Calc3: Partial Derivative
Functions of Several Variables
Functions of Two Variables
A function of 2 variables f(x,y)f(x,y)f(x,y) where (x,y)∈D⊂R2(x,y) \in D \sub R^2(x,y)∈D⊂R2. Let z=f(x,y)z =f(x,y)z=f(x,y). We can draw the graph on the 3d coordinate.
Contour Curves
The contour curve, or level curves, of a function f of two variables are the curve with equation f(x,y)=kf(x,y) = kf(x,y)=k where k is a real number.
When we have dense level curves, it means the height changes rapidly.
E.x.
We have a function f(x,y)=6−3x−2yf(x,y)=6-3x-2yf(x,y)=6−3x−2y. Sketch the contour curve.
Sol:
First set z=kz = kz=k and solve the equation. We get one straight lines for one fixed k.
Limits of Two-Variable Functions
The limit of f(x,y)f(x,y)f(x,y) as (x,y)(x,y)(x,y) approach (a,b)(a,b)(a,b) is L if it holds that for every ϵ>0\epsilon > 0ϵ>0, there exists δ>0\delta > 0δ>0 such that ∣f(x,y)−L∣<ϵ|f(x,y) - L| < \epsilon∣f(x,y)−L∣<ϵ whenever 0<(x−a)2+(y−b)2<δ0 < \sqrt{(x-a)^2+(y-b)^2} < \delta0<(x−a)2+(y−b)2<δ.
E.x.
Show that lim(x,y)→(0,0)x2−y2x2+y2lim_{(x,y)\rightarrow(0,0)}\frac{x^2-y^2}{x^2+y^2}lim(x,y)→(0,0)x2+y2x2−y2 does not exist.
Sol:
We approach the point from different directions and obtain different limits.
Let y=0, the limit is 1. Let x=0, the limit is -1. They are not equal.
Some other potential directions include: y=mx, y=mx^2, etc.
Limit Laws
If we know lim(x,y)→(a,b)f(x)lim_{(x,y)\rightarrow(a,b)} f(x)lim(x,y)→(a,b)f(x) and lim(x,y)→(a,b)g(x)lim_{(x,y)\rightarrow(a,b)} g(x)lim(x,y)→(a,b)g(x), then
lim(x,y)→(a,b)[f(x)+g(x)]lim(x,y)→(a,b)[f(x)⋅g(x)]lim(x,y)→(a,b)[f(x)/g(x)],g(x)≠0lim_{(x,y)\rightarrow(a,b)} [f(x)+g(x)] \\\\ lim_{(x,y)\rightarrow(a,b)} [f(x) \cdot g(x)] \\\\ lim_{(x,y)\rightarrow(a,b)} [f(x)/g(x)], g(x)\ne 0 \\\\ lim(x,y)→(a,b)[f(x)+g(x)]lim(x,y)→(a,b)[f(x)⋅g(x)]lim(x,y)→(a,b)[f(x)/g(x)],g(x)=0
all exists.
E.x.
Find lim(x,y)→(0,0)3x2yx2+y2lim_{(x,y)\rightarrow(0,0)}\frac{3x^2y}{x^2+y^2}lim(x,y)→(0,0)x2+y23x2y if it exists.
Sol:
Notice that x2≤x2+y2x^2 \le x^2+y^2x2≤x2+y2, we have
∣3x2yx2+y2∣≤3∣y∣≤3x2+y2≤3δ|\frac{3x^2y}{x^2+y^2}| \le 3|y| \le 3 \sqrt{x^2+y^2} \le 3 \delta ∣x2+y23x2y∣≤3∣y∣≤3x2+y2≤3δ
We set δ=ϵ/10\delta = \epsilon/10δ=ϵ/10. Therefore, ∣3x2yx2+y2∣<ϵ|\frac{3x^2y}{x^2+y^2}| < \epsilon∣x2+y23x2y∣<ϵ.
This will satisfy the definition of the limit.
Partial Derivatives
Partial Derivative and Geometric Meaning
We can freeze one variable y=by=by=b and let g(x)=f(x,b)g(x) = f(x,b)g(x)=f(x,b). Define a partial derivative of fff with respect to x at (a,b).
fx(a,b)=g′(a)=limh→0g(a+h)−g(a)h=limh→0f(a+h,b)−f(a,b)hf_x(a,b) = g'(a) \\\\ = lim_{h\rightarrow0} \frac{g(a+h)-g(a)}{h} \\\\ = lim_{h\rightarrow0} \frac{f(a+h,b)-f(a,b)}{h} \\\\ fx(a,b)=g′(a)=limh→0hg(a+h)−g(a)=limh→0hf(a+h,b)−f(a,b)
First, we plot the function in 3d coordinates. We cut our graph by a vertical plane that passes through the point P and perpendicular to y-axis. The intersection is given by the graph of the function f(x,b)f(x,b)f(x,b).
Implicit Functions
E.x.
Find ∂z∂x\frac{\partial z}{\partial x}∂x∂z for x3+y3+z3+6xyz=1x^3+y^3+z^3+6xyz=1x3+y3+z3+6xyz=1.
Sol:
The variable z can be considered as z(x,y)z(x,y)z(x,y). We differential both sides to get
3x2+0+3z2∂z∂x+6yz+6xy∂z∂x=03x^2+0+3z^2\frac{\partial z}{\partial x}+6yz+6xy\frac{\partial z}{\partial x} = 0 3x2+0+3z2∂x∂z+6yz+6xy∂x∂z=0
Solve the equation
∂z∂x=−3x2+6yz3z2+6xy\frac{\partial z}{\partial x} = -\frac{3x^2+6yz}{3z^2+6xy} ∂x∂z=−3z2+6xy3x2+6yz
Higher Degree Derivative
fxx=f11=∂2f∂x2fyy=f22=∂2f∂y2fxy=∂2f∂y∂xfyx=∂2f∂x∂yf_{xx} = f_{11} = \frac{\partial^2 f}{\partial x^2} \\\\ f_{yy} = f_{22} = \frac{\partial^2 f}{\partial y^2} \\\\ f_{xy} = \frac{\partial^2 f}{\partial y \partial x} \\\\ f_{yx} = \frac{\partial^2 f}{\partial x \partial y} \\\\ fxx=f11=∂x2∂2ffyy=f22=∂y2∂2ffxy=∂y∂x∂2ffyx=∂x∂y∂2f
Clairaut’s Theorem:
If fff has continuous second order partial derivatives, then fxy=fyxf_{xy} = f_{yx}fxy=fyx.
Partial Differential Equations
E.x.
We have a function u(x,y)=exsin(y)u(x,y) = e^xsin(y)u(x,y)=exsin(y), prove it satisfies the Laplace equation.
Sol:
Δu=0,uxx+uyy=0\Delta u = 0, u_{xx} + u_{yy} = 0 Δu=0,uxx+uyy=0
Tangent Planes and Linear Approximations
Tangent Planes
Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface z=f(x,y)z = f(x,y)z=f(x,y) at the point P=(x0,y0,z0)P = (x_0, y_0, z_0)P=(x0,y0,z0) is
fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)−(z−z0)=0f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) - (z-z_0) = 0 fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)−(z−z0)=0
Linear Approximations
In the example z=2x2+y2z=2x^2+y^2z=2x2+y2 at the origin, the tangent plane is z=0z=0z=0. Among all planes passing through the origin, it approximates the surface best when x and y are small.
We say that
z=fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)+z0z = f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) + z_0 z=fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)+z0
is the linearization of f at (a,b)(a,b)(a,b).
Differentiability of Function of Several Variables
Denote Δz=f(x0+Δx,y0+Δy)−f(x0,y0)\Delta z = f(x_0+\Delta x, y_0+\Delta y) - f(x_0,y_0)Δz=f(x0+Δx,y0+Δy)−f(x0,y0). We say that f is diff at (x0,y0)(x_0,y_0)(x0,y0) if Δz=fx(x0,y0)Δx+fy(x0,y0)Δy+ϵ1Δx+ϵ2Δy\Delta z = f_x(x_0,y_0)\Delta x + f_y(x_0,y_0)\Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta yΔz=fx(x0,y0)Δx+fy(x0,y0)Δy+ϵ1Δx+ϵ2Δy where ϵ1,ϵ2→0\epsilon_1, \epsilon_2 \rightarrow 0ϵ1,ϵ2→0 as $ \Delta x, \Delta y \rightarrow 0$.
If the first derivatives exist at the point, are we always able to show that f is diff at that point?
No. There are some counter examples.
But if the first derivates exist and are continuous, then f is diff at that point.
E.x.
We have a function f(x,y)=xexyf(x,y) = xe^{xy}f(x,y)=xexy is diff at (1,0), find its linearization. Use it to approximate f(1.1,-0.1).
Sol:
fx=exy+xyexy,fx(1,0)=1fy=x2exy,fy(1,0)=1L(x,y)=f(1,0)+1(x−x0)+1(y−y0)=x+yL(1.1,−0.1)=1f_x = e^{xy} + xye^{xy}, f_x(1,0)=1 \\\\ f_y = x^2e^{xy}, f_y(1,0)=1 \\\\ L(x,y) = f(1,0) + 1(x-x_0) + 1(y-y_0) = x+y \\\\ L(1.1, -0.1) = 1 fx=exy+xyexy,fx(1,0)=1fy=x2exy,fy(1,0)=1L(x,y)=f(1,0)+1(x−x0)+1(y−y0)=x+yL(1.1,−0.1)=1
Differentials
We have a function z = f(x,y) and independent variables dx, dy. The (total) differential is defined as
dz=fxdx+fydydz≈Δzdz = f_x dx + f_y dy \\\\ dz \approx \Delta z dz=fxdx+fydydz≈Δz
E.x.
We have a function z=f(x,y)=x2+3xy−y2z = f(x,y) = x^2+3xy-y^2z=f(x,y)=x2+3xy−y2, find its differential
Sol:
fx=2x+3y,fy=3x−2ydz=(2x+3y)dx+(3x−2y)dyf_x = 2x+3y, f_y = 3x-2y \\\\ dz = (2x+3y)dx + (3x-2y)dy fx=2x+3y,fy=3x−2ydz=(2x+3y)dx+(3x−2y)dy
Chain Rule
We have a function z = f(x,y), and x=g(t), y=h(t).
dzdt=∂z∂xdxdt+∂z∂ydtdt\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dt}{dt} dtdz=∂x∂zdtdx+∂y∂zdtdt
We have a function z = f(x,y), and x=g(s,t), y=h(s,t).
∂z∂t=∂z∂x∂x∂t+∂z∂y∂t∂t\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial t}{\partial t} ∂t∂z=∂x∂z∂t∂x+∂y∂z∂t∂t
Implicit Derivatives
We have a function y=g(x) that is implicitly defined via F(x,y)=0. Compute its derivative. Write it as F(x, g(x))=0, and then differentiate both sides.
∂F∂xdxdx+∂F∂ydydx=0dydx=−∂F∂x/∂F∂y\frac{\partial F}{\partial x}\frac{dx}{dx} + \frac{\partial F}{\partial y}\frac{dy}{dx} = 0 \\\\ \frac{dy}{dx} = -\frac{\partial F}{\partial x}/\frac{\partial F}{\partial y} ∂x∂Fdxdx+∂y∂Fdxdy=0dxdy=−∂x∂F/∂y∂F
E.x.
y=g(x) is implicitly defined via x3+y3=6xyx^3+y^3=6xyx3+y3=6xy. Compute its derivative.
sol:
Write it as F(x,y)=x3+y3−6xyF(x,y)=x^3+y^3-6xyF(x,y)=x3+y3−6xy. Therefore,
F(x,y)=0dydx=−∂F∂x/∂F∂y=−3x2−6y3y2−6xF(x,y)=0 \\\\ \frac{dy}{dx} = -\frac{\partial F}{\partial x}/\frac{\partial F}{\partial y} = -\frac{3x^2-6y}{3y^2-6x} F(x,y)=0dxdy=−∂x∂F/∂y∂F=−3y2−6x3x2−6y
Implicit Partial Derivatives
We have a function z=g(x,y) that is implicitly defined via F(x,y,z)=0. Compute its derivative. Write it as F(x, y, g(x,y))=0, and then differentiate both sides.
sol:
∂F∂xdxdx+∂F∂ydydx+∂F∂zdzdx=0∂F∂x+∂F∂zdzdx=0∂z∂x=−∂F∂x/∂F∂z\frac{\partial F}{\partial x}\frac{dx}{dx} + \frac{\partial F}{\partial y}\frac{dy}{dx} + \frac{\partial F}{\partial z}\frac{dz}{dx} = 0 \\\\ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial z}\frac{dz}{dx} = 0 \\\\ \frac{\partial z}{\partial x} = -\frac{\partial F}{\partial x}/\frac{\partial F}{\partial z} ∂x∂Fdxdx+∂y∂Fdxdy+∂z∂Fdxdz=0∂x∂F+∂z∂Fdxdz=0∂x∂z=−∂x∂F/∂z∂F
Directional Derivative
Whenever we talk about directional derivative, we always assume that u⃗\vec uu is a unit vector.
Theorem:
f is a differentiable function of x and y. The directional derivative of f along the direction u⃗=(u1,u2)\vec u = (u_1, u_2)u=(u1,u2) is given by
Du⃗f(x,y)=∂f∂xu1+∂f∂yu2D_{\vec u} f(x,y) = \frac{\partial f}{\partial x} u_1 + \frac{\partial f}{\partial y}u_2 Duf(x,y)=∂x∂fu1+∂y∂fu2
proof:
We define g(h)=f(x+hu1,y+hu2)g(h)=f(x+hu_1, y+hu_2)g(h)=f(x+hu1,y+hu2). The main observation is that
g′(o)=limh→0g(h)−g(o)h=limh→01h[f(x+hu1,y+hu2)−f(x,y)]=Du⃗f(x,y)g'(o) = lim_{h\rightarrow 0} \frac{g(h)-g(o)}{h} \\\\ = lim_{h\rightarrow 0} \frac 1 h [f(x+hu_1, y+hu_2)-f(x,y)]\\\\ = D_{\vec u} f(x,y) g′(o)=limh→0hg(h)−g(o)=limh→0h1[f(x+hu1,y+hu2)−f(x,y)]=Duf(x,y)
and from the chain rule
g′(h)=∂f∂xu1+∂f∂yu2=Du⃗f(x,y)g'(h) = \frac{\partial f}{\partial x} u_1 + \frac{\partial f}{\partial y}u_2 = D_{\vec u} f(x,y) g′(h)=∂x∂fu1+∂y∂fu2=Duf(x,y)
E.x.
Find the directional derivative of f(x,y)=x3−3xy+4y2f(x,y) = x^3-3xy+4y^2f(x,y)=x3−3xy+4y2, u⃗=(cosθ,sinθ)\vec u = (cos \theta, sin \theta)u=(cosθ,sinθ) with θ=π/6\theta=\pi/6θ=π/6.
sol:
Du⃗f(x,y)=cosθ(3x2−3y)+sinθ(−3x+8y)=32(3x2−3y)+12θ(−3x+8y)D_{\vec u} f(x,y) = cos \theta(3x^2-3y) + sin \theta(-3x+8y) \\\\ = \frac{\sqrt 3}{2}(3x^2-3y) + \frac{1}{2} \theta(-3x+8y) Duf(x,y)=cosθ(3x2−3y)+sinθ(−3x+8y)=23(3x2−3y)+21θ(−3x+8y)
Gradient Vector
We define the gradient of f to be ∇f=<∂f∂x,∂f∂y>\nabla f = <\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}>∇f=<∂x∂f,∂y∂f>. It is possible to expand it to 3d.
With this notation, the directional derivative can be expressed as
Du⃗f(x,y)=∇f(x,y)⋅u⃗D_{\vec u} f(x,y) = \nabla f(x,y) \cdot \vec u Duf(x,y)=∇f(x,y)⋅u
Theorem:
Along the direction u⃗\vec uu that is the same as the gradient, the value changes the fastest.
proof:
Assume that the angle between the gradient the unit vector is α\alphaα.
Du⃗f(x,y)=∣∇f(x,y)∣⋅∣u⃗∣⋅cosαD_{\vec u} f(x,y) = |\nabla f(x,y)| \cdot |\vec u| \cdot cos \alpha \\\\ Duf(x,y)=∣∇f(x,y)∣⋅∣u∣⋅cosα
If we let the angle equal to 0, then the directional derivative attains its max.
E.x.
We have f(x,y)=xeyf(x,y) = xe^yf(x,y)=xey, find the rate of change of f at the point P=(2,0), in the direction of PQ where Q=(0.5,2). Also, find the direction that leads to the max rate of change in f.
sol:
First, compute its gradient vector and the unit directional vector.
∇f(x,y)=<ey,xey>u⃗=v⃗∣v⃗∣=<−3/5,−4/5>Du⃗f(x,y)=−3/5(ey)+(−4/5)xeyDu⃗f(2,0)=1\nabla f(x,y) = <e^y, xe^y> \\\\ \vec u = \frac{\vec v}{|\vec v|} = <-3/5, -4/5> \\\\ D_{\vec u} f(x,y) = -3/5(e^y)+(-4/5)xe^y \\\\ D_{\vec u} f(2,0) = 1 ∇f(x,y)=<ey,xey>u=∣v∣v=<−3/5,−4/5>Duf(x,y)=−3/5(ey)+(−4/5)xeyDuf(2,0)=1
Then, we evaluate the gradient vector.
∇f(2,0)=<1,2>v⃗′=<1/5,2/5>\nabla f(2,0) = <1,2> \\\\ \vec v' =<1/\sqrt 5, 2/\sqrt 5> ∇f(2,0)=<1,2>v′=<1/5,2/5>
Level Surfaces
Assume that S is given by the level surface F(x,y,z)=k. Take a curve C on S as r⃗(t)=<x(t),y(t),z(t)>\vec r(t) = <x(t), y(t),z(t)>r(t)=<x(t),y(t),z(t)>. We differentiate the equation on both sides to get
∂F∂xx′(t)+∂F∂yy′(t)+∂F∂zz′(t)=0\frac{\partial F}{\partial x}x'(t) + \frac{\partial F}{\partial y}y'(t) + \frac{\partial F}{\partial z}z'(t) = 0 \\\\ ∂x∂Fx′(t)+∂y∂Fy′(t)+∂z∂Fz′(t)=0
The tangent direction of the curve is always perpendicular to the gradient vector. We can compute tangent plane by ∇F\nabla F∇F, which is one normal direction of the tangent plane.
Maximum and Minimum Values
Local Min/Max
local max: A function of two variables has a local max at (a,b) if f(x,y)≤f(a,b)f(x,y) \le f(a,b)f(x,y)≤f(a,b) for every point satisfying dist((x,y),(a,b))<rdist((x,y), (a,b)) < rdist((x,y),(a,b))<r. (r>0)
Theorem:
If f attains a local max/min at (a,b), and f is diff at (a,b), then
∂f∂x(a,b)=∂f∂y(a,b)=0\frac{\partial f}{\partial x} (a,b) = \frac{\partial f}{\partial y} (a,b) =0 ∂x∂f(a,b)=∂y∂f(a,b)=0
proof:
Assume f(a,b) is local max. The partial derivate
∂f∂x(a,b)=limh→0f(a+h,b)−f(a,b)h\frac{\partial f}{\partial x} (a,b) = lim_{h\rightarrow0} \frac{f(a+h,b)-f(a,b)}{h} \\\\ ∂x∂f(a,b)=limh→0hf(a+h,b)−f(a,b)
If h→0+h\rightarrow0+h→0+, the result is a non-positive number. If h→0−h \rightarrow 0-h→0−, the result is a non-negative number. Therefore, the result for the above formula is 0.
Or, we can restrict f(x,y) to a function depending on one variable by putting y = b. Define g(x) = f(x,b). The function g attains local max at x = a. We know that g’(a)=0, which is exactly the partial derivative of f w.r.t. x.
Critical Point
(a,b) is a critical point of f if
- f is not diff at (a,b)
- all partial derivative is 0
E.x.
We have f(x,y)=−x2+y2f(x,y)=-x^2+y^2f(x,y)=−x2+y2, find its critical point
sol:
fx=−2x=0,x=0fy=2y=0,y=0f_x = -2x = 0, x=0 \\\\ f_y = 2y = 0, y=0 fx=−2x=0,x=0fy=2y=0,y=0
Thus, (0,0) is the only critical point. If we freeze y and let x change, f(x,0)≤f(0,0)f(x,0) \le f(0,0)f(x,0)≤f(0,0) for every x. If we freeze x and let y change, f(0,y)≤f(0,0)f(0,y) \le f(0,0)f(0,y)≤f(0,0) for every y. This is a saddle point.
Second Derivative Test
Suppose f has a critical point at (a,b). Let us denote
D=∂2f∂x2∂2f∂y2(a,b)−[∂2f∂x∂y(a,b)]2D = \frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}(a,b) - [\frac{\partial^2 f}{\partial x \partial y}(a,b)]^2 D=∂x2∂2f∂y2∂2f(a,b)−[∂x∂y∂2f(a,b)]2
D > 0 and fxx>0f_{xx} > 0fxx>0, then a local min. For example, f(x,y)=x2+y2f(x,y)=x^2+y^2f(x,y)=x2+y2 at the origin.
D > 0 and fxx<0f_{xx} < 0fxx<0, then we have a local max.
D < 0, then a saddle point. For example, f(x,y)=x2−y2f(x,y)=x^2-y^2f(x,y)=x2−y2 at the origin.
D = 0, we cannot tell.
E.x.
Find the shortest distance from the point P=(1,0,-2) to the plane x+2y+z=4.
sol:
Compute the distance to a random point Q on the plane.
∣PQ∣=(x−1)2+y2+(z+2)2=(x−1)2+y2+(6−x−2y)2|PQ| = \sqrt{(x-1)^2+y^2+(z+2)^2} \\\\ = \sqrt{(x-1)^2+y^2+(6-x-2y)^2} \\\\ ∣PQ∣=(x−1)2+y2+(z+2)2=(x−1)2+y2+(6−x−2y)2
Find its critical points and then use the second derivative test to find its min.
E.x.
A rectangular box without a lid is made of 12 m2m^2m2. Find the max volume of such a box.
sol:
Our goal is to maximize xyz under the constraint that 2xz+2yz+xy=122xz+2yz+xy=122xz+2yz+xy=12. We express z in terms of x and y, z=(12−xy)/(2x+2y)z=(12-xy)/(2x+2y)z=(12−xy)/(2x+2y).
Global Min/Max
Compute the value of f at all critical points with the boundary points.
We need the notion of closed sets when dim > 2 (pointset topology). Closed sets are the sets that contain all its boundary points.
The set (x,y):x2+y2≤1{(x,y): x^2+y^2 \le 1}(x,y):x2+y2≤1 is closed, whereas the set (x,y):x2+y2<1{(x,y): x^2+y^2 < 1}(x,y):x2+y2<1 is not.
bounded sets: A bounded set is a set that is contained within some disk.
Extreme Value Theorem (for functions of two variable)
If Ω\OmegaΩ is a set that is closed and bounded and f is a continuous function defined on Ω\OmegaΩ, then f attains an global max/min at some point in Ω\OmegaΩ.
Lagrange Multipliers
With the constraint of g(x,y,z)=0g(x,y,z) = 0g(x,y,z)=0, find the max/min value of f(x,y,z)f(x,y,z)f(x,y,z).
Let ∇f=λ∇g\nabla f = \lambda \nabla g∇f=λ∇g, and solve the equation. Evaluate f at all the above points. Besides, remember to check the cases when x, y, z equals 0.
proof:
Assume that one extreme value is attained at (x0,y0,z0)(x_0, y_0, z_0)(x0,y0,z0), and let a function r⃗(t)\vec r (t)r(t) passing through the point at t=t0t=t_0t=t0. Consider the function h(t)=f(x(t),y(t),z(t))h(t) = f(x(t), y(t), z(t))h(t)=f(x(t),y(t),z(t)). We conclude that h(t)h(t)h(t) has an extreme value at t=t0t=t_0t=t0.
0=h′(t0)=∇f(x0,y0,z0)⋅r⃗(t0)0 = h'(t_0) = \nabla f(x_0, y_0, z_0) \cdot \vec r(t_0) 0=h′(t0)=∇f(x0,y0,z0)⋅r(t0)
We know that the gradient of f is perpendicular to the tangent plane of the surface g(x,y,z)=0g(x,y,z) = 0g(x,y,z)=0.
E.x.
Max f(x,y,z)=xyz when g(x,y,z)=2xz+2yz+xy-12.
sol:
g=2xz+2yz+xy−12=0∇f=<yz,xz,xy>∇g=<2z+y,2z+x,2x+2y>∇f=λ∇gg = 2xz+2yz+xy-12 = 0 \\\\ \nabla f = <yz, xz, xy> \\\\ \nabla g = <2z+y, 2z+x, 2x+2y> \\\\ \nabla f = \lambda \nabla g g=2xz+2yz+xy−12=0∇f=<yz,xz,xy>∇g=<2z+y,2z+x,2x+2y>∇f=λ∇g
if λ=0\lambda = 0λ=0, then yz=xz=xy=0. This violates that g = 0. Thus, this is impossible.
if z = 0, then f = 0. Not a max.
if x=y=0.5z, plug into g to get x=0.75.
Reference
- Multivariable_Calculus_8th_Edition (14.1-14.8), James Stewart
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