in two dimensions. It has no y dependence and it is linear in x. Its gradient is given by:
| ∇ƒ | = | ∂/∂x (3x + 3)i + ∂/∂y (3x + 3)j |
| = | 3i + 0j |
As would be expected the gradient has zero component in the y direction and its component in the x direction is constant (3).
Quiz 4: Select a point from the answers below at which the scalar field ƒ(x,y,z) = x2yz − xy2z decreases in the y direction.
(a)(1, −1, 2) Incorrect - please try again!
(b)(1, 1, 1) Correct - well done!
(c)(−1, 1, 2) Incorrect - please try again!
(d)(1, 0, 1) Incorrect - please try again!
Solution: The partial derivative of the scalar function ƒ(x,y,z) = x2yz − xy2z with respect to y is:
∂ƒ/∂y (x,y,z) = x2z − 2xyz
Evaluating it at the point (1, 1, 1) gives:
∂ƒ/∂y(1, 1, 1) = 12 × 1 − 2 × 1 × 1 × 1 = 1 − 2 = −1
This is negative and therefore the function ƒ decreases in the y direction at this point.
It may be verified that the function does not decrease in the y direction at any of the other three points.
Definition: if  is a unit vector, then ⋅ ∇ƒ is called the directional derivative of ƒ in the direction . The directional derivative is the rate of change of ƒ in the direction . |
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Example 3: Find the directional derivative of ƒ (x,y,z) = x2yz in the direction 4i − 3k at the point (1, −1, 1).
Solution: The vector 4i − 3k has magnitude √42 + (−3)2 = √25 = 5. The unit vector in the direction 4i−3k is thus 1/5(4i −3k).
The gradient of ƒ is:
| ∇ƒ | = | ∂/∂x (x2yz)i + ∂/∂y (x2yz)j + ∂/∂z (x2yz)k |
| = | 2xyzi + x2zj + x2yk |
and so the required directional derivative is:
| ⋅ ∇ƒ | = | 1/5 (4i − 3k) ⋅ (2xyzi + x2zj + x2yk) |
| = | 1/5 [4 × 2xyz + 0 − 3 × x2y] |
At the point (1, −1, 1) the desired directional derivative is thus:
⋅ ∇ƒ = 1/5 [8 × (−1) − 3 × (−1)] = −1
Exercise 2: Calculate the directional derivative of the following functions in the given directions and at the stated points:
(a)ƒ = 3x2 − 3y2 in the direction j at (1, 2, 3).
Solution: The directional derivative of the function ƒ = 3x2 − 3y2 in the unit vector j direction is given by the scalar product j ⋅ ∇.
The gradient of the function ƒ is
∇ƒ = 6xi − 6yj
Therefore the directional derivative in the unit vector j direction is:
j ⋅ ∇ƒ = j ⋅ (6xi − 6yj) = −6y
and at the point (1, 2, 3) it has the value −6 × 2 = −12.
(b)ƒ = √x2 + y2 in the direction 2i + 2j + k at (0, −2, 1).
Solution: The directional derivative of the function ƒ = √x2 + y2 in the direction defined by the vector 2i + 2j + k is given by the scalar product ⋅ ∇ƒ, where the unit vector is:
= 2i + 2j + k / √22 + 22 + 12 = 2i + 2j + k / √9 = 2/3i + 2/3j + 1/3k
The gradient of the function ƒ is:
∇ƒ = x/ √x2 + y2 i + y/ √x2 + y2 j + 0k = xi + yi/ √x2 + y2
Therefore the required directional derivative is:
⋅ ∇ƒ = ( 2/3i + 2/3j + 1/3k ) ⋅ ( xi + yi/ √x2 + y2 ) = 2/3 x + y/ √x2 + y2
At the point (0, −2, 1) it is equal to:
2/3 0 − 2/ √02 + (−2)2 = 2/3 × −2/2 = − 2/3
(c)ƒ = sin (x) + cos (y) + sin (z) in the direction πi + πj at (π, 0, π).
Solution: The directional derivative of the function ƒ = sin (x) + cos (y) + sin (z) in the direction defined by the vector πi + πj is given by the scalar product ⋅ ∇ƒ, where the unit vector is:
= πi + πj / √π2 + π2 = i + j / √2
The gradient of the function ƒ is:
∇ƒ = cos (x)i − sin (y)j + cos (z)k
Therefore the directional derivative is:
⋅ ∇ƒ = ( i + j / √2 ) ⋅ [cos (x)i − sin (y)j + cos (z)k] = cos (x) − sin (y) / √2
and at the point (π, 0, π) it becomes cos (π) − sin (0) / √2 = − 1 / √2
We now state, without proof, two useful properties of the directional derivative and gradient:
- The maximal directional derivative of the scalar field ƒ (x,y,z) is in the direction of the gradient vector ∇ƒ.
- If a surface is given by ƒ (x,y,z) = c where c is a constant, then the normals to the surface are the vectors ±∇ƒ.
Example 4: Consider the surface xy3 = z + 2. To find its unit normal at (1, 1, −1), we need to write it as : ƒ = xy3 −z = 2 and calculate the gradient of ƒ:
∇ƒ = y3i + 3xy2j − k
At the point (1, 1, −1) this is ∇ƒ = i + 3j − k. The magnitude of this maximal rate of change is √12 + 33 + (−1)2 = √11
Thus the unit normals to the surface are: 1/√11(i + 3j − k).
Quiz 5: Which of the following vectors is normal to the surface x2yz = 1 at (1, 1, 1)?
(a) 4i + j + 17k Incorrect - please try again!
(b) 2i + j + 2k Incorrect - please try again!
(c) i + j + k Incorrect - please try again!
(d) −2i − j − k Correct - well done!
Explanation: The surface is defined by the equation:
x2yz = 1
To find its unit normal at (1, 1, 1) we need to evaluate the gradient of the function ƒ (x,y,z) = x2yz:
∇ƒ = 2xyzi + x2zj + x2yk
At the point (1, 1, 1) this is:
∇ƒ = i + j + k
Thus the required normals to the surface are ±(i + j + k). Hence (d) is a normal vector to the surface.
Quiz 6: Which of the following vectors is a unit normal to the surface cos (x)yz = −1 at (π, 1, 1)?
(a) − 1/ √2j + 1/ √2k Incorrect - please try again!
(b) πi + j + 2/ √πk Incorrect - please try again!
(c) i Incorrect - please try again!
(d) − 1/ √2j − 1/ √2k Correct - well done!
Explanation: The surface is defined by the equation:
cos (x)yz = −1
To find its unit normal at the point (π, 1, 1), we need to evaluate the gradient of ƒ = cos (x)yz:
∇ƒ = −sin (x)yzi + cos (x)zj + cos (x)yk
At the point (π, 1, 1) this is:
∇ƒ = 0i + (−1)j + (−1)k = −j −k
The magnitude of this vector is:
√(−1)2 + (−1)2 = √2
Therefore the unit normal is:
= − 1/ √2 j − 1/ √2 k
Quiz 7: Select a unit normal to the (spherically symmetric) surface at x2 + y2 + z2 = 169 at (5, 0, 12):
(a) i + 1/6 j − 1/6 k Incorrect - please try again!
(b) 1/3 i + 1/3 j + 1/3 k Incorrect - please try again!
(c) 5/13 i + 12/13 k Correct - well done!
(d) − 5/13 i + 12/13 k Incorrect - please try again!
Explanation: The surface is defined by the equation:
x2 + y2 + z2 = 169
To find its unit normal at point (5, 0, 12) we need to evaluate the gradient of ƒ = x2 + y2 + z2:
∇ƒ = 2xi + 2yj + 2zk
At the point (5, 0, 12) this is:
∇ƒ = 2 × 5i + 0 × j + 2 × 12k = 10i + 24k
The magnitude of this vector is:
√(2 × 5)2 + (2 × 12)2 = √4 × (25 + 144) = 2 √169 = 2 × 13
Therefore the unit normal is:
= 5/13 j + 12/13 k
Choose the solutions from the options given
1. What is the gradient of ƒ (x, y, z) = xyz−1?
(a) i + j + z−2k
(b) y/zi + x/zj − xy/z2k
(c) yz−1 i + xz−1 j + xyz−2 k
(d) − 1/z2
2.If n is a constant, choose the gradient of ƒ (r) = 1 ⁄ rn, where r = |r| and r = xi + yj + zk.
(a)0
(b)− n/2 i + j + k/rn+1
(c) − nr/ rn+2
(d) − n/2 r/rn+2
3.Select the unit normals to the surface of revolution, z = 1/2x2 + 1/2y2, at the point (1, 1, 4)
(a) ±1/ √3 (i + j − k)
(b) ±1/ √3 (i + j + k)
(c) ±1/ √2 (i + j)
(d) ±1/ √2 (2i + 2j − 4k)
