Answer to Question #176388 in Calculus for Nikhil Rawat

Question #176388

Let D: {(x,y)| x>0, y>0}. Consider two function f and g from D to R, defined by:

f(x,y) = Inx - Iny and g(x,y)= x^2+ 3y^2/(2xy)

Show that the necessary condition for the functional dependence of f and g is satisfied. Also find a functional relation between f and g


1
Expert's answer
2021-04-14T07:12:07-0400

As we know, two functions "f(x,y)" and "g(x,y)" are dependent if the Jacobi matrix is equal to zero :



"J=\\frac{\\partial\\left(f,g\\right)}{\\partial\\left(x,y\\right)}=\n\\left|\\begin{array}{cc}\nf'_x&f'_y\\\\[0.3cm]\ng'_x&g'_y\n\\end{array}\\right|"

In our case,



"f(x,y)=\\ln x-\\ln y\\longrightarrow\\left\\{\n\\begin{array}{l}\nf'_x=\\displaystyle\\frac{1}{x}\\\\[0.3cm]\nf'_y=-\\displaystyle\\frac{1}{y}\n\\end{array}\\right.\\\\[0.3cm]\ng(x,y)=\\frac{x^2+3y^2}{2xy}\\equiv\\frac{x}{2y}+\\frac{3y}{2x}\\longrightarrow\\left\\{\n\\begin{array}{l}\ng'_x=\\displaystyle\\frac{1}{2y}-\\displaystyle\\frac{3y}{2x^2}\\\\[0.3cm]\ng'_y=-\\displaystyle\\frac{x}{2y^2}+\\displaystyle\\frac{3}{2x}\n\\end{array}\\right.\\\\[0.3cm]"

Then,



"J=\\left|\\begin{array}{cc}\nf'_x&f'_y\\\\[0.3cm]\ng'_x&g'_y\n\\end{array}\\right|=\\left|\\begin{array}{cc}\n\\displaystyle\\frac{1}{x} &-\\displaystyle\\frac{1}{y}\\\\[0.5cm]\n\\displaystyle\\frac{1}{2y}-\\displaystyle\\frac{3y}{2x^2}&\n-\\displaystyle\\frac{x}{2y^2}+\\displaystyle\\frac{3}{2x}\n\\end{array}\\right|=\\\\[0.3cm]\n=\\frac{1}{x}\\cdot\\left(-\\frac{x}{2y^2}+\\frac{3}{2x}\\right)-\n\\left(-\\frac{1}{y}\\right)\\cdot\\left(\\frac{1}{2y}-\\frac{3y}{2x^2}\\right)=\\\\[0.3cm]\n=-\\frac{1}{2y^2}+\\frac{3}{2x^2}+\\frac{1}{2x^2}-\\frac{3}{2x^2}\\equiv0\\longrightarrow\\\\[0.3cm]\n\\boxed{\\text{Conclusion,}\\,\\,\\,J\\left(f,g\\right)\\equiv0,\\,\\,\\forall (x,y)\\in D=\\left\\{(x,y)\\left|x>0,y>0\\right.\\right\\}}"

To find the functional relationship between "f(x,y)" and "g(x,y)" we will do the following :



"z=f(x,y)=\\ln x-\\ln y\\equiv\\ln\\left(x\/y\\right)\\\\[0.3cm]\nu=g(x,y)=\\frac{x^2+3y^2}{2xy}=\\frac{1}{2}\\cdot\\left(\\frac{x}{y}+3\\cdot\\frac{y}{x}\\right)\\\\[0.3cm]\nu=\\frac{1}{2}\\cdot\\left(\\frac{x}{y}+3\\cdot\\left(\\frac{x}{y}\\right)^{-1}\\right)"

Our task is reduced to finding a function "\\varphi" between variables "z" and "u" .

As we know



"e^z=e^{\\ln\\left(x\/y\\right)}=\\frac{x}{y}\\\\[0.3cm]\ne^{-z}=e^{-\\ln\\left(x\/y\\right)}=e^{\\ln\\left(y\/x\\right)}=\\frac{y}{x}\\\\[0.3cm]\n\\varphi(z)=\\frac{1}{2}\\cdot\\left(e^z+3e^{-z}\\right)"

Or in the original notation



"\\boxed{\\varphi\\left(f(x,y)\\right)=\\frac{1}{2}\\cdot\\left(e^{f(x,y)}+3e^{-f(x,y)}\\right)\\equiv g(x,y)}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS