Multivariable functions #5 - Questions and Answers

Question 1 Relative extrema of multivariables: f(x,y) = xy / 7 find critical points and relative extrema, given an open region. Answer We have, f(x,y) = xy / 7 To determine the pivotal point: Equivalent to zero the partial derivative: Now, Differentiate f(x,y) with respect to x and equate to zero fₓ(x,y) = y / 7 = 0 ⇒ y = 0 Differentiate f(x,y) with respect to y and equate to zero fᵧ(x,y) = x / 7 = 0 ⇒ x = 0 Therefore, the critical point is (0,0) Now we have to find the relative extrema. Use: D(x,y) = fₓₓ(x,y) fᵧᵧ(x,y) – (fₓᵧ(x,y))² Find D(0,0): Now fₓₓ(0,0) = 0 fᵧᵧ(0,0) = 0 fₓᵧ(0,0) = 1 / 7 Hence, D(0,0) = 0*0 - (1/7)² = -1/49 < 0 Hence, by the 2nd derivative test: f(x,y) neither max. nor min at (0,0) which means (0,0) is saddle point. Question 2 Use polar coordinates to find the limit. [Hint: Let x = rcos(θ) and y= rsin(θ), and note that (x, y) → (0, 0) implies r → 0.] lim(x,y)→(0,0) (x² - y²) / √x² + y² Answer Given the multivariable limit function: lim(x,y)→(0,0) (x² - y²) / √x² + y² For polar coordinate system: x = rcosθ, y = rsinθ lim(x,y)→(0,0) (x² - y²) / √x² + y² = lim(r→0) (r²cos²θ - r²sin²θ) / √(r²cos²θ + r²sin²θ) = lim(r→0) (r²(cos²θ - sin²θ)) / √(r²(cos²θ + sin²θ)) We know the trigonometry identities (cos²θ - sin²θ) = cos 2θ (cos²θ + sin²θ) = 1 lim(r→0) (r²cos 2θ) / √r² = lim(r→0) (r²cos 2θ) / r lim(r→0) rcos 2θ = 0 lim(x,y)→(0,0) (x² - y²) / √x² + y²= 0 Hence the limit of the given function is zero. Question 3 a) Find the functions f(x, y) = 1 / √(16 - x² - y²) b) Evaluate the range for the given function. Range for 16 - x² - y² is (0, 16]. c) Determine the boundary of the domain using the domain for f(x, y). The domain for f(x,y) is (x,y) ∈ ℝ²: x² + y² < 16. Answer a) Determine the domain of the multivariable function. f(x, y) = 1 / √(16 - x² - y²) 16 - x² - y² ≥ 0 -x² - y² ≥ -16 −1 × (-x² - y²) ≤ −1 × −16 x² + y² ≤ 16 Also, 16 - x² - y² > 0 16 - x² - y² > 0 x² + y² < 16 From, x² + y² ≤ 16 and x² + y² < 16 The domain for f(x,y) is (x,y) ∈ ℝ²: x² + y² < 16. b) Evaluate the range for the given function. Range for 16 - x² - y² is (0, 16]. Therefore, range for f(x,y) is f(x, y) = 1 / √(16 - x² - y²) As 16 - x² - y² → 0, f(x, y) → ∞ As 16 - x² - y² → 16, f(x, y) → 1/4 Therefore, range for f(x, y) ∈ [1/4, ∞). c) Determine the boundary of the domain using the domain for f(x, y). The domain for f(x,y) is (x,y) ∈ ℝ²: x² + y² < 16. The equation x² + y² < 16 is a circle. The boundary for the domain is points lying in the circle x² + y² = 16. d) The domain for the function f(x, y) do not contain the boundary points. Therefore, the domain for the function is an open region. e) The domain for the function is enclosed in the circle. Hence, the domain is bounded in the circle of equation x² + y² < 16. Question 4 Let z(x, y) = e^(3xy), x(p,q) = p/q and y(p,q) = q/p are functions. Use multivariable chain rule of partial derivatives to find (i) ∂z/∂p (ii) ∂z/∂q Answer (i) ∂z/∂p = (∂z/∂x)(∂x/∂p) + (∂z/∂y)(∂y/∂p) (1) ⇒ ∂z/∂x = e^(3xy) * 3y = 3ye^(3xy) ⇒ ∂z/∂y = e^(3xy) * 3x = 3xe^(3xy) ∂x/∂p = 1/q and ∂y/∂p = -q/p² Using these values equation (1) becomes: ∂z/∂p = (3ye^(3xy) / q) + (3xe^(3xy) * (-q / p²)) ∂z/∂p = (3ye^(3xy) / q) - (3xqe^(3xy) / p²) ∂z/∂p = e^(3xy) * (3y / q - 3xq/p²) ∂z/∂p = e^(3xy) * (3p²y - 3xq²) / pq² (ii) ∂z/∂q = (∂z/∂x)(∂x/∂q) + (∂z/∂y)(∂y/∂q) ∂x/∂q = -p/q² and ∂y/∂q = 1/p ∂z/∂q = (3ye^(3xy) * (-p/q²)) + (3xe^(3xy) * 1/p) ∂z/∂q = e^(3xy)(-3yp/q² + 3x/p) ∂z/∂q = e^(3xy) * (-3yp² + 3xq²) / pq²