Question #105386
(1) Verify Euler’s relation for the function f(x,y)=(x^3+y^3)/(x+y).
(2) Check whether the following functions are differentiable at the point given
against them:
i) f(x,y)=y^3+y Sin2x+e^(x+y) at (1,–1)
ii) f(x,y)=|x=1| at (1,0)
(3) Find the range of the function f defined by f(x,y)=10–x^2–y^2 for all (x,y)
for which x^2+y^2 ≤9. Sketch two of its level curve
1
Expert's answer
2020-03-18T16:37:28-0400

1.f(x,y)=x3+y3x+y=(x+y)(x2xy+y2)x+y=x2xy+y2f(x,y)=\frac{x^3+y^3}{x+y}=\frac{(x+y)(x^2-xy+y^2)}{x+y}=x^2-xy+y^2

Degree is 2.

According to Euler's relation:

xfx+yfy=nf(x,y)x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)

xfx+yfy=x(2xy)+y(2yx)=2(x2xy+y2)=nf(x,y)x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=x(2x-y)+y(2y-x)=2(x^2-xy+y^2)=nf(x,y)

(Proved)

3.f(x,y)=10x2y2f(x,y)=10-x^2-y^2

0x2+y290 \leqslant x^2+y^2 \leqslant9

9(x2+y2)0-9 \leqslant -(x^2+y^2) \leqslant0

10910(x2+y2)0+1010-9 \leqslant 10-(x^2+y^2) \leqslant0+10

1f(x,y)101 \leqslant f(x,y)\leqslant10 is the range of the function.

Level Curve in the image.

2.f(x,y)=y3+ySin2x+ex+yf(x,y)=y^3+y Sin2x+e^{x+y}

fx(x,y)=2ycos2x+ex+yf_x(x,y)=2ycos2x+e^{x+y}

fxx(x,y)=4ySin2x+ex+yf_{xx}(x,y)=-4y Sin2x+e^{x+y}

fy(x,y)=3y2+Sin2x+ex+yf_y(x,y)=3y^2+ Sin2x+e^{x+y}

fyy(x,y)=6y+ex+yf_{yy}(x,y)=6y+ e^{x+y}

Partial derivatives are continuous and differentiable as sin (2x),cos(2x),ex+y,y2,y3,y are continuous and differentiable.

So,f(x,y) is differentiable at (1,-1).

(ii)f(x,y)=x1at(1,0)f(x,y)=|x-1| at (1,0)

fy(x,y)=0f_y(x,y)=0

fx(1,0)=limh0f((1+h),0)f(1,0)hf_x(1,0)=lim_{h\to0}\frac{f((1+h),0)-f(1,0)}{h}

fx(1,0)=limh0h0h=limh0hhf_x(1,0)=lim_{h\to0}\frac{|h|-0}{h}=lim_{h\to0}\frac{|h|}{h}

limh0+hh=1lim_{h\to0^+}\frac{h}{h}=1 whereas limh0hh=1lim_{h\to0^-}\frac{-h}{h}=-1

So,limit does not exist,partial derivative fx(x,y) does not exist at (1,0) .Thus, f(x,y) is not differentiable.


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