Directional Derivatives and the Gradient

The Essentials

A directional derivative is the slope of the plane tangent to a function in a given direction. The partial derivative of a function with respect to \( x \) is just the directional derivative in the \( x \) direction. The formula for the directional derivative in the direction of the unit vector \( \vec{u} \) is:

\[ D_{\vec{u}}f(x,y)=f_x\cos(\theta)+f_y\sin(\theta) \]

The gradient is an arbitrary directional derivative, or a directional derivative that hasn't been given a direction yet:

\[ \nabla f = f_x\hat{i}+f_y\hat{j} \]

The directional derivative of a function a direction is the dot product of the gradient of the function and the unit vector of the direction:

\[ D_{\vec{u}}f=\nabla f\cdot\vec{u} \]

The maximum value of the directional derivative occurs when the direction of the directional derivative is the same as the direction of the gradient. The maximum value of the directional derivative is the magnitude of the gradient.

The minimum value of the directional derivative occurs when the direction of the directional derivative is the opposite as the direction of the gradient. The minimum value of the directional derivative is the negative magnitude of the gradient.

The directional derivative and the gradient can be generalized to more dimensions:

\[ D_{\vec{u}}f(x,y,z)=f_x\cos(\alpha)+f_y\cos(\beta)+f_z\cos(\gamma) \]

Where the angles \( \alpha,\beta,\gamma \) determine the unit vector.

\[ \nabla f=f_x\hat{i}+f_y\hat{j}+f_z\hat{k} \]

Example

Find the unit vector in the direction for which the directional derivative of the function \( f(x,y)=\frac{7}{2}x^2-3xy+3y^2 \) at the point (1, 1). Find the maximum value.

The gradient of \( f \) is:

\[ \nabla f=\langle 7x-3y, -3x+6y \rangle \]

The direction of the gradient at the point (1, 1) is:

\[ \nabla f(1,1)=\langle 4, 3\rangle \]
\[ \frac{\langle 4, 3\rangle}{5} \]
\[ \langle \frac{4}{5}, \frac{3}{5}\rangle \]

The maximum value of the gradient is 5.

Practice

1) Find the direction in which the value of the directional derivative of the function \( 6x^2-2xy+13y^2 \) at the point (1, 1) is a maximum.

2) Find the directional derivative of the function \( f(x,y)=x^2-y^2 \) at the point (2, 1) in the direction \( \vec{u}=\langle2,3\rangle \)

Solutions:

1) \( \langle \frac{5}{13},\frac{12}{13} \rangle \)

2) 2

More Resources

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