Examples from Derivatives and level-set geometry All examples from Derivatives and level-set geometry.
[ EXAMPLE ] Let
f ( x , y , z ) = x 2 y + 2 x z + cos ( z 3 ) . f(x,y,z)=x^2y+2xz+\cos(z^3). f ( x , y , z ) = x 2 y + 2 x z + cos ( z 3 ) . Holding the other variables fixed gives
f x = 2 x y + 2 z , f y = x 2 , f z = 2 x − 3 z 2 sin ( z 3 ) . f_x=2xy+2z,
\qquad
f_y=x^2,
\qquad
f_z=2x-3z^2\sin(z^3). f x = 2 x y + 2 z , f y = x 2 , f z = 2 x − 3 z 2 sin ( z 3 ) . Therefore
∇ f ( x , y , z ) = ( 2 x y + 2 z x 2 2 x − 3 z 2 sin ( z 3 ) ) . \nabla f(x,y,z)
=
\begin{pmatrix}
2xy+2z\\
x^2\\
2x-3z^2\sin(z^3)
\end{pmatrix}. ∇ f ( x , y , z ) = 2 x y + 2 z x 2 2 x − 3 z 2 sin ( z 3 ) . At a specified point this vector also predicts directional change. For
example,
∇ f ( 1 , 0 , 0 ) = ( 0 1 2 ) , \nabla f(1,0,0)
=\begin{pmatrix}0\\1\\2\end{pmatrix}, ∇ f ( 1 , 0 , 0 ) = 0 1 2 , so a small displacement h \mathbf h h ( 1 , 0 , 0 ) (1,0,0) ( 1 , 0 , 0 ) f f f ∇ f ( 1 , 0 , 0 ) ⋅ h \nabla f(1,0,0)\cdot\mathbf h ∇ f ( 1 , 0 , 0 ) ⋅ h
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Consider
f ( x , y ) = e x y + x 2 y . f(x,y)=e^{xy}+x^2y. f ( x , y ) = e x y + x 2 y . Treating the other coordinate as constant gives
f x = y e x y + 2 x y , f y = x e x y + x 2 . f_x=ye^{xy}+2xy,
\qquad
f_y=xe^{xy}+x^2. f x = y e x y + 2 x y , f y = x e x y + x 2 . Differentiate f x f_x f x y y y f y f_y f y x x x
f x y = e x y + x y e x y + 2 x = f y x . f_{xy}=e^{xy}+xye^{xy}+2x
=f_{yx}. f x y = e x y + x y e x y + 2 x = f y x . The two mixed partials agree, as Clairaut's theorem guarantees because they
are continuous everywhere. The other second partials are
f x x = y 2 e x y + 2 y f_{xx}=y^2e^{xy}+2y f xx = y 2 e x y + 2 y f y y = x 2 e x y f_{yy}=x^2e^{xy} f y y = x 2 e x y
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
f ( x , y , z ) = x ln z y 2 , f(x,y,z)=\frac{x\ln z}{y^2}, f ( x , y , z ) = y 2 x ln z , the domain requires y ≠ 0 y\ne0 y = 0 z > 0 z>0 z > 0
f x = ln z y 2 , f y = − 2 x ln z y 3 , f z = x y 2 z , f_x=\frac{\ln z}{y^2},
\qquad
f_y=-\frac{2x\ln z}{y^3},
\qquad
f_z=\frac{x}{y^2z}, f x = y 2 ln z , f y = − y 3 2 x ln z , f z = y 2 z x , and the second partial derivatives are
f x x = 0 , f x y = − 2 ln z y 3 , f x z = 1 y 2 z , f_{xx}=0,
\qquad
f_{xy}=-\frac{2\ln z}{y^3},
\qquad
f_{xz}=\frac{1}{y^2z}, f xx = 0 , f x y = − y 3 2 ln z , f x z = y 2 z 1 , f y y = 6 x ln z y 4 , f y z = − 2 x y 3 z , f z z = − x y 2 z 2 . f_{yy}=\frac{6x\ln z}{y^4},
\qquad
f_{yz}=-\frac{2x}{y^3z},
\qquad
f_{zz}=-\frac{x}{y^2z^2}. f y y = y 4 6 x ln z , f y z = − y 3 z 2 x , f z z = − y 2 z 2 x . [RIGHT CLICK] [ASK AI] [ EXAMPLE ] The unit circle is the level set
F ( x , y ) = x 2 + y 2 − 1 = 0. F(x,y)=x^2+y^2-1=0. F ( x , y ) = x 2 + y 2 − 1 = 0. Its gradient is
∇ F ( x , y ) = ( 2 x 2 y ) . \nabla F(x,y)
=
\begin{pmatrix}
2x\\
2y
\end{pmatrix}. ∇ F ( x , y ) = ( 2 x 2 y ) . At
x 0 = ( cos ε , sin ε ) , \mathbf{x}_0=(\cos\varepsilon,\sin\varepsilon), x 0 = ( cos ε , sin ε ) , a normal vector is
∇ F ( x 0 ) = ( 2 cos ε 2 sin ε ) . \nabla F(\mathbf{x}_0)
=
\begin{pmatrix}
2\cos\varepsilon\\
2\sin\varepsilon
\end{pmatrix}. ∇ F ( x 0 ) = ( 2 cos ε 2 sin ε ) . Any non-zero scalar multiple is also normal, so the radius
( cos ε , sin ε ) (\cos\varepsilon,\sin\varepsilon) ( cos ε , sin ε )
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Consider
x 2 + y 2 − x y z = 1. x^2+y^2-xyz=1. x 2 + y 2 − x y z = 1. Set
F ( x , y , z ) = x 2 + y 2 − x y z − 1. F(x,y,z)=x^2+y^2-xyz-1. F ( x , y , z ) = x 2 + y 2 − x y z − 1. Then
∇ F ( x , y , z ) = ( 2 x − y z 2 y − x z − x y ) . \nabla F(x,y,z)
=
\begin{pmatrix}
2x-yz\\
2y-xz\\
-xy
\end{pmatrix}. ∇ F ( x , y , z ) = 2 x − y z 2 y − x z − x y . The point ( 1 , 2 , 2 ) (1,2,2) ( 1 , 2 , 2 ) 1 2 + 2 2 − ( 1 ) ( 2 ) ( 2 ) = 1 1^2+2^2-(1)(2)(2)=1 1 2 + 2 2 − ( 1 ) ( 2 ) ( 2 ) = 1
∇ F ( 1 , 2 , 2 ) = ( − 2 2 − 2 ) . \nabla F(1,2,2)
=
\begin{pmatrix}
-2\\
2\\
-2
\end{pmatrix}. ∇ F ( 1 , 2 , 2 ) = − 2 2 − 2 . The parallel vector ( 1 , − 1 , 1 ) (1,-1,1) ( 1 , − 1 , 1 ) ( 1 , 2 , 2 ) (1,2,2) ( 1 , 2 , 2 )
( 1 , − 1 , 1 ) ⋅ ( x − 1 y − 2 z − 2 ) = 0 , (1,-1,1)\cdot
\begin{pmatrix}
x-1\\
y-2\\
z-2
\end{pmatrix}
=0, ( 1 , − 1 , 1 ) ⋅ x − 1 y − 2 z − 2 = 0 , or
x − y + z = 1 . {x-y+z=1}. x − y + z = 1 . [RIGHT CLICK] [ASK AI] Derivatives and level-set geometry Leads to Constrained extrema and 3 more.