Vector or Cross Product r The area A due to a vector product

b. Vector or Cross Product r The area A due to a vector product

A =× xy , (82) can be written in the form

A = | || | sinθ kxy , (83) r

rr where k denotes the unit vector in a plane normal to that containing the vectors x and y , and

θ the angle between these two vectors. The vector product yields an area vector in a direction normal to the plane containing the two vectors.

Consider the circular motion of an object around an origin in a plane. The force due to that object in the plane r r

F = iF x + jF y = iF cos θ + jF sin θ , (84) where r θ denotes the angle between the force and an arbitrary x–wise coordinate at any instant, r

and i and j denote unit vectors in the x– and y– directions, respectively. The torque exerted about the center

B =×= F r ( iF cos θ + jF sin ) ( θ × ix + jy ) = kyF ( cos θ + xF sin ) θ , (85) r r

where i × i = 0, i ×= j k , and j ×= i − k . When a screw is loosened from a flat surface by rotating it in the counter clockwise direction, it emerges outward normal to the surface, say, in the z–direction. To place the screw back into the surface, it must be rotated in the clockwise direction, i.e., it may be visualized as moving towards the origin of the z–direction. The rotation is caused by an applied torque that is a vector. If the term (F cos θ y – Fsinθ x) = 0 in Eq.(87), then there is no rotation around the

z–axis. In general, a force has three spatial components, i.e., r r

F = iF x + jF y + kF z , (86) and the torque is described by the relation

B =×= F r iFzFy ( y + z ) + jFxFz ( z + x ) + kFyFx ( x + y ) , i.e.,

(87) r

r there are rotational components in the x– and y– directions also. If r r r F and r are parallel to

r each other, e.g., F = iF x and r = ix , then r r r

B =×=0 F r .

c. Gradient of a Scalar Consider a one–dimensional heat transfer problem in which the temperature T is only

a function of one spatial coordinate, say, y, i.e., T = T(y). In this case T(y) is a point or scalar function of y, since its value is fixed once y is specified. In general, the gradient of T is defined as a function of one spatial coordinate, say, y, i.e., T = T(y). In this case T(y) is a point or scalar function of y, since its value is fixed once y is specified. In general, the gradient of T is defined as

∇ T = ( i ∂∂ / x + j ∂∂ / yk + ∂∂ / zT ) , (88) which for the one–dimensional problem assumes the form

rr ∇= T jT ∂ / ∂ y ,

The x–z plane contains isotherms, since T ≠T(x,z), and ∇ T is a vector along normal to the isotherms in the y–direction.

Consider, now, the temperature profile in an infinite cylindrical rod. Assume that the temperature is constant along the axial direction z, once a cross–sectional location (x,y) is specified, i.e., T=T(x,y), and T ≠T(z). Assume an axisymmetric problem for which the iso-

therms are circular in the x–y plane and form cylindrical surfaces. In this case, r ∇

(90) r

dT = ( r ∂T/∂x)dx + (∂T/∂y)dy = ∇ T·

ds ,

rr where , ∇ T = iT ∂ / ∂ x + jT ∂ / ∂ y . Therefore, r

(91) the gradient dT/ds varies, depending upon the direction of the gradient between any two iso- r

dT/ds = r ∇ T·

ds /ds, i.e.,

r therms. Along any circular isotherm r ∇ T· ds = 0 according to Eq. (93), since ∇ T and ds are

normal to each other. In general, if T=T(x,y,z) then isotherms form surfaces that lie in all three (x,y,z) coor- r dinates, and, at any location, ∇ T represents a vector that lies normal to a scalar surface on which T is constant.