Math 209, Fall 2009 Homework 3

Transcription

1 Math 209, Fall 2009 Homework 3 () Find equations of the tangent plane and the normal line to the given surface at the specified point: x 2 + 2y 2 3z 2 = 3, P (2,, ). Solution Using implicit differentiation we obtain: At P (2,, ) : 2x 6zz x = 0, 4y 6zz y = z x = 0, 4 6z y = 0, that is, z x = 2/3, z y = 2/3. Hence, an equation of the tangent plane is z = 2(x 2) 2 (y + ), or 2x 2y 3z 3 = 0. Parametric equations of 3 3 the normal line: x = 2 + 2t, y = 2 t, z = t. 3 3 (2) Find df dt, if f(x, y, z) = 2yex ln z, where Solution By the Chain Rule: x = ln(t 2 + ), y = arctan t, z = e t. df dt = f dx x dt + f dy y dt + f dz z dt = = 2ye x 2t t ex + t 2 z et = = 4t arctan t +. (3) If f and g are twice differentiable functions of a single variable, show that the function u(x, y) = xf(x + y) + yg(x + y) satisfies the equation u xx 2u xy + u yy = 0. Solution Let s = x + y. Hence, u = xf(s) + yg(s). Using the Chain Rule and the fact that s = s =, we get: x y u x = f(s) + x df s ds x + y dg s ds x = f(s) + xf (s) + yg (s), u y = +x df s ds y dg s + g(s) + y ds y = xf (s) + g(s) + yg (s), u xx = df s ds x + f (s) + x d[f (s)] s ds x + y d[g (s)] s ds x = 2f (s) + xf (s) + yg (s),

2 u yy = x d[f (s)] s ds y + dg s ds y + g (s) + y d[g (s)] s ds y = xf (s) + 2g (s) + yg (s), u xy = df s ds y +xd[f (s)] s ds y +g (s)+y d[g (s)] s ds y = f (s)+xf (s)+g (s)+yg (s). Hence, u xx 2u xy +u yy = 2f (s)+xf (s)+yg (s) 2[f (s)+xf (s)+g (s)+yg (s)]+ +xf (s) + 2g (s) + yg (s) = 0. (4) If u = e a x +a 2 x 2 + +a nx n, where a 2 + a a 2 n =, show that 2 u + 2 u u x 2 x 2 2 x 2 n = u. Solution: Let s = a x + a 2 x a n x n. Then u = e s. Hence, u = du s = e s a i = a i u, i n. x i ds x i Differentiating one more time with respect to x i, we get: 2 u x 2 i = x i ( u x i ) = a i u x i = a 2 i u. Finally, 2 u + 2 u u x 2 x 2 2 x 2 n = a 2 u + a 2 2u + + a 2 nu = (a 2 + a a 2 n)u = u. (5) If z = f(u, v), where u = xy and v = y, and f has continuous partial x derivatives of the second order, show that: x 2 2 z x 2 z 2 y2 y = 4uv 2 z z + 2v 2 u v v ( ). Solution: Using the Chain Rule: z x = z u u x + z v v x = z u y + z v ( y x ), 2 z y = z u u y + z v v y = z u x + z v x.

3 Differentiating one more time: 2 z x = 2 x ( z x ) = y x ( z u )+ 2y x 3 z v +( y x 2 ) x ( z v ) = y[ 2 z u 2 y+ 2 z v u ( y x 2 )]+ + 2y z x 3 v + ( y x )[ 2 z 2 u v y + 2 z v ( y 2y z )] = 2 x2 x 3 v + 2 z y2 u 2y2 2 z 2 x 2 u v + y2 2 z x 4 v. 2 Similarly: 2 z y = 2 y ( z y ) = x y ( z u ) + x Substituting 2 z x 2 x 2 2 z x 2 y2 2 z y 2 = 2y x y ( z v ) = x[ 2 z u x + 2 z 2 v u x ]+ + x [ 2 z u v x + 2 z v 2 x ] = x2 2 z u z u v + x 2 2 z v 2. and 2 z y 2 z v +x2 y 2 2 z (6) Let w = x 2 + y 2 z 2, where into the left hand side of the equation, we get: 2 z 2 z u 2 2y2 u v +y2 x 2 v 2 y2 x 2 2 z 2 z 2 z u 2 2y2 u v y2 x 2 v = 2 = 4uv 2 z z + 2v u v v ( ). x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. Use the Chain Rule to show that w satisfies the equation: Solution: Using the Chain Rule: w θ = w x x θ + w y y θ + w z z θ w ρ = w x x ρ + w y y ρ + w z z ρ Therefore, w θ w 2ρ ρ = cos(2φ). = 2x( ρ sin φ sin θ)+2y(ρ sin φ cos θ) 2z 0 = 0. = 2x sin φ cos θ+2y sin φ sin θ 2z cos θ = = = 2ρ(sin 2 φ cos 2 φ). w θ w 2ρ ρ = 0 2ρ 2ρ(sin2 φ cos 2 φ) = cos 2 φ sin 2 φ = cos(2φ).

4 (7) Find the directional derivative of the function at the point P in the direction of the vector v : (a) f(x, y, z) = z 3 x 2 y, P (, 6, 2), v = (3, 4, 2); (b) g(x, y, z) = xe yz + xye z, P ( 2,, ), v = i 2 j + 3 k. (a) Solution: f(x, y, z) = ( f x, f y, f z ) = ( 2xy, x2, 3z 2 ) = f(, 6, 2) = ( 2,, 2). Since the unit vector in the direction of the vector v is u = ( 3, 4, 2), we get: D uf(, 6, 2) = f(, 6, 2) u = 8. v = v (b) g(x, y, z) = ( g x, g y, g z ) = (eyz + ye z, xze yz + xe z, xy(e yz + e z )) = g( 2,, ) = (2e, 4e, 4e). Since the unit vector in the direction of the vector v is u = 4 (, 2, 3), we get: v = v D u G( 2,, ) = G( 2,, ) u = e 4. 7 (8) Consider the function f(x, y, z) = xy + z.. Find the rate of change of f at the point P (0,, 2) in the direction of the vector v = i + j k. 2. In what direction does f increase most rapidly at P? Find the maximum rate of increase. 3. Find all directions ( if any) in which the rate of change of f at P is 0?

5 Solution:. f(x, y, z) = ( f x, f y, f z ) = (y, x, ) = f(0,, 2) = (, 0, ). Since the unit vector in the direction of the vector v is u = v = v ( 3, 3, 3 ), the rate of change of f at P (0,, 2) in the direction of v is : D u f(0,, 2) = f(0,, 2) u = The function f increases most rapidly at P in the direction of f(0,, 2) = (, 0, ) 3. Let w = (a, b, c) be an arbitrary vector. Then and a u = ( a2 + b 2 + c, 2 b a2 + b 2 + c 2, c a2 + b 2 + c 2 ), D u f(0,, 2) = 0 = f(0,, 2) u = 0 = a + c = 0, b R. Hence, w = (a, b, a), a, b R. (9) The temperature T at some point in a metal ball is inversely proportional to the distance between the point and the center of the ball, which we take to be the origin. The temperature et the point P (, 2, 2) is 20. (a) Find the rate of change of T at P in the direction toward the point Q(2,, 3). (b) Show that at any point in the ball the direction of the greatest increase in temperature is given by a vector that points toward the origin. Solution: Let T (x, y, z) denotes the temperature at some point (x, y, z) in the ball. Then K T (x, y, z) = x2 + y 2 + z, 2 where K is a constant. T x, y, z) = x. 2 +y 2 +z 2 Since T (, 2, 2) = 20, we get K =. Thus,

6 (a) The unit vector in the direction toward the point Q(2,, 3) is: (b) Since (0) u = P Q = (,, ). P Q 3 Also, T (x, y, z) = (x, y, z) and T (, 2, 2) = 40 (, 2, 2). (x 2 +y 2 +z 2 ) 3/2 3 Hence, the rate of change of T in the direction of u: D u T (, 2, 2) = T (, 2, 2) u = = T (x, y, z) = (x, y, z) = ( x, y, z) = (x 2 + y 2 + z 2 ) 3/2 d3 P O, d 3 where d denotes the distance between (x, y, z) and (0, 0, 0), the direction of the greatest increase of T is toward the origin. a) Find the point(s) on the ellipsoid x 2 + 2y 2 + 3z 2 =, where the tangent plane is parallel to the plane 3x y + 3z =. b) The plane y + z = 3 intersects the cylinder x 2 + y 2 = 5 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point (, 2, ). Solution: (a) Let F (x, y, z) = x 2 + 2y 2 + 3z 2. Then the ellipsoid is a level surface of F : (S) F (x, y, z) x 2 + 2y 2 + 3z 2 =. Let P (x 0, y 0, z 0 ) be a point on the ellipsoid we are looking for. Then the normal vector N = F (x 0, y 0, z 0 ) = (2x 0, 4y 0, 6z 0 ) of the tangent plane to the surface at P must be parallel to the normal vector n = (3,, 3) of the given plane, that is N = t n. This implies x 0 = 2t, y 3 0 = t, z 4 0 = t. 2 On the other hand, the point P (x 0, y 0, z 0 ) is on the surface (S), that is, x y z0 2 =. This implies t = 2 2, that is, there are or t = two solutions: t = 2 2 gives the point P 5 ( 3 2, 2, 2 ) and t = 2 2, gives P ( 3 2 5, 2 0, 2 5 ).

7 (b) Let (S ) F (x, y, z) x 2 + y 2 = 5. (S 2 ) G(x, y, z) y + z = 3. Let v denotes a direction vector of the tangent line to the curve of intersection of (S ) and (S 2 ) at the point (, 2, ). Then v is orthogonal to both normal vectors F (, 2, ) and G(, 2, ) to surfaces (S ) and (S 2 ) respectively. Hence, we can take v = F (, 2, ) G(, 2, ) = (2, 4, 0) (0,, ) = (4, 2, 2). Finally, parametric equations of the tangent line to the ellipse at the point (, 2, ) are: x = + 4t, y 2t + 2 z = 2t +. () For what value of the constant α is the sphere: x 2 + y 2 + z 2 αy 2z + α 2 = 0, orthogonal to the paraboloid 3x 2 + 2y 2 2z = at the point P (,, 2). Solution: Let (S ) F (x, y, z) x 2 + y 2 + z 2 αy 2z + α 2 = 0. (S 2 ) G(x, y, z) 3x 2 + 2y 2 2z =. Note that (S ) is orthogonal to (S 2 ) at the point P if and only if tangent planes to (S ) and (S 2 ) at P are orthogonal, equivalently, if and only if the normal vectors F (,, 2) and G(,, 2) are orthogonal. Hence, F (,, 2) G(,, 2) = 0. Since F (,, 2) = (2, 2 α, 2) and G(,, 2) = (6, 4, 2) we get α = 4.

8 (2) Show that the sum of the x, y, and z intercepts of any tangent plane to the surface x + y + z = c, (c a constant), is a constant. Solution: Let (S) F (x, y, z) x + y + z = c. Let P (x 0, y 0, z 0 ) be any point on the surface (S). Then x 0 > 0, y 0 > 0, z 0 > 0. The normal vector for the tangent plane to (S) at P is given by: N = F (x 0, y 0, z 0 ) = ( 2, x 0 Hence, the equation of the tangent plane is: or 2 y 0, 2 z 0 ). 2 (x x 0 ) + x 0 2 (y y 0 ) + y 0 2 (z z 0 ) = 0, z 0 or 2 x + x 0 2 y + y 0 2 z = z 0 2 x 0 x + 2 y 0 y + 2 z 0 z = x0 + y 0 + z 0, 2 c 2. Taking y = z = 0, we get x intercept, x int = c x 0. Similarly, x = z = 0 gives y intercept, y int = c y 0 and x = y = 0 gives z intercept, z int = c z0. The sum of the intercepts: x int + y int + z int = c x 0 + c y 0 + c z 0 = c c = c = const., that is, the sum of intercepts does not depend on the choice of the point on the surface.