Suppose that we know that if f (x, y) is differentiable, then the rate at which f changes with respect to t along a differentiable curve x = g (t), y = h (t) is

At any point P₀ (x ₀, y ₀) = P₀ (g (t ₀), h (t ₀)), this equation gives the rate of change of f with respect to increasing t. Therefore, it depends on the direction of motion along the curve. This fact is important when the curve is a straight line and t is the arc length parameter along the line measured from P₀ in the direction of a given unit vector u. This implies that df /dt is the rate of change of f with respect to distance in its domain in the direction of u. By varying u, we will find the rates at which f changes with respect to distance as we move through P₀ in different directions. These are what we will call directional derivatives.

Suppose that the function f (x, y) is defined throughout a region R in the xy-plane. (See figure 1) Let P₀ (x ₀, y ₀) is a point in R, and u = u ₁i + u ₂ j be a unit vector.

figure 1

Then the equations x = x ₀ + su ₁, y = y ₀ + su ₂ parameterize the line through P₀ parallel to u. The parameter s measures the arc length from P₀ in the direction of u by calculating df /ds at P₀.

DEFINITION:

The derivative of f at P₀ (x ₀, y ₀) in the direction of the unit vector u = u₁i + u ₂ j is the number

provided this limit exists,

This directional derivative is also denoted by the following notation:

Let the equation z = f (x, y) represent a surface S in space. If z ₀ = f (x ₀, y ₀), then the point P (x ₀, y ₀, z ₀) lies on S. (See figure 2)

figure 2

The vertical plane that passes through P and P₀ (x ₀, y ₀) parallel to u intersects S in a curve C. The rate of change of f in the direction of u is the slope of the tangent line to C at P.

When u = i, the directional derivative at P₀ is ¶ f /¶ x evaluated at (x ₀, y ₀). When u = j_, the directional derivative at P₀ is ¶ f /¶ y evaluated at (x ₀, y ₀). Therefore, the directional derivative generalizes to two partial derivatives, and we can now find the rate of change of f in any direction u.

To avoid using the definition to find the directional derivative, let us start with the line through P₀ (x ₀, y ₀) parameterized with the arc length parameter s increasing in the direction of the unit vector u = u₁i + u₂ j. Here is the equation for the line: x = x ₀ + su ₁, y = y ₀ + su ₂.

Then

DEFINITION:

The gradient vector (gradient) of f (x, y) at a point P₀ (x ₀, y ₀) is the vector

obtained by evaluating the partial derivatives of f at P₀.

THEOREM:

If the partial derivatives of f (x, y) are defined at P₀ (x ₀, y ₀). Then

the scalar product of the gradient f at P₀ and u.

 EXAMPLE 1:

 Find the gradient of the following function at the given point, and then sketch the level curve and the gradient together.

SOLUTION:

This is the graph of the level curve through P₀ and the associated gradient vector.

EXAMPLE 2:

Find the derivative of the function f (x, y) = 2xy - 3y ² at P₀ (5, 5) in the direction of A = 4i + 3j,

SOLUTION:

EXAMPLE 3:

Find the derivative of the function f (x, y) = x ² - xy + y ² at P₀ (1, 1) in the direction of A = i - 3 j.

SOLUTION:

When evaluating the dot product in the formula

D _u f = Ñ f · u = | Ñ f || u | cos q = | Ñ f | cos q ,

it reveals the following properties.

1.

The function f increases most rapidly when cos q = 1, or when u is in the direction of Ñf. (f increase most rapidly in the direction of the gradient vector Ñ f at P.) The derivative in this direction is D _u f = | Ñ f | cos 0 = | Ñ f |.

2.

f decreases most rapidly in the direction of -Ñ f.

D _u f = | Ñ f | cos p = - | Ñ f |

3.

Any direction u orthogonal to the gradient is a direction of zero change in f because q = p /2 and D _u f = | Ñ f | cos (p /2) = | Ñ f |(0) = 0.

EXAMPLE 4:

Find the direction in which f (x, y) = x ² + xy + y ² increases and decreases most rapidly at P₀ (-1, 0).

SOLUTION:

Most rapid increase is in the direction of u.

Most rapid decrease is in the direction of -u.

 If a differentiable function f (x, y) has a constant value c along a smooth curve r = g (t) i + h(t) j (making the curve a level curve of f), then f (g (t), h (t)) = c. Differentiating both sides of this equation with respect to t leads to the following equations.

This implies that Ñ f is normal to the tangent vector dr/dt, therefore it is normal to the curve.

 FACT:

 At every point (x ₀, y ₀) in the domain of f (x, y), the gradient of f is normal to the level curve through (x ₀, y ₀).

FACT:

The tangent lines to a level curves are lines normal to the gradients.

FACT:

The line through a point P₀ (x ₀, y ₀) normal to a vector N = Ai + Bj has the equation A(x - x ₀) + B(y - y ₀) = 0.

If N is the gradient (Ñ f)_{(x0, y0)} = f _x (x ₀, y ₀)i + f _y (x ₀, y ₀)j, then the equation of the line through the point (x ₀, y ₀) normal to the gradient becomes

f _x (x ₀, y ₀) (x - x ₀) + f _y (x ₀, y ₀) (y - y ₀) = 0.

 EXAMPLE 5:

Sketch the curve f (x, y) = c together with Ñ f and the tangent line at the given point. Then write the equation for the tangent line.

x ² - xy + y ² = 7 (-1, 2)

SOLUTION:

f _x = 2x - y

f _x (-1, 2) = -2 - 2 = -4

Ñ f = -4i + 5j

TL: -4(x + 1) + 5(y - 2) = 0

® -4x - 4 + 5y - 10 = 0

® -4x + 5y = 14

f _y = -x + 2y

f _y (-1, 2) = 1 + 4 = 5

We will obtain the three variable formulas by adding the z-terms to the two-variable formulas.

 FACT:

For a differentiable function f (x, y, z) and a unit vector u = u ₁i + u ₂ j + u ₃ k in space, we have Ñ f = f _x i +f _y j + f _z k and

D_u f = Ñ f · u = |Ñ f || u | cos q = |Ñ f | cos q .

 All the properties hold for the three-variable case.

 EXAMPLE 6:

Find Ñ f of f (x, y, z) = (x ² + y ² + z ²) ^{-1/ 2} + ln (xyz) at (-1, 2, -2).

SOLUTION:

EXAMPLE 7:

Find the derivative of the function f (x, y, z) = 3e ^x cos yz at P₀ (0, 0, 0) in the direction of A = 2i + j - 2k.

SOLUTION:

 EXAMPLE 8:

Find the directions in which the function h (x, y, z) = ln (x ² + y ² - 1) + y + 6z increase and decrease most rapidly at P₀ (1, 1, 0). Then find the derivatives of the functions in these directions. 

SOLUTION:

The most rapid increase is in the direction of u. D _u h = |Ñ h | cos 0 = 7

The most rapid decrease is in the direction of -u.

D _u h = |Ñ h | cos p = -7

 If r = g (t) i + h (t) j + k (t) k us a smooth curve on the level surface f (x, y, z) = c of a differentiable function of f, then f (g (t), h (t), k (t)) = c. Now, differentiate both sides with respect to t.

DEFINITION:

The tangent plane at the point P₀ (x ₀, y₀, z ₀) on the level surface f(x,y,z) = c is the plane through P₀ normal to Ñ f |_Po. The normal line of the surface at P₀ is the line through P₀ parallel to Ñ f |_Po.

Normal line eq: x = x ₀ + f _x (P₀)t, y = y ₀ + f _y (P₀)t, z = z ₀ + f _z (P₀)t

Tangent plane eq: f _x (P₀)(x - x ₀) + f _y (P₀)(y - y ₀) + f _z (P₀)(z - z ₀)

 EXAMPLE 9:

Find the equations for the (a) tangent plane and (b) normal line at the point (0, 1, 2) on the surface cos p x - x ²y + e ^{x z} + yz = 4. 

SOLUTION:

f _x = -p sin p x -2xy + ze ^{x z}

f _x (0, 1, 2) = -p sin 0 -2(0)(1) + 2e ⁽⁰⁾⁽²⁾ = 2

f _y = x ² + z

f _y (0, 1, 2) = 0 + 2 = 2

f _z =xe ^{x z} + y

f _z (0, 1, 2) = 0 + 1 = 1

Ñ f = 2i + 2j + k

Tangent plane: 2(x - 0) + 2(y - 1) + 1(z - 2) = 0 ® 2x + 2y + z = 4

Normal line: x = 2t, y = 1 + 2t, z = 2 + t

EXAMPLE 10:

Find the parametric equations for the line tangent to the curve of intersection of the surfaces x ³ + 3x ² y ² + y ³ + 4xy - z ² = 0 and x ² + y ² + z ² = 11 at the point (1, 1, 3).

SOLUTION:

Let g: x ³ + 3x ² y ² + y ³ + 4xy - z ² = 0 and f: x ² + y ² + z ² = 11.

 g _x = 3x ² + 6xy ² + 4y

g _x (1, 1, 3) = 3 + 6 + 4 = 13

g _y = 6x ²y + 3y ² + 4x

g _y (1, 1, 3) = 6 + 3 + 4 = 13

g _z = -2z

g _z (1, 1, 3) = -6

Ñ g = 13i + 13j - 6k

f _x = 2x

f _x (1, 1, 3) = 2

f _y = 2y

f _y (1, 1, 3) = 2

f _z = 2z

f _z (1, 1, 3) = 6

Ñ f = 2i + 2j + 6k

FACT: The tangent line is orthogonal to both Ñ f and Ñ g at P₀, therefore it is parallel to v = Ñ f ´ Ñ g.

Tangent line: x = 1 + 90t, y = 1 - 90t, z = 3

To find the equation for the plane tangent to the surface z = f (x, y) at a point P₀ (x ₀, y ₀, z ₀) where z ₀ = f (x ₀, y ₀), we will have to first observe the following.

z = f (x, y) if and only if f (x, y) - z = 0 = F(x,y)

Therefore, the tangent plane equation is the following.

F _x (P₀)(x - x ₀) + F _y (P₀)(y - y ₀) - (z - z ₀) = 0

 EXAMPLE 11:

Find the equation for the plane that is tangent to the surface z = ln(x ² + y²) at the point (1, 0, 0).

SOLUTION:

F = ln (x ² + y ²) - z

 The directional derivative plays a role of an ordinary derivative when we want to estimate how much a function f changes if we move a small distance ds from a point P₀ to another point nearby. If f where a function of a single variable, then df = f ' (P₀)ds. For a function of two or more variables, df = (Ñ f |_Po · u)ds where u is the direction of motion away from P₀.

 EXAMPLE 12:

By about how much will g (x, y, z) = x + x cos z - y sin z + y change if the point P (x, y, z) moves from P₀ (2, -1, 0) a distance of ds = 0.2 units toward the point P₁ (0, 1, 2)?

SOLUTION:

First, I must find the direction we are moving by find the vector P₀P₁.

P₀P₁ = (0 - 2)i +(1 + 1) j + (2 - 0)k = -2i + 2j + 2k

Now to find the length of this vector.

If we know the gradients for two functions f and g, we can find the gradients of their constant multiples, sum, difference, product, and quotient.

1.

Constant multiple rule:

Ñ (k f) = k Ñ f

2.

Sum rule:

Ñ (f + g) = Ñ f + Ñ g

3.

Difference rule:

Ñ (f - g) = Ñ f - Ñ g

4.

Product rule:

Ñ (fg) = f Ñ g + g Ñ f

5.

Quotient rule: