484 CHAPTER 22. DERIVATIVE OF A FUNCTIONS OF MANY VARIABLES

Example 22.9.1 Find the equation of the tangent plane to the level surface

f (x,y,z,w) = 6

of the function f (x,y,z) = x2 +2y2 +3z2 +w at the point (1,1,1,0).

First note that (1,1,1,0) is a point on this level surface. To find the desired plane itsuffices to find the normal vector to the proposed plane. But ∇ f (x,y,z,w) = (2x,4y,6z,1)and so ∇ f (1,1,1,0) = (2,4,6,1). Therefore, from this problem, the equation of the planeis (2,4,6,1) · (x−1,y−1,z−1,w) = 0 or in other words, 2x−12+4y+6z+w = 0. Notethat this is a three dimensional plane because there are three free variables. Indeed, it is ofthe form w = 12−4y−6z−2x.

Example 22.9.2 The point(√

3,1,4)

is on both the surfaces, z = x2 + y2 and z = 8 −(x2 + y2

). Find the cosine of the angle between the two tangent planes at this point.

Recall this is the same as the angle between two normal vectors. Of course there issome ambiguity here because if n is a normal vector, then so is −n and replacing n with−n in the formula for the cosine of the angle will change the sign. We agree to look forthe acute angle and its cosine rather than the obtuse angle. The normals are

(2√

3,2,−1)

and(2√

3,2,1). Therefore, the cosine of the angle desired is (2

√3)

2+4−1

17 = 1517 .

Example 22.9.3 The point(1,√

3,4)

is on the surface z = x2 + y2. Find the line perpen-dicular to the surface at this point.

All that is needed is a direction vector for this line. The surface is the level surfacex2 + y2 − z = 0. The normal to this surface is given by the gradient at this point. Thus thedesired line is

(1,√

3,4)+ t(2,2

√3,−1

).

22.10 Exercises1. Find the gradient at the indicated point if f =

(a) x2y+ z3, (1,1,2)(b) zsin

(x2y)+2x+y, (1,1,0)

(c) u ln(x+ y+ z2 +w

), (x,y,z,w,u)

= (1,1,1,1,2)

(d) sin(xy)+ z3, (1,π,1)

(e) ln(x+ y2

)z

(f) z ln(4+ sin(xy)), (0,π,1)

2. Find the directional derivatives of f at the indicated point in the direction(

12 ,

12 ,

1√2

).

(a) x2y+ z3 at (1,1,1)(b) zsin

(x2y)+2x+y at (1,1,0)

(c) xy+ z2 +1 at (1,2,3)(d) sin(xy)+ z at (0,1,1)

(e) xy + z at (1,1,1).

(f) sin(sin(x+ y)) + z at the point(1,0,1).

3. Find the directional derivatives of the given function at the indicated point in theindicated direction.

484 CHAPTER 22. DERIVATIVE OF A FUNCTIONS OF MANY VARIABLESExample 22.9.1 Find the equation of the tangent plane to the level surfaceFf (X,y,z,w) = 6of the function f (x,y,z) = x* +2y? +32 +w at the point (1,1,1,0).First note that (1,1,1,0) is a point on this level surface. To find the desired plane itsuffices to find the normal vector to the proposed plane. But Vf (x, y,z,w) = (2x, 4y, 6z, 1)and so Vf (1,1,1,0) = (2,4,6,1). Therefore, from this problem, the equation of the planeis (2,4,6,1)-(x—1,y—1,z—1,w) =0 or in other words, 2x — 12+ 4y+6z+w = 0. Notethat this is a three dimensional plane because there are three free variables. Indeed, it is ofthe form w = 12 —4y—6z—2x.Example 22.9.2 The point (V3, 1,4) is on both the surfaces, z = x* + y” and z= 8—(x? + y’). Find the cosine of the angle between the two tangent planes at this point.Recall this is the same as the angle between two normal vectors. Of course there issome ambiguity here because if n is a normal vector, then so is —n and replacing n with—n in the formula for the cosine of the angle will change the sign. We agree to look forthe acute angle and its cosine rather than the obtuse angle. The normals are (2V3 52, —1); ye (2V3) +4-1 45and (23,2, 1). Therefore, the cosine of the angle desired is 4. =_}3.Example 22.9.3 The point (1 V3, 4) is on the surface z= x? +y*. Find the line perpen-dicular to the surface at this point.All that is needed is a direction vector for this line. The surface is the level surfacex* +y* —z=0. The normal to this surface is given by the gradient at this point. Thus thedesired line is (1, /3,4) +1 (2,2V3,—1).22.10 Exercises1. Find the gradient at the indicated point if f =(a) x*y +23, (1, 1,2) (d) sin(xy)+2°, (1,2, 1)(b) zsin (x?y) +2*, (1, 1,0)(c) uln(xt+y+z?+w), (x,y,z,w,u)= (1,1,1,1,2) (f) zIn(4+sin(xy)), (0,2, 1)(e) In(x+y")z2. Find the directional derivatives of f at the indicated point in the direction (5. 5, Js) .(a) x*y +23 at (1,1,1) (e) x” +zat (1,1,1).(b) zsin (x?y) +2** at (1, 1,0)(c) xy+z? + Lat (1,2,3) (f) sin(sin(w+y)) +z at the point(d) sin (xy) +z at (0, 1,1) (1,0, 1).3. Find the directional derivatives of the given function at the indicated point in theindicated direction.