Chapter 16 Partial Differentiation

16.5 Directional Derivatives 381origin, and as we get farther from the source, the temperature decreases. The gradient is−2T 0 x∇T = 〈(1+x 2 +y 2 +z 2 ) 2 +=−2T 0(1+x 2 +y 2 +z 2 ) 2〈x,y,z〉.−2T 0 x(1+x 2 +y 2 +z 2 ) 2 +−2T 0 x(1+x 2 +y 2 +z 2 ) 2〉The gradient points directly at the origin from the point (x,y,z)—by moving directlytoward the heat source, we increase the temperature as quickly as possible.EXAMPLE 16.5.5 Find the points on the surface defined by x 2 +2y 2 +3z 2 = 1 wherethe tangent plane is parallel to the plane defined by 3x−y +3z = 1.Two planes are parallel if their normals are parallel or anti-parallel, so we want tofind the points on the surface with normal parallel or anti-parallel to 〈3,−1,3〉. Let f =x 2 +2y 2 +3z 2 ; the gradient of f is normal to the level surface at every point, so we arelooking for a gradient parallel or anti-parallel to 〈3,−1,3〉. The gradient is 〈2x,4y,6z〉; ifit is parallel or anti-parallel to 〈3,−1,3〉, then〈2x,4y,6z〉 = k〈3,−1,3〉for some k. This means we need a solution to the equations2x = 3k 4y = −k 6z = 3kbut this is three equations in four unknowns—we need another equation. What we haven’tused so far is that the points we seek are on the surface x 2 + 2y 2 + 3z 2 = 1; this is thefourth equation. If we solve the first three equations for x, y, and z and substitute intothe fourth equation we get( ) 2 ( ) 2 ( 3k −k 3k1 = +2 +32 4 6( 9=4 + 216 + 3 )k 24= 258 k2(so k = ± 2√ 25 . The desired points are 3 √ √ √ ) (2 2 25 ,− 10 , and − 3√ √ √ )2 2 25 5 , 10 ,− . You5can see the surface and all three planes in the Java applet.) 2