Differentials and Partial Derivatives

19

Step 1: Use the chain rule: dw/dt = (∂w/∂x)(dx/dt) + (∂w/∂y)(dy/dt). Step 2: Find partial derivatives: ∂w/∂x = 2xy + 2y², ∂w/∂y = x² + 4xy. Step 3: Find dx/dt = 2t and dy/dt = 3t². Step 4: At t = 1: x = 1, y = 1. ∂w/∂x = 2(1)(1) + 2(1) = 4. ∂w/∂y = 1 + 4(1)(1) = 5. dx/dt = 2, dy/dt = 3. Step 5: dw/dt = 4(2) + 5(3) = 8 + 15 = 23. Wait - recalculate. At t=1, x=1, y=1. ∂w/∂x = 2(1)(1)+2(1)² = 2+2 = 4. ∂w/∂y = (1)²+4(1)(1) = 1+4=5. dw/dt = 4·2+5·3 = 8+15=23. Recheck with direct substitution: w = t⁴·t³+2t²·t⁶ = t⁷+2t⁸. dw/dt = 7t⁶+16t⁷. At t=1: 7+16=23. Corrected answer is 23. However given options

F is homogeneous of degree 1 and x·Fₓ + y·F_y = F

Step 1: Test homogeneity: F(λx,λy) = (λ³x³+λ³y³)/(λ²x²+λ²y²) = λ³(x³+y³)/[λ²(x²+y²)] = λ·(x³+y³)/(x²+y²) = λ·F(x,y). Step 2: So F is homogeneous of degree 1 (p = 1). Step 3: By Euler's theorem for homogeneous functions of degree p: x·(∂F/∂x) + y·(∂F/∂y) = p·F. Step 4: With p = 1, we get x·Fₓ + y·F_y = 1·F = F. Step 5: This confirms the first option. A common error is confusing the degree by looking only at the numerator (degree 3) and forgetting to subtract the degree of the denominator (degree 2).

L(x,y) = x - 2y + 5

Step 1: Linear approximation formula: L(x,y) = u(x₀,y₀) + uₓ(x₀,y₀)(x-x₀) + u_y(x₀,y₀)(y-y₀) at (x₀,y₀)=(1,1). Step 2: u(1,1) = 1 - 1 + 1 + 2 - 3 + 5 = 5. Step 3: uₓ = 2x - y + 2; uₓ(1,1) = 2 - 1 + 2 = 3. Step 4: u_y = -x + 2y - 3; u_y(1,1) = -1 + 2 - 3 = -2. Step 5: L(x,y) = 5 + 3(x-1) + (-2)(y-1) = 5 + 3x - 3 - 2y + 2 = 3x - 2y + 4. Recheck: 5 + 3x - 3 - 2y + 2 = 3x - 2y + 4. This matches none exactly - let me recheck option A: x-2y+5 would require uₓ=1. Re-examine: uₓ=2(1)-1+2=3. So L = 3x-2y+4. The answer should be 3x - 2y + 4.