Question Number 115201 by mohammad17 last updated on 24/Sep/20

Commented bymohammad17 last updated on 24/Sep/20

help me sir please

Answered by 1549442205PVT last updated on 24/Sep/20

Q4  1a)The equation of tangent plane at  P(1;0;1) for the function z=xe^(−y)  is:  z−z_0 =((∂z/∂x))_P (x−x_0 )+((∂z/∂y))_P (y−y_0 )  ⇔z−1=1.(x−1)+(−1)(y−0)  ⇔z−1=x−y−1⇔x−y−z=0  b)The equation of the normal line  is  ((x−x_0 )/(((∂z/∂x))_P ))=((y−y_0 )/(((∂z/∂y))_P ))=((z−z_0 )/(−1))  ⇔((x−1)/1)=((y−0)/(−1))=((z−1)/(−1))=t or in the form  of the parameter:   { ((x=t+1)),((y=−t)),((z=−t+1)) :}  2)The function will  increase most rapidly  in the direction of the gradien vector   of the function f(x,y)=(x/(x+y))=1−(y/(x+y))  ((∂f(x,y))/∂x)=(y/((x+y)^2 )),((∂f(x,y))/∂y)=((−x)/((x+y)^2 )).Hence  grad f(x,y)=((∂f(x,y))/∂x)i+((∂f(x,y))/∂y)j.Therefore,  the unit vector we need find is:  e^→ =(1/( (√((((∂f(x,y))/∂x)+((∂f(x,y))/∂y))^2 ))))(((∂f(x,y))/∂x)i+((∂f(x,y))/∂y)j)  At the point P(0,2) we have:  ((∂f(x,y))/∂x)=(1/2),(((∂f(x,y))/∂y))_((0,2)) =0  e^→ =i⇒ α=(e,Ox)^(�) =0⇒cosα=1  sinα=0.Therefore,the rate change   of change of the function equal to  v=(((∂f(x,y))/∂e^→ ))_((0,2)) =(((∂f(x,y))/∂x)c)_((0,2)) cosα+(((∂f(x,y))/∂y))_((0,2)) sinα  =(1/2)×1=(1/2)  second way:  (((∂f(x,y))/∂e^(→) ))_(max(0,2)) =∣grad f(x,y)∣_((0,2))   =(√((((∂f(x,y))/∂x))^2 +(((∂f(x,y))/∂y))^2 ))  =(√(((1/2))^2 +0^2 ))=(1/2)