Problem Overview
We evaluate the photometric illuminance distribution I ( x , y ) = 4000 ( x 2 + y 2 + 1 ) − 1 I(x, y) = 4000(x^2 + y^2 + 1)^{-1} I ( x , y ) = 4000 ( x 2 + y 2 + 1 ) − 1 P ( 3 , 4 ) P(3, 4) P ( 3 , 4 ) Q ( 9 , 12 ) Q(9, 12) Q ( 9 , 12 ) ∇ I ( 3 , 4 ) \nabla I(3, 4) ∇ I ( 3 , 4 ) P Q → \overrightarrow{PQ} P Q D u ⃗ I ( 3 , 4 ) D_{\vec{u}} I(3, 4) D u I ( 3 , 4 )
Step-by-Step Mathematical Solution
1. Compute and Evaluate the Illuminance Gradient
State the Gradient Vector Formula (defined in The Gradient Vector ):
∇ I ( x , y ) = ⟨ ∂ I ∂ x , ∂ I ∂ y ⟩ \nabla I(x, y) = \left\langle \frac{\partial I}{\partial x}, \frac{\partial I}{\partial y} \right\rangle ∇ I ( x , y ) = ⟨ ∂ x ∂ I , ∂ y ∂ I ⟩ Using the single-variable power/chain rule on I ( x , y ) = 4000 ( x 2 + y 2 + 1 ) − 1 I(x, y) = 4000(x^2 + y^2 + 1)^{-1} I ( x , y ) = 4000 ( x 2 + y 2 + 1 ) − 1
I x = − 4000 ( x 2 + y 2 + 1 ) − 2 ⋅ ( 2 x ) = − 8000 x ( x 2 + y 2 + 1 ) 2 I_x = -4000(x^2 + y^2 + 1)^{-2} \cdot (2x) = \frac{-8000x}{(x^2 + y^2 + 1)^2} I x = − 4000 ( x 2 + y 2 + 1 ) − 2 ⋅ ( 2 x ) = ( x 2 + y 2 + 1 ) 2 − 8000 x I y = − 4000 ( x 2 + y 2 + 1 ) − 2 ⋅ ( 2 y ) = − 8000 y ( x 2 + y 2 + 1 ) 2 I_y = -4000(x^2 + y^2 + 1)^{-2} \cdot (2y) = \frac{-8000y}{(x^2 + y^2 + 1)^2} I y = − 4000 ( x 2 + y 2 + 1 ) − 2 ⋅ ( 2 y ) = ( x 2 + y 2 + 1 ) 2 − 8000 y Evaluate at P ( 3 , 4 ) P(3, 4) P ( 3 , 4 ) x 2 + y 2 + 1 = 3 2 + 4 2 + 1 = 9 + 16 + 1 = 26 x^2 + y^2 + 1 = 3^2 + 4^2 + 1 = 9 + 16 + 1 = 26 x 2 + y 2 + 1 = 3 2 + 4 2 + 1 = 9 + 16 + 1 = 26
( x 2 + y 2 + 1 ) 2 = 26 2 = 676 (x^2 + y^2 + 1)^2 = 26^2 = 676 ( x 2 + y 2 + 1 ) 2 = 2 6 2 = 676 I x ( 3 , 4 ) = − 8000 ( 3 ) 676 = − 24000 676 = − 6000 169 ≈ − 35.503 lx/m I_x(3, 4) = \frac{-8000(3)}{676} = -\frac{24000}{676} = -\frac{6000}{169} \approx -35.503\text{ lx/m} I x ( 3 , 4 ) = 676 − 8000 ( 3 ) = − 676 24000 = − 169 6000 ≈ − 35.503 lx/m I y ( 3 , 4 ) = − 8000 ( 4 ) 676 = − 32000 676 = − 8000 169 ≈ − 47.337 lx/m I_y(3, 4) = \frac{-8000(4)}{676} = -\frac{32000}{676} = -\frac{8000}{169} \approx -47.337\text{ lx/m} I y ( 3 , 4 ) = 676 − 8000 ( 4 ) = − 676 32000 = − 169 8000 ≈ − 47.337 lx/m ∇ I ( 3 , 4 ) = ⟨ − 6000 169 , − 8000 169 ⟩ \nabla I(3, 4) = \left\langle -\frac{6000}{169}, -\frac{8000}{169} \right\rangle ∇ I ( 3 , 4 ) = ⟨ − 169 6000 , − 169 8000 ⟩ 2. Find and Normalize the Displacement Vector
State the Displacement & Normalization Formulas :
P Q → = Q − P = ⟨ 9 − 3 , 12 − 4 ⟩ = ⟨ 6 , 8 ⟩ \overrightarrow{PQ} = Q - P = \langle 9 - 3, \, 12 - 4 \rangle = \langle 6, 8 \rangle P Q = Q − P = ⟨ 9 − 3 , 12 − 4 ⟩ = ⟨ 6 , 8 ⟩ ∥ P Q → ∥ = 6 2 + 8 2 = 36 + 64 = 100 = 10 m \|\overrightarrow{PQ}\| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ m} ∥ P Q ∥ = 6 2 + 8 2 = 36 + 64 = 100 = 10 m u ⃗ = P Q → ∥ P Q → ∥ = ⟨ 6 , 8 ⟩ 10 = ⟨ 0.6 , 0.8 ⟩ = ⟨ 3 5 , 4 5 ⟩ \vec{u} = \frac{\overrightarrow{PQ}}{\|\overrightarrow{PQ}\|} = \frac{\langle 6, 8 \rangle}{10} = \langle 0.6, 0.8 \rangle = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle u = ∥ P Q ∥ P Q = 10 ⟨ 6 , 8 ⟩ = ⟨ 0.6 , 0.8 ⟩ = ⟨ 5 3 , 5 4 ⟩ 3. Compute the Directional Derivative
State the Directional Derivative Formula (defined in Directional Derivatives ):
D u ⃗ I ( P ) = ∇ I ( P ) ⋅ u ⃗ D_{\vec{u}} I(P) = \nabla I(P) \cdot \vec{u} D u I ( P ) = ∇ I ( P ) ⋅ u Evaluate the dot product:
D u ⃗ I ( 3 , 4 ) = ⟨ − 6000 169 , − 8000 169 ⟩ ⋅ ⟨ 3 5 , 4 5 ⟩ D_{\vec{u}} I(3, 4) = \left\langle -\frac{6000}{169}, -\frac{8000}{169} \right\rangle \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle D u I ( 3 , 4 ) = ⟨ − 169 6000 , − 169 8000 ⟩ ⋅ ⟨ 5 3 , 5 4 ⟩ D u ⃗ I ( 3 , 4 ) = ( − 6000 ) ( 3 ) + ( − 8000 ) ( 4 ) 169 ⋅ 5 = − 18000 − 32000 845 = − 50000 845 = − 10000 169 ≈ − 59.17 lux/m D_{\vec{u}} I(3, 4) = \frac{(-6000)(3) + (-8000)(4)}{169 \cdot 5} = \frac{-18000 - 32000}{845} = \frac{-50000}{845} = -\frac{10000}{169} \approx -59.17\text{ lux/m} D u I ( 3 , 4 ) = 169 ⋅ 5 ( − 6000 ) ( 3 ) + ( − 8000 ) ( 4 ) = 845 − 18000 − 32000 = 845 − 50000 = − 169 10000 ≈ − 59.17 lux/m 4. Lighting Automation & Control Analysis
Threshold Evaluation: The rate of illuminance loss is − 59.17 lux/m -59.17\text{ lux/m} − 59.17 lux/m − 50 lux/m -50\text{ lux/m} − 50 lux/m ∣ − 59.17 ∣ > 50 |-59.17| > 50 ∣ − 59.17∣ > 50 Control Application: Entertainment engineers feed this directional derivative into automated tracking software. Because the rate of drop along u ⃗ \vec{u} u 59.17 lux 59.17\text{ lux} 59.17 lux v ⃗ performer \vec{v}_{\text{performer}} v performer d I d t = D u ⃗ I ⋅ ∥ v ⃗ performer ∥ \frac{dI}{dt} = D_{\vec{u}} I \cdot \|\vec{v}_{\text{performer}}\| d t d I = D u I ⋅ ∥ v performer ∥ 150 lux 150\text{ lux} 150 lux