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Let $g(u, v, w)=\left(u e^{v w}+v e^{w w}+w e^{u t}\right) /\left(u^{2}+v^{2}+w^{2}\right)$. Compute \(g(1,2,3)\) and \(g(3,2,1)\).

Short Answer

Expert verified
\(g(1,2,3)=\frac{1\cdot e^{6} + 2\cdot e^{9} + 3\cdot e^{t}}{14}\) and \(g(3,2,1)=\frac{3\cdot e^{2} + 2\cdot e^{1} + 1\cdot e^{3t}}{14}\).
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Step 1: Understanding the function g(u, v, w)

The function \(g(u, v, w)\) is given as: $$g(u, v, w)=\frac{ue^{vw} + ve^{ww} + we^{ut}}{u^2+v^2+w^2}$$ Let's break down the expression. The numerator is the sum of three product terms containing exponentials, and the denominator is the sum of squares of \(u\), \(v\), and \(w\).

Step 2: Compute g(1,2,3)

We are asked to find \(g(1,2,3)\). That means substituting \(u=1\), \(v=2\), and \(w=3\) in the function: $$g(1,2,3)=\frac{1\cdot e^{(2)(3)} + 2\cdot e^{(3)(3)} + 3\cdot e^{(1)(t)}}{1^2+2^2+3^2}$$ Now we will simplify the expression: 1. Calculate exponentials: $$g(1,2,3)=\frac{1\cdot e^{6} + 2\cdot e^{9} + 3\cdot e^{t}}{1^2+2^2+3^2}$$ 2. Simplify denominator: $$g(1,2,3)=\frac{1\cdot e^{6} + 2\cdot e^{9} + 3\cdot e^{t}}{1+4+9}$$ 3. Final simplification: $$g(1,2,3)=\frac{1\cdot e^{6} + 2\cdot e^{9} + 3\cdot e^{t}}{14}$$ So, \(g(1,2,3)=\frac{1\cdot e^{6} + 2\cdot e^{9} + 3\cdot e^{t}}{14}\).

Step 3: Compute g(3,2,1)

Now, we are asked to find \(g(3,2,1)\). That means substituting \(u=3\), \(v=2\), and \(w=1\) in the function: $$g(3,2,1)=\frac{3\cdot e^{(2)(1)} + 2\cdot e^{(1)(1)} + 1\cdot e^{(3)(t)}}{3^2+2^2+1^2}$$ Now we will simplify the expression: 1. Calculate exponentials: $$g(3,2,1)=\frac{3\cdot e^{2} + 2\cdot e^{1} + 1\cdot e^{3t}}{3^2+2^2+1^2}$$ 2. Simplify denominator: $$g(3,2,1)=\frac{3\cdot e^{2} + 2\cdot e^{1} + 1\cdot e^{3t}}{9+4+1}$$ 3. Final simplification: $$g(3,2,1)=\frac{3\cdot e^{2} + 2\cdot e^{1} + 1\cdot e^{3t}}{14}$$ So, \(g(3,2,1)=\frac{3\cdot e^{2} + 2\cdot e^{1} + 1\cdot e^{3t}}{14}\). To summarize our results: - \(g(1,2,3)=\frac{1\cdot e^{6} + 2\cdot e^{9} + 3\cdot e^{t}}{14}\) - \(g(3,2,1)=\frac{3\cdot e^{2} + 2\cdot e^{1} + 1\cdot e^{3t}}{14}\)

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