Section 3.6 Directional Derivative Gradient

3.6 Directional Derivative Gradient

In previous section, we learn take derivative of a function [latex]z=f(x,y)[/latex] with respect to [latex]x[/latex] or with respect to [latex]y[/latex]. We can think about the partial derivative with respect to the direction [latex]\lt 1,0\gt[/latex] and [latex]\lt 0,1\gt[/latex]. In this section we learn to take derivative at any direction. 

Definition: Directional Derivative

Suppose [latex]z=f(x,y)[/latex] is a function of two variables with a domain of [latex]D[/latex]. Let [latex](x_{0},y_{0})[/latex] be a point in the domain of [latex]z[/latex] and [latex]\overrightarrow{u}=\lt a,b\gt[/latex] be an unit vector. Then the directional derivative of [latex]f[/latex] in the direction of [latex]\overrightarrow{u}[/latex] is given by

\[D_{\overrightarrow{u}}f(x_{0},y_{0})=\lim_{h\rightarrow0}\frac{f(x_{0}+ah,y_{0}+bh)-f(x_{0},y_{0})}{h}\]

provided the limit exists.

 

Theorem: Directional Derivative of a Function of Two Variables 

Let [latex]z=f(x,y)[/latex] be a function of two variables [latex]x[/latex] and [latex]y[/latex], and assume that [latex]f_{x}(x,y)[/latex] and [latex]f_{y}(x,y)[/latex] exist. Then the directional derivative of [latex]f[/latex] in the direction of unit vector [latex]\overrightarrow{u}=\lt a,b\gt[/latex] is given by 

\[D_{\overrightarrow{u}}f(x,y)=f_{x}(x,y)a+f_{y}(x,y)b.\]

Proof: chain rule. 

 

 

 

Example 1: Finding the directional derivative of [latex]z=f(x,y)=x^{2}y-xy^{2}+2x+y[/latex] in the direction of [latex]\overrightarrow{u}=\lt \frac{1}{2},\frac{\sqrt{3}}{2}\gt[/latex].

 

 

 

Exercise 1: Finding the directional derivative of [latex]z=f(x,y)=2x^{2}+x^{2}y-y^{3}[/latex] in the direction of [latex]\overrightarrow{u}=\lt \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\gt[/latex].

 

 

 

Remark: Any unit vector [latex]\overrightarrow{u}=\lt a,b\gt[/latex] can be rewritten as [latex]\overrightarrow{u}=\lt \text{cos}(\theta),\text{sin}(\theta)\gt[/latex]. The directional derivative can only be defined for an unit vector. 

 

 

 

Example 2: Finding the directional derivative of [latex]z=f(x,y)=x^{3}-2xy+y^{2}+2x-y[/latex], in the direction of [latex]\overrightarrow{u}=\lt 3,4\gt[/latex].

 

 

 

Exercise 2: Finding the directional derivative of [latex]z=f(x,y)=x^{2}+3x^{2}y-y^{2}+4x+y[/latex], in the direction of [latex]\overrightarrow{u}=\lt 3,3\sqrt{3}\gt[/latex].

 

 

 

Remark: The directional derivative formula can be rewritten as \[D_{\overrightarrow{u}=\lt a.b\gt }f(x,y)=f_{x}(x,y)a+f_{y}(x,y)b\] \[=\lt f_{x}(x,y),f_{y}(x,y)\gt \cdot\lt a,b\gt .\]

Definition: 

Let [latex]z=f(x,y)[/latex] be a function of [latex]x[/latex] and [latex]y[/latex] such that [latex]f_{x}[/latex] and [latex]f_{y}[/latex] exist. The vector [latex]\nabla f(x,y)[/latex] is called the gradient of [latex]f[/latex] and is defined as 

\[\nabla f(x,y)=\lt f_{x}(x,y),f_{y}(x,y)\gt \] 

The vector [latex]\nabla f(x,y)[/latex] is also written as \textquotedblleft grad [latex]f[/latex] .\textquotedblright{}

 

 

 

Example3: Find the gradient [latex]\nabla f(x,y)[/latex] where [latex]f(x,y)=\text{sin}(2x)e^{y}[/latex].

 

 

 

Exercise 3: Find the gradient [latex]\nabla f(x,y)[/latex] where [latex]f(x,y)=\text{ln}(y)\text{cos}(2x)[/latex].

 

 

 

Theorem: Properties of the Gradient

Suppose the function [latex]z=f(x,y)[/latex] is differentiable at [latex](x_{0},y_{0})[/latex].

i. If [latex]\nabla f(x_{0},y_{0})=0[/latex], then [latex]D_{\overrightarrow{u}}f(x_{0},y_{0})=0[/latex] for any unit vector [latex]\overrightarrow{u}[/latex]. \\

ii. If [latex]\nabla f(x_{0},y_{0})\neq0[/latex], then [latex]D_{\overrightarrow{u}}f(x_{0},y_{0})[/latex] is maximized when [latex]\overrightarrow{u}[/latex] points in the same direction as [latex]\nabla f(x_{0},y_{0})[/latex]. The maximum value of [latex]D_{\overrightarrow{u}}f(x_{0},y_{0})[/latex] is [latex]||\nabla f(x_{0},y_{0})||[/latex].

 

 

 

Example 4: Find the direction for which the directional derivative of [latex]f(x,y)=3x^{2}+2xy+y^{2}+5y-11[/latex] at [latex](-2,3)[/latex] is a maximum. What is the maximum value? 

 

 

 

Exercise 4: Find the direction for which the directional derivative of [latex]f(x,y)=x^{2}-2xy+3y^{2}+3y-1[/latex] at [latex](1,-1)[/latex] is a maximum. What is the maximum value? 

 

 

 

Theorem: Gradient Is Normal to the Level Curve 

Suppose the function [latex]z=f(x,y)[/latex] has continuous first-order partial derivatives in an open disk centered at a point [latex](x_{0},y_{0})[/latex]. If [latex]\nabla f(x_{0},y_{0})\neq0[/latex], , then [latex]\nabla f(x_{0},y_{0})[/latex], is normal to the level curve of [latex]f[/latex] at [latex](x_{0},y_{0})[/latex].

 

 

 

Example 5: For the function [latex]f(x,y)=x^{2}-4x+y^{2}+2y+4[/latex], find a tangent vector to the level curve at point [latex](1,2)[/latex]. Graph the level curve corresponding to [latex]f(x,y)=9[/latex] and draw in [latex]\nabla f(1,2)[/latex] and a tangent vector.

 

 

 

Exercise 5: For the function [latex]f(x,y)=x^{2}+2x+y^{2}-6y+4[/latex], find a tangent vector to the level curve at point [latex](2,1)[/latex]. Graph the level curve corresponding to [latex]f(x,y)=7[/latex] and draw in [latex]\nabla f(2,1)[/latex] and a tangent vector.

 

 

 

Definition: Directional Derivative

Suppose [latex]w=f(x,y,z)[/latex] is a function of three variables with a domain of [latex]D[/latex]. Let [latex](x_{0},y_{0},z_{0})[/latex] be a point in the domain of [latex]w[/latex] and [latex]\overrightarrow{u}=\lt a,b,c\gt[/latex] be an unit vector. Then the directional derivative of [latex]f[/latex] in the direction of [latex]\overrightarrow{u}[/latex] is given by 

\[D_{\overrightarrow{u}}f(x_{0},y_{0},z_{0})=\lim_{h\rightarrow0}\frac{f(x_{0}+ah,y_{0}+bh,z_{0}+ch)-f(x_{0},y_{0},z_{0})}{h}\]

 provided the limit exists.

 

Theorem: Directional Derivative of a Function of Three Variables 

Let [latex]w=f(x,y,z)[/latex] be a function of three variables [latex]x[/latex], [latex]y[/latex] and [latex]z[/latex], and assume that [latex]f_{x}(x,y,z)[/latex], [latex]f_{y}(x,y,z)[/latex] and [latex]f_{z}(x,y,z)[/latex] exist. Then the directional derivative of [latex]f[/latex] in the direction of unit vector [latex]\overrightarrow{u}=\lt a,b.c\gt[/latex] is given by 

\[D_{\overrightarrow{u}}f(x,y,z)=f_{x}(x,y,z)a+f_{y}(x,y,z)b+f_{z}(x,y,z)c.\]

 

 

Example 6: Calculate [latex]D_{\overrightarrow{u}}f(1,-2,1)[/latex] in the direction of [latex]\overrightarrow{u}=\lt 2,3,4\gt[/latex] for the function [latex]f(x,y,z)=x^{2}y-y^{2}z+2xyz[/latex].

 

 

 

Exercise 6: Calculate [latex]D_{\overrightarrow{u}}f(0,1,2)[/latex] in the direction of [latex]\overrightarrow{u}=\lt 1,2,3\gt[/latex] for the function [latex]f(x,y,z)=x^{3}y+xy^{2}z+2yz^{2}[/latex].

 

 

 

Example 7: The temperature [latex]T[/latex] in a metal sphere is inversely proportional to the distance from the center of the sphere (the origin: [latex](0,0,0)[/latex]). The temperature at point [latex](1,2,2)[/latex] is [latex]120{^\circ}C.[/latex]

a. Find the rate of change of the temperature at point [latex](1,2,2)[/latex] in the direction toward point [latex](2,1,3)[/latex].

b. Show that, at any point in the sphere, the direction of greatest increase in temperature is given by a vector that points toward the origin.

 

 

 

Group work:

1. The electrical potential (voltage) in a certain region of space is given by the function [latex]V(x,y,z)=5x^{2}-3xy+xyz.[/latex] 

a. Find the rate of change of the voltage at point [latex](3,4,5)[/latex] in the direction of the vector [latex]\lt 1,1,-1\gt[/latex]. 

b. In which direction does the voltage change most rapidly at point [latex](3,4,5)[/latex] ? 

c. What is the maximum rate of change of the voltage at point (3, 4, 5)?

 

2. Find the gradient vector [latex]f(x,y)=xe^{y}-\text{ln}(x),\text{ at }P(-3,0)[/latex]. 

 

3. Find the gradient vector [latex]f(x,y,z)=x\sqrt{y^{2}+z^{2}}[/latex], at [latex]P(-2,-1,-1)[/latex].

 

4. Find the derivative of the function of [latex]f(x,y)=e^{xy}[/latex] at point [latex](6,7)[/latex] in the direction the function increases most rapidly. 

 

5. Find the derivative of the function of [latex]f(x,y)=\text{arctan}(\frac{y}{x})[/latex] at point [latex](-9,9)[/latex] in the direction the function increases most rapidly. 

 

6. Find the maximum rate of change of [latex]f=xe^{-y}[/latex] at [latex](-1,0)[/latex] and the direction in which it occurs.

 

7. Find the maximum rate of change of [latex]f=\text{cos}(3x+2y)[/latex] at [latex](\frac{\pi}{6},-\frac{\pi}{8})[/latex] and the direction in which it occurs.

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