Mathematical Physics Vol 1

4.6 Examples

121

Solution Let r = x i + y j + z k be the position vector of point P ( x , y , z ) on this surface. Then d r = d x i + d y j + d z k belongs to the tangent plane of the surface in point P . Given that d f = d x + d y + d z = d ( c )= 0 or k · ( d x i + d y j + d z k )= 0 , that is, ∇ f · d r = 0, it follows that these two vectors (grad f and d r ) are mutually orthogonal. ∂ f ∂ x ∂ f ∂ y ∂ f ∂ z d f = ∂ f ∂ x i + ∂ f ∂ y j + ∂ f ∂ z

Problem36 Find the equation of the tangent to the plane 2 xz 2 − 3 xy − 4 x = 7 in point A ( 1 , − 1 , 2 ) .

Solution The equation of the plane through point A is ( r − r A ) · n = 0 , where n is the unit vector of the orthogonal vector of that plane in point A . Thus, in order to determine the equation of the plane we need the vector orthogonal to that plane. According to the previous example, vector ∇ f = ∇ ( 2 xz 2 − 3 xy − 4 x )=( 2 z 2 − 3 y − 4 ) i − 3 x j + 4 xz k , is the vector in the direction of the normal to the observed plane. The unit vector of this vector ( n = ∇ f / | ∇ f | ), in point A ( 1 , − 1 , 2 ) , is n = 7 i − 3 j + 8 k . The equation of the tangent plane, in this case, is [( x i + y j + z k ) − ( i − j + 2 k )] · ( 7 i − 3 j + 8 k ) √ 122 = 0 , that is 7 ( x − 1 ) − 3 ( y + 1 )+ 8 ( z − 2 )= 0 , 7 x − 3 y ++ 8 z − 26 = 0 . Problem37 Let T(x,y,z) and T ( x + ∆ x , y + ∆ y , z + ∆ z ) be temperature values in two close points P ( x , y , z ) and Q ( x + ∆ x , y + ∆ y , z + ∆ z ) . a) Give a physical interpretation of the value ∆ T ∆ s = T ( x + ∆ x , y + ∆ y , z + ∆ z ) − T ( x , y , z ) ∆ s