Engng Math 145 Multi-variable Functions

Domain & range of f(x,y)

Domain of f(x,y) is its undefined spots. D = {(x,y) ∈ ℝ² | …}
Range of f(x,y) is {f(x,y) | (x,y) ∈ D}

Graph of f(x,y) & sketching

Set of all (x, y, z) ∈ ℝ³, such that z = f(x,y).

Sketch LOC for different heights k

Level of curves of f(x,y)

f(x,y) = k, where k is an ℤ.

Turns f(x,y) into ay = k - bx form which is easy to graph.

Plane & sphere on f(x,y) & sketching

• Linear plane is of form f(x,y) = ax + by + c

  • Determine intercepts & draw plane through these points.

• Sphere is of form r² = x² + y² + z²

  • z = sqrt(r²-x²-y²) is top half & z = -sqrt(r²-x²-y²) is bottom half.

Limits and f(x,y)

lim_{(x,y) → (a,b)} f(x,y) = L.

Solve using standard methods.

Continuity and f(x,y)

• Continuous if lim_{(x,y) → (a,b)} f(x,y) = f(a,b)

• Solving strategy:

  • If (a,b) is in domain then directly substitute

  • If not in domain then:

    • Polar substitution if terms of form x²+y²

    • Simplify function by using factorisation, cancelling & identities

    • Test limit by using path method & prove by using squeeze theorem

Path method to show undefined limit

Let C₁ be [choose], then substitute into f(x,y) and compute given limit. Next let C₂ be [choose], then substitute into f(x,y) and compute given limit. If answers not equal then limit does not exist.

C₁ & C₂ can be as simple as y=0 & x=0, but if that doesn’t work then choose options that will make answer not equal to each other.

eg. options that cause either x or y to cancel out and give constant that is not 0.

Squeeze theorem with f(x,y)

If g(x,y) < f(x,y) < h(x,y) for all (x,y) ∈ D, and lim_{(x,y) → (a,b)} g(x,y) = L = lim_{(x,y) → (a,b)} h(x,y), then lim_{(x,y) → (a,b)} f(x,y) = L.

Polar substitution and f(x,y)

lim_{(x,y) → (0,0)} f(x,y) = lim_{r → 0⁺} f(rcosθ, rsinθ), where r > 0.

If a value of θ makes limit undefined then polar substitution is invalid.

If polar limit in intermediate form use L’Hopital

Partial derivatives and f(x,y)

• f_x & f_y also written as df/dx & df/dy, D₁f & D₂f and D_xf & D_yf.

  • when determining f_x, treat y as a constant.

  • when determining f_y, treat x as a constant.

  • If >2 variable function, then treat all other variables as constants.

Clairaut’s theorem

If f_xy & f_yx are both continuous, then f_xy = f_yx

Implicit differentiation and f(x,y)

If f(x,y) cannot be written explicitly.

1. eg. given: z = f(x,y) & want to determine dz/dx

2. d/dx * LHS = d/dx * RHS

3. solve for dz/dx

Chain rule and f(x,y)

• Case 1: z=f(x,y) differentiable function of x & y, and x=g(t) & y=h(t) differentiable functions of t, then:

  • dz/dt = (dz/dx * dx/dt) + (dz/dy * dy/dt)

• Case 2: w=f(x,y) differentiable function of x & y, and x=g(s,t) & y=h(s,t) differentiable functions of s & t, then w=w(s,t) and:

  • dw/ds = (dw/dx * dx/ds) + (dw/dy * dy/ds)

  • dw/dt = (dw/dx * dx/dt) + (dw/dy * dy/dt)

• General case: If u differentiable function of n variables x₁, …, x_n AND x_k (k=1,2...,n) differentiable function of m variables t₁, …, t_m, then:

  • du/dt_i = (du/dx₁ * dx/dt_i) + (du/dx₂ * dx/dt_i) + … + (du/dx_n * dx/dt_i), for i ∈ {1,2,…,m}.

Higher derivatives and f(x,y)

Critical points of f(x,y)

(a,b) is critical point of f if:

• f_x(a,b) = 0 & f_y(a,b) = 0 OR f_x(a,b) & f_y(a,b) does not exist

Second derivative test & discriminant with f(x,y)

Optimisation problems

1. Make sketch if possible

2. Choose appropriate variables & set up equation to be maximised or minimised

3. Find relationship between variables & eliminate

4. Create function of one/two variables & determine domain

5. Determine max of function

Useful formulas:

• distance = sqrt[(x₁-x₂)² + (y₁-y₂)² + (z₁+z₂)²]

• Volume = xyz

• Area = 2xy + 2yz + 2xz

Absolute max & min of f(x,y)

1. Find critical points in given domain

2. Find critical points & end points on boundary of domain

3. Largest value found from previous steps is abs max & smallest value found from previous steps is abs min

Extreme value theorem

If a function f(x, y) is continuous on a closed and bounded region R ⊂ ℝ², then f(x, y) attains both an absolute maximum value and an absolute minimum value at some points within R.