2011 solution A5 - University of Toronto Scarborough...

This preview shows page 1 - 2 out of 5 pages.

University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT B41H 2011/2012 Solutions #5 1. (a) To find the equation of the tangent plane to the surface given by x 3 z + x 2 y 2 + sin( yz ) = - 3 at ( - 1 , 0 , 3) we regard the surface as a level set of g ( x, y, z ) = x 3 z + x 2 y 2 + sin( yz ) + 3 = 0. Now g = ( 3 x 2 z + 2 xy 2 , 2 x 2 y + z cos( yz ) , x 3 + y cos( yz ) ) ; hence, a normal vector to the surface at ( - 1 , 0 , 3) is g ( - 1 , 0 , 3) = ( 9 , 3 , - 1 ) . The equation of the tangent plane is given by g ( - 1 , 0 , 3) · ( ( x, y, z ) - ( - 1 , 0 , 3) ) = 0 ⇐⇒ (9 , 3 , - 1) · ( x +1 , y, z - 3) = 0 ⇐⇒ 9( x +1)+3 y - ( z - 3) = 0 which we can rewrite as 9 x + 3 y - z = - 12. (b) Since the line is orthogonal to the surface, a direction vector for the line is (9 , 3 , - 1), a normal vector to the surface at ( - 1 , 0 , 3). Hence a parametric de- scription of the line is ( - 1 , 0 , 3) + t (9 , 3 , - 1), t R . If ( x, y, z ) is a point on this line we have ( x, y, z ) = ( - 1 , 0 , 3) + t (9 , 3 , - 1) = ( - 1 + 9 t, 3 t, 3 - t ) for some t . Solving for t in each component we get ( t = ) x + 1 9 = y 3 = 3 - z , which is the required rectangular description of the line. The line meets the xy –plane when the z –component is 0. Hence we have 3 - t = 0 or t = 3. The required point of intersection is (26 , 9 , 0). 2. For f ( x, y ) = x y - y x , we have ∂f ∂x = y x x y - y x ln y and ∂f ∂y = x y ln x - x y y x . At a point ( a, b ), a normal vector n ( a,b ) for the tangent plane of f is parenleftbigg ∂f ∂x ( a, b ) , ∂f ∂y ( a, b ) , - 1 parenrightbigg . Hence, n (1 , 2) = (2(1 - ln 2) , - 1 , - 1) and n (2 , 1) = (1 , 2(ln 2 - 1) , - 1). The line of intersection lies in both tangent planes so its direction vector v is orthogonal to both n (1 , 2) and n (2 , 1) . Thus v = n (1 , 2) × n (2 , 1) = (2(1 - ln 2) , - 1 , - 1) × (1 , 2(ln 2 - 1) , - 1) = (2 ln 2 - 1 , 1 - 2 ln 2 , 1 - 4 (ln 2 - 1) 2 ). 3. We know that the direction of maximum increase is the gradient direction; so the tem- perature will increase fastest in direction T = - 80 (1 + x 2 + 2 y 2 + 3 z 2 ) 2 parenleftbigg 2 x, 4 y, 6 z parenrightbigg = - 160 (1 + x 2 + 2 y 2 + 3 z 2 ) 2 parenleftbigg x, 2 y, 3 z parenrightbigg . At the point (1 , 1 , - 2), the temperature will in- crease fastest in direction T (1 , 1 , - 2) = - 160 256 parenleftbigg 1 , 2 , - 6 parenrightbigg = - 5 8 parenleftbigg 1 , 2 , - 6 parenrightbigg . The maximum rate of increase is the magnitude of the gradient, bardbl∇ T (1 , 1 , - 2) bardbl = 5 8 vextenddouble vextenddouble vextenddouble vextenddouble parenleftbigg 1 , 2 , - 6 parenrightbiggvextenddouble vextenddouble vextenddouble vextenddouble = 5 41 8 . The maximum rate of increase of temperature

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture