# 1. Introduction (Vectors)

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.  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) = 3x2y + cos (3z)

Solution: The gradient of the function ƒ (x,y,z) = 3x2y + 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/ √y j − 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.

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/∂z ∂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) = sin (x) ey ∂/∂z (ln(z)) = sin (x) ey  1/z

ƒ (x,y,z) = cos (x) ey ln(z)i + sin (x) ey ln(z)j + sin (x) ey1/z k

# 3. Directional Derivatives

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) = x2yzxy2z 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) = x2yzxy2z 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 .

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).

 ∇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 = xy3z = 2 and calculate the gradient of ƒ:

ƒ = y3i + 3xy2jk

At the point (1, 1, −1) this is ƒ = i + 3jk. 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 + 3jk).

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) −2ijk 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 / 2 j + 1 / 2 k Incorrect - please try again!
(b) πi + j + 2 / π k Incorrect - please try again!
(c) i Incorrect - please try again!
(d) −  1 / 2 j1 / 2 k 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 = −jk

The magnitude of this vector is:

(−1)2 + (−1)2 = 2

Therefore the unit normal is: = − 1 / 2  j1 / 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 j1 /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

# 5. Quiz on Gradients and Directional Derivatives

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/zjxy/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 + jk)
(b) ±1 / 3  (i + j + k)
(c) ±1 / 2  (i + j)
(d) ±1 / 2  (2i + 2j − 4k)