"b) Find the equation of the tangent plane to f (x,y) at(4/5,3/5)a) Here we have to find the equation of the tangent planeto Z = f ( x , y ) at ( 4/5,3/5)Solution: The partial derivative forAt (4/5, 3/5) are:1 (4/5,3/5) = (2x) at (4/5,3/5)2 2 ? + ? = { 1/ ( 4/5)^2 +(3/5)^2}*2(4/5) = { 1/(16/25)+(9/25)}*(8/5)=(8/5)/(16+9)/25=(8/5)/25/25=8/5(4/5,3/5) = {1/ ( x^2 + y^2 ) * ( 2y) }at (4/5,3/5)= { 1/ ( 4/5)^2 +(3/5)^2}*2(3/5) = { 1/(16/25)+(9/25)}*(6/5)=(6/5)/(16+9)/25=(6/5)/25/25=6/5Thus (grad f) at (4/5, 3/5) is (8/5)i + ((6/5)j------( 1 )Now the equation of the tangent plane is (8/5)i + ((6/5)j { ( x-(4/5) i+(y-(3/5)j)}=08/5( x-(4/5) )+6/5(y-(3/5)) = 0(8/5)*(5x-4)/5+(6/5)*(5y-3)/5=0(40x-32)/25+(30y-18)/25=040x+30y-50= 0 Dividing throughout by 10 we get4x+3y-5=0 or 4x+3y= 5 is the required equation .b) In which direction is f (x,y) decreasing most rapidly atthe point ( 4/5, 3/5 ) Solution: The greatest decrease at (4/5, 3/5) is in the direction ofthe negative gradient Now grad f =From (1) of part ( b ) we know that (grad f) at (4/5 ,3/5)is (8/5)i + ((6/5)jHence (-grad f) at (4/5, 3/5) is (-8/5)i + (-6/5)j 5) a)Solution :Here we have to prove that F is a conservative vectorfieldWe know that in a conservative field F , curl F = 0 Curl F = ? X F = 0Here F = F1 + F2 And curl F = ? X F=-i ( ?F2 / ?z) +j ( ?F1/ ?z)+ k {( ?F2 / ?x) – ( ?F1/ ?y)} Therefore n.curlF = k.k({( ?F2 / ?x) – ( ?F1/ ?y)}Since the unit normal vector on the X-Y plane is kWe have n.curl F ={( ?F2 / ?x) – ( ?F1/ ?y)}Here F = F1i+F2jTherefore F1 = x / ( x^2 + y ^ 2 ) andF2 = y / ( x^2 + y ^2 ) Hence ( ?F1/ ?y) = 1 / ( x^2 + y ^ 2 ) ^2( - x ( 2y )) = - 2xy/ ( x^2 + y ^ 2 )^2and ( ?F2/ ?x) = 1 / ( x^2 + y ^ 2 ) ^2( - y( 2x )) = - 2xy/ ( x^2 + y ^ 2 )^2Therefore ? {( ?F2 / ?x) – ( ?F1/ ?y)}becomes-2xy/(x^2+y^2)^2-(-2xy/(x^2+ y^2)^2 )-2xy/(x^2+y^2)^2+2xy/(x^2+ y^2)^2 )2xy-2xy / ( x^2+y^2)^2 = 0Hence n. curl F = 0 ? curl F = 0Hence the given function F is a conservative vector fieldHence the proof.b)Solution :we have to find a function f such that F = ?f r2 r2F = ? f results ?F.dr = ?df = f ( r2 – r1 )r1 r1Hence f ( r ) = ?F.dr = ? x /( x ^ 2 + y ^ 2) dx + ? y/ ( x ^ 2 + y ^ 2 ) dy]Multiplying and dividing by 2 we get½ { 2x/( x^2 + y^2} +1/2{ 2y(x^2+y^2)}½ { log ( x^2+ y^2 )log ?( x^2 + y^2 )Hence f(x, y ) = log ?( x^2 + y^2 ) "