Question #219988

If the temperature on metal sheet is defined by f (x, y,z) = e2x

cos(y − 2z) then find the maximum

rate of change of the function at the (4, −2, 0) and the direction in which this maximum rate of change

of temperature occurs.


1
Expert's answer
2021-07-25T16:43:15-0400
f=fxi+fyj+fzk\nabla f=\dfrac{\partial f}{\partial x}\vec i +\dfrac{\partial f}{\partial y}\vec j +\dfrac{\partial f}{\partial z}\vec k




=(2e2xcos(y2z))i+(e2xsin(y2z))j+(2e2xsin(y2z))k=(2e^{2x}\cos(y-2z))\vec i +(-e^{2x}\sin(y-2z))\vec j +(2e^{2x}\sin(y-2z))\vec k



(4,2,0)=2e8cos(2)i+e8sin(2)j2e8sin(2)k\nabla|_{(4, −2, 0)}=2e^{8}\cos(2)\vec i +e^{8}\sin(2)\vec j -2e^{8}\sin(2)\vec k

The maximum rate of change is the magnitude of the gradient



(4,2,0)=(2e8cos(2))2+(e8sin(2))2+(2e8sin(2))2|\nabla|_{(4, −2, 0)}|=\sqrt{(2e^{8}\cos(2))^2+(e^{8}\sin(2))^2+(-2e^{8}\sin(2))^2}

=e84+sin2(2)=e^8\sqrt{4+\sin^2(2)}

The fastest increase is in the direction of the gradient


(4,2,0)=2e8cos(2)i+e8sin(2)j2e8sin(2)k\nabla|_{(4, −2, 0)}=2e^{8}\cos(2)\vec i +e^{8}\sin(2)\vec j -2e^{8}\sin(2)\vec k


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