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Higher Mathematics

Differential Calculus of Several Variables

This chapter systematically studies limits, continuity, partial derivatives, total differentials, directional derivatives, gradients, tangent and normal planes, Taylor’s formula, extrema, and constrained extrema for multivariable functions.

Basic Concepts of Multivariable Functions

Definition of Multivariable Functions

  • z=f(x,y)z = f(x, y), w=f(x,y,z)w = f(x, y, z), etc.
  • Geometric meaning of a function of two variables: a surface in space

Limits and Continuity

  • Definition and properties of limits
  • Properties of continuous functions on closed and bounded regions

Partial Derivatives and Total Differentials

Partial Derivatives

  • fx(x,y)=fxf_x(x, y) = \frac{\partial f}{\partial x}, fy(x,y)=fyf_y(x, y) = \frac{\partial f}{\partial y}
  • Geometric meaning: slope of the tangent in the direction of the coordinate axes

Higher-Order Partial Derivatives

  • 2fx2\frac{\partial^2 f}{\partial x^2}, 2fxy\frac{\partial^2 f}{\partial x \partial y}
  • Schwarz’s theorem: order of mixed partial derivatives can be interchanged (if continuous)

Total Differential

  • df=fxdx+fydydf = f_x dx + f_y dy
  • Necessary and sufficient condition for the existence of the total differential

Differentiation of Composite, Implicit, and Parametric Functions

  • Chain rule
  • Implicit function partial derivatives
  • Parametric equations partial derivatives

Directional Derivatives and Gradient

  • Directional derivative: Dlf=fxcosα+fycosβD_{\vec{l}}f = f_x \cos\alpha + f_y \cos\beta
  • Gradient: f=(fx,fy)\nabla f = (f_x, f_y), the maximum value of the directional derivative equals the modulus of the gradient, and the maximum direction is the direction of the gradient

Tangent Lines, Normal Planes, Tangent Planes, and Normals

  • For the surface z=f(x,y)z = f(x, y) at (x0,y0,z0)(x_0, y_0, z_0), the equation of the tangent plane: zz0=fx(x0,y0)(xx0)+fy(x0,y0)(yy0)z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)
  • Normal vector: (fx,fy,1)(f_x, f_y, -1)

Taylor’s Formula

  • Taylor expansion for functions of two variables

Extrema and Constrained Extrema

  • Criteria for extrema: Δ=fxxfyy(fxy)2\Delta = f_{xx}f_{yy} - (f_{xy})^2
  • Lagrange multiplier method: f=λg\nabla f = \lambda \nabla g

Exercises

  1. Find the partial derivatives fx,fyf_x, f_y of f(x,y)=x2y+y3f(x, y) = x^2y + y^3 at the point (1,2)(1, 2).
  2. Determine whether f(x,y)=xyx2+y2f(x, y) = \frac{xy}{x^2 + y^2} is continuous at (0,0)(0, 0).
  3. Find the equation of the tangent plane to z=x2+y2z = x^2 + y^2 at the point (1,2,5)(1, 2, 5).
  4. Let u=x2+y2,x=rcosθ,y=rsinθu = x^2 + y^2, x = r\cos\theta, y = r\sin\theta, find ur\frac{\partial u}{\partial r}.
  5. Find the extrema and extreme values of f(x,y)=x2+y22x4yf(x, y) = x^2 + y^2 - 2x - 4y.
  6. Use the Lagrange multiplier method to find the extrema of f(x,y)=x2+y2f(x, y) = x^2 + y^2 under the constraint x+y=1x + y = 1.
Reference Answers

1. Find the partial derivatives fx,fyf_x, f_y of f(x,y)=x2y+y3f(x, y) = x^2y + y^3 at (1,2)(1, 2)

fx=2xy, fy=x2+3y2f_x = 2xy,\ f_y = x^2 + 3y^2

At (1,2)(1, 2): fx=2×1×2=4f_x = 2\times1\times2 = 4, fy=12+3×4=1+12=13f_y = 1^2 + 3\times4 = 1 + 12 = 13


2. Determine whether f(x,y)=xyx2+y2f(x, y) = \frac{xy}{x^2 + y^2} is continuous at (0,0)(0, 0)

Approaching along the xx-axis, f(x,0)=0f(x, 0) = 0; along y=xy = x, f(x,x)=x22x2=12f(x, x) = \frac{x^2}{2x^2} = \frac{1}{2}, so the limits are different, not continuous.


3. Find the equation of the tangent plane to z=x2+y2z = x^2 + y^2 at (1,2,5)(1, 2, 5)

fx=2x, fy=2yf_x = 2x,\ f_y = 2y, at (1,2)(1, 2), fx=2,fy=4f_x = 2, f_y = 4

Tangent plane: z5=2(x1)+4(y2)z - 5 = 2(x - 1) + 4(y - 2)


4. Let u=x2+y2,x=rcosθ,y=rsinθu = x^2 + y^2, x = r\cos\theta, y = r\sin\theta, find ur\frac{\partial u}{\partial r}

u=r2u = r^2, ur=2r\frac{\partial u}{\partial r} = 2r


5. Find the extrema and extreme values of f(x,y)=x2+y22x4yf(x, y) = x^2 + y^2 - 2x - 4y

Set partial derivatives to zero: 2x2=0,2y4=02x - 2 = 0, 2y - 4 = 0, so x=1,y=2x = 1, y = 2

Extremum at (1,2)(1, 2), minimum value f(1,2)=1+428=5f(1, 2) = 1 + 4 - 2 - 8 = -5


6. Use the Lagrange multiplier method to find the extrema of f(x,y)=x2+y2f(x, y) = x^2 + y^2 under the constraint x+y=1x + y = 1

Construct L=x2+y2λ(x+y1)L = x^2 + y^2 - \lambda(x + y - 1)

Lx=2xλ=0\frac{\partial L}{\partial x} = 2x - \lambda = 0

Ly=2yλ=0\frac{\partial L}{\partial y} = 2y - \lambda = 0

Lλ=x+y1=0\frac{\partial L}{\partial \lambda} = x + y - 1 = 0

So x=y=12x = y = \frac{1}{2}, minimum value f(12,12)=12f(\frac{1}{2}, \frac{1}{2}) = \frac{1}{2}