Worked examples · Sheet 3

Directional derivatives and direction of steepest ascent

Theory used: Directional derivative Gradient formula Direction of steepest ascent

Exercise 1

Compute the directional derivative of

f (x, y) = x^{2} y + y^{3}

at the point $(1, 2)$ along the unit vector $v = (\frac{3}{5}, \frac{4}{5})$ .

f_{x} = 2 x y, f_{y} = x^{2} + 3 y^{2}

\nabla f (1, 2) = (2 \cdot 1 \cdot 2, 1 + 3 \cdot 4) = (4, 13)

D_{v} f (1, 2) = 4 \cdot \frac{3}{5} + 13 \cdot \frac{4}{5} = \frac{12 + 52}{5} = \frac{64}{5}

The function is a polynomial, so the partial derivatives are continuous on all of $R^{2}$ , and by the total differential theorem $f$ is differentiable at every point. The gradient formula applies, therefore the directional derivative is the dot product of the gradient at the point and the unit vector. The given vector has norm $\frac{9}{25} + \frac{16}{25} = 1$ , so it is already a unit vector and the dot product directly gives $\frac{64}{5}$ .

Exercise 2

Compute the directional derivative of

f (x, y) = e^{x} cos y

at the point $(0, \frac{π}{2})$ in the direction of the vector $w = (1, - 1)$ .

v = \frac{w}{∣ w ∣} = (\frac{1}{2}, - \frac{1}{2})

f_{x} = e^{x} cos y, f_{y} = - e^{x} sin y

\nabla f (0, \frac{π}{2}) = (e^{0} cos \frac{π}{2}, - e^{0} sin \frac{π}{2}) = (0, - 1)

D_{v} f (0, \frac{π}{2}) = 0 \cdot \frac{1}{2} + (- 1) (- \frac{1}{2}) = \frac{1}{2} = \frac{2}{2}

The vector $w = (1, - 1)$ has norm $2$ , so one first normalizes it by dividing by its norm: the directional derivative is defined along a unit vector. The function is the product of an exponential and a cosine, with continuous partial derivatives, therefore it is differentiable and the gradient formula applies. At the given point the cosine vanishes and the sine equals $1$ , so the gradient is $(0, - 1)$ . The dot product with the unit vector gives $\frac{1}{2}$ , which after rationalizing is written $\frac{2}{2}$ .

Exercise 3

Determine the direction of steepest ascent of

f (x, y) = ln (1 + x^{2} + 2 y^{2})

at the point $(1, 1)$ and the maximum value of the directional derivative at that point.

f_{x} = \frac{2 x}{1 + x ^{2} + 2 y ^{2}}, f_{y} = \frac{4 y}{1 + x ^{2} + 2 y ^{2}}

\nabla f (1, 1) = (\frac{2}{4}, \frac{4}{4}) = (\frac{1}{2}, 1)

\nabla f (1, 1) = \frac{1}{4} + 1 = \frac{5}{2}

v = \frac{\nabla f ( 1 , 1 )}{∣\nabla f ( 1 , 1 ) ∣} = (\frac{1}{5}, \frac{2}{5}), v max D_{v} f (1, 1) = \frac{5}{2}

The level curves

x^{2} + 2 y^{2} = 3

and

x^{2} + 2 y^{2} = 6

(dashed), where

f

equals

ln 4

and

ln 7

: at

(1, 1)

the gradient

(\frac{1}{2}, 1)

is orthogonal to the tangent and points toward higher levels

The function is differentiable because its partial derivatives are continuous, since the denominator $1 + x^{2} + 2 y^{2}$ is always greater than or equal to $1$ . At the point $(1, 1)$ the denominator equals $4$ , so the gradient is $(\frac{1}{2}, 1)$ . By the theorem on the direction of steepest ascent, the directional derivative is maximal along the unit vector of the gradient and its maximum value is the norm of the gradient. Normalizing $(\frac{1}{2}, 1)$ , which has norm $\frac{5}{2}$ , gives the unit vector $(\frac{1}{5}, \frac{2}{5})$ ; the maximum rate of increase equals $\frac{5}{2}$ .