Suppose that you are climbing a hill whose shape is given by z = 426 - 0.1 x^2 -

Question

Suppose that you are climbing a hill whose shape is given by z = 426 - 0.1 x^2 - 0.01 y^2, and that you are at the point (30, 60, 300). In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest? If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)? The temperature at a point (x, y, z) is given by T(x, y, z) = 200e^-x^2 -y^2/4 - z^2/9, where T is measured in degrees Celsius and x, y, and z in meters. There are lots of places to make silly errors in this problem; just try to keep track of what needs to be a unit vector. Find the rate of change of the temperature at the point (1, -1, 1) in the direction toward the point (-1, -1, 5). In which direction (unit vector) does the temperature increase the fastest at (1, -1, 1)? What is the maximum rate of increase of T at (1, -1, 1)?

Explanation / Answer

1) given f(x,y)=z=426-0.1x2-0.01y2

f=(-0.2x)i +(-0.02y)j

at x =30, y =60

f=(-0.2*30)i +(-0.02*60)j

f=-6i -1.2j

unit vector =f/|f|

unit vector =(-6i -1.2j)/[(-6)2+(-1.2)2]

unit vector =(-6i -1.2j)/37.44

unit vector =-(6/37.44)i -(1.2/37.44)j

unit vector =-0.98i -0.196j

unit vector =<-0.98,-0.196>

slope =|f|

=tan-1(37.44)

=80.72o

initially climbing at 80.72o above horizontal

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