8|Multivariable Calculus 8 As a check, the dimensions are correct (are they?). At time zero, this vanishes, and that’s what I expect because at the beginning of the motion you ….. restarting to move in the direction perpendicular to the direction in which the temperature is changing. The farther you go, the more nearly parallel to the direction of the radius you’removing. If you are moving exactly parallel to the radius, this time-derivative is easier to calculate; it’s then almost a problem in a single variable. dT dt dT…

8|Multivariable Calculus 8 As a check, the dimensions are correct (are they?). At time zero, this vanishes, and that’s what I expect because at the beginning of the motion you’restarting to move in the direction perpendicular to the direction in which the temperature is changing. The farther you go, the more nearly parallel to the direction of the radius you’removing. If you are moving exactly parallel to the radius, this time-derivative is easier to calculate; it’sthenalmosta problem in a single variable. dT dt dT dr dr dt 2 T 1 r a2 v 0 2 T 1 V 0 t a2 v 0 So the approximate and the exact calculation agree. In fact they agree so well that you should try to ndoutifthisisa lucky coincidence or if there some special aspect of the problem that you might have seen from the beginning and that would have made the whole thing much simpler. 8.5 Gradient The equation ( 8.13 ) for thedierential has another geometric interpretation. Fora function such asf ( x;y ) = x2 +4 y2, the equations representing constant values off describe curves in thex-y plane. In this example, they are ellipses. If you start from any xedpointinthe plane and start to move away from it, the rate at which the value off changes will depend on the direction in which you move. If you move along the curvedened by f = constant then fwon’tchangeat all. If you move perpendicular to that direction then fmaychangea lot. The gradient off at a point is the vector pointing in the direction in whichfis increasing most rapidly and the component of the gradient along that direction is the derivative off with respect to the distance in that direction. To relate this to the partial derivatives that we’vebeenusing, and to understand how to compute and touse the gradient, return to Eq. ( 8.13 ) and write it in vector form. Use the common notation for the basis: ^ xand ^ y. Then let d~r = dx ^ x + dy ^ y and ~ G= @f @x y ^ x + @f @y x ^ y (8 : 15) The equation for thedierential is now df = df ( x;y;dx;dy ) = ~ G. d~r (8 : 16) ~ G d~r Because you know the properties of the dot product, you know that this is Gdr cos and it is largest when the directions ofd~rand of ~ Garethesame. It’szerowhentheyare perpendicular. You also know that df is zero whend~risinthe direction along the curve where fisconstant.

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