Mathcad - Lecture4C Directional Derivatives are the slopes of surfaces in specific directions at a...

GRADIENT

Directional Derivatives are the slopes of surfaces in specific directions at a particular point on that surface. Example # 1: Find the directional derivative, " D f " at the point: " P " in the direction: "u

→ ".

f x y,( ) 2x 3 y⋅+= P 0 0,( ) u→

1 i→⋅=

dzx

z∂

∂

⎛⎜⎝

⎞⎟⎠

dx⋅y

z∂

∂

⎛⎜⎝

⎞⎟⎠

dy⋅+=

dz 2( ) dx⋅ 3( ) dy⋅+=

dy 0=

dz 2( ) dx⋅=

xzd

d2=

Di

→ f 0 0,( )⋅ 2=

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yx

z

Direction along x-axis

Example # 2: Find the directional derivative, " D f " at the point: " P " in the direction: "u

→".

f x y,( ) 2 x⋅ 3 y⋅+= P 0 0,( ) u→

1 j→⋅=

dzx

z∂

∂

⎛⎜⎝

⎞⎟⎠

dx⋅y

z∂

∂

⎛⎜⎝

⎞⎟⎠

dy⋅+=

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dz 2( ) dx⋅ 3( ) dy⋅+=

dx 0=

dz 3( ) dy⋅=

yzd

d3=

Dj

→ f 0 0,( )⋅ 3=

y

z

x

Direction along y-axis

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→".

f x y,( ) 2x 3 y⋅+= P 0 0,( )

u→ 2

13i

→⋅

313

j→⋅+=

dzx

z∂

∂

⎛⎜⎝

⎞⎟⎠

dx⋅y

z∂

∂

⎛⎜⎝

⎞⎟⎠

dy⋅+=

dz 2( ) dx⋅ 3( ) dy⋅+=

ds dx( )2 dy( )2+=

dzds

2( ) dxds⋅ 3( ) dy

ds⋅+=

szd

d2 dx

dx( )2 dy( )2+⋅ 3 dy

dx( )2 dy( )2+⋅+=

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uxdx

dx( )2 dy( )2+= uy

dy

dx( )2 dy( )2+=

szd

d2 ux⋅ 3 uy⋅+=

Du f⋅ 2( ) 213

⋅ 3( ) 313

⋅+⎡⎢⎣

⎤⎥⎦

= 3.606=

Du→ f 0 0,( )⋅ 3.606=

y x

z

Direction along "u"

Extend this approach to find the directional derivative of curved surfaces.

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szd

d xz∂

∂

⎛⎜⎝

⎞⎟⎠

ux⋅y

z∂

∂

⎛⎜⎝

⎞⎟⎠

uy⋅+= P x0 y0,( )

u→

ux i→⋅ uy j

→⋅+=


→".

f x y,( ) x2 y2−= P 1 1−,( )

u→ 1−

2i

→⋅

12

j→⋅+=

Du→ f⋅

xf∂

∂

⎛⎜⎝

⎞⎟⎠

ux⋅y

f∂

∂

⎛⎜⎝

⎞⎟⎠

uy⋅+=

xf∂

∂2 x⋅=

yf∂

∂2− y⋅=

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xf 1 1−,( )∂

∂2=

yf 1 1−,( )∂

∂2=

uy12

=ux1−2

=

Du→ f 1 1−,( )⋅ 2 1−

2⎛⎜⎝

⎞⎟⎠

⋅ 2 12

⎛⎜⎝

⎞⎟⎠

⋅+= 0=

y

z

x

f(x,y) with Contours

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The Directional Derivative at P 1 1−,( ) in the

direction: u→ 1−

2i

→⋅

12

j→⋅+= was found to be

zero. Verify this result graphically using the following contour plot of " f x y,( ) ".

Contour Plot

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→".

z f x y,( )= P x0 y0,( )

u→

ux i→⋅ uy j

→⋅+=

Du f fx x0 y0,( )⎡⎣ ⎤⎦ ux⋅ fy x0 y0,( )⎡⎣ ⎤⎦ uy⋅+=

Definition: This is the GRADIENT of " f ".

Δf( )→⎯

xf∂

∂

⎛⎜⎝

⎞⎟⎠

i→⋅

yf∂

∂

⎛⎜⎝

⎞⎟⎠

j→⋅+=

The Directional Derivative " Du f " is the Dot Product of the Gradient with the unit vector in the specified direction.

Du→ f Δf( )

→⎯u→⋅=

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Recall this formula for the Dot Product.

a→

b→⋅ a

→b→⋅ cos θ( )=

Δf( )→⎯

u→⋅ Δf( )

→⎯u→⋅ cos θ( )⋅=

Evidently, " f " has its steepest slope when θ 0= π, . Conversely, ALL Directional Derivatives are zero at a point where the Gradient is zero.


→".

f x y,( ) x y⋅ ey⋅= P 1 1,( )

u→

j→−=

Δf( )→⎯

xf∂

∂

⎛⎜⎝

⎞⎟⎠

i→⋅

yf∂

∂

⎛⎜⎝

⎞⎟⎠

j→⋅+=

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fx 1 1,( ) e2

= 1.36=x

f∂

∂

12

yx

⋅ ey⋅=

fy 1 1,( ) 3 e⋅2

= 4.08=y

f∂

∂

12

xy

⋅ x y⋅+⎛⎜⎝

⎞⎟⎠

ey⋅=

Δf( )→⎯

1.36( ) i→⋅ 4.08( ) j

→⋅+=

Δf( )→⎯

u→⋅ 1.36( ) i

→⋅ 4.08( ) j

→⋅+⎡⎣ ⎤⎦ j

→−( )⋅= 4.08−=

Du→ f 1 1,( ) 4.08−=

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Example # 7: Sketch the Level Curve of " f " that passes thru the point "P" and draw the Gradient Vector at "P".

f x y,( ) x2 4 y2⋅+= P 2− 0,( )

f 2− 0,( ) 4=

x2 4 y2⋅+ 4=

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Δf→⎯

xf∂

∂

⎛⎜⎝

⎞⎟⎠

i→⋅

yf∂

∂

⎛⎜⎝

⎞⎟⎠

j→⋅+=

yf∂

∂8 y⋅=

xf∂

∂2 x⋅=

fx 2− 0,( ) 4−= fy 2− 0,( ) 0=

Δf 2− 0,( )→⎯⎯⎯

4−( ) i→⋅ 0( ) j

→⋅+= 4− i

→⋅=

This, the Gradient Vector at the given point is normal or perpendicular to the level curve passing thru that point because the Directional Derivative in the direction of the gradient vector is a maximum.

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Gradient Vector@(-2,0)

Example # 8: Sketch the surface, " f x y,( ) ", and the tangent line that passes thru the point: " 2 0, 2,( ) " and has the steepest slope.

f x y,( ) 4 x2− y2−=

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z

xy

Steepest Slope Tangent

Infinity of tangents at point, all with differentslopes depending on chosen direction. Tangent in Gradient Direction is illustrated in the figure. Note that the level curves of the surface are concentric circles centered at the origin AND that the Gradient Direction is perpendicular to the level curve at the point.

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yx

z

Gradient Direction Illustrated

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Mathcad - Lecture4C Directional Derivatives are the slopes of surfaces in specific directions at a...

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