Question

Let f(0, -2) = 4 and 0f (0, -2) = 6,23). a. Write the parametric equations of the line in the xy-plane that is perpendicular to the contour f(x,y) = 4 at ( 0-2). b. Write an equation for the tangent plane to the surface z = f(x,y) at (0, -2,4).

Answer 1

a ..o to co. Given, f(0, -2)=4, Vf(0,-2) = ). Let the parametric equations of the reduired line be, a=Xqt + kg; y = y ₂ t + J2; where, Ky, K2, I, and ya are constants and t is the parameter, Varrying from Also let when t=0, we have , X=0, y=-2 So, X =2;0 +22=0 » X2 ZO Y=Y, 07 / 2 -2 => Z2 = -2 X = azt ; Y = Ist 2; Are the parametric eduations. Differentiating x and y so have, we W.8. to t dy = Y. da dt. as i dt Y dr X So, tangent of the line is dy at)-(Ferty. We know that of (x, y) = (26 - 1 so of (0,-2) 41.200) *143) of dy 2 (o - ax 11-0. y=-2 Y2-2

fxt dy zo ty 4 + zo 5 dy 5 3 dx If the required line, be a perpendicular to Parametric form of the line will be, flx, y)= 4: to a we have Differentining war da At Point 10,-2), we have. 12 dy 25 da So dy da (0,-2) मा 봤 x f(x,y)=4, = هادی) / have : on the Curve at the point. 10,-2). f(x,y)=4 at 20,-2), then we 632-4 9 3 2 5 -00 <teo. x=5t y z3t-2

b) Let gla, 2) = -f(x,y) so ag . . 22 ax 1 ay oz og esta JX 21 ag af ag At Point 10, 2, 4) we have. 잎 160,-2,4) 1(0,2) ag of gy 0,-3,4) ag 21 So the direction ratios of the normal of g(x, y, z)=0, at the point (0, -2,4) will be 1 - 1 / 1 / 1 Let (a, Y, 2) be any point on the tangent plane of g (1,2,2)-0 at Co, -2,4). 12 25 of 20,-2) 22 120,-2,4)

Eduation of the tangent Hame will be (2-0, 4+2, 2-4). 61 -0 7-4-264+2)+(2-9) 20 1 > 20x+127 - -20% -127 +252-100 20 -25% + 100 20