Suppose you are climbing a hill whose shape is given by the equation z = 1300 -

Question

Suppose you are climbing a hill whose shape is given by the equation z = 1300 - 0.05x^2 - 0.01y^2, where x, y, and z are measured in meters, and you are standing at a point with coordinates (60,40,1266). The positive x-axis points east and the positive y-axis points north. If you walk due south, will you start to ascend or descend? At what rate? Vertical meters per horizontal meter if you walk northwest, will you answer to two stat to ascend or descend? At what rate? (round your answer to two decimal places. ) vertical meters per horizontal meter At what angle above the horizontal does the path in that direction begin? (Round your answer to two decimal places.)

Explanation / Answer

given z=1300-0.005x2-0.01y2

partial derivatives are

zx=-0.005*2x=-0.01x

zy=-0.01*2y=-0.02y

z =<-0.01x ,-0.02y>

at (60,40)

z =<-0.6,-0.8>

a)

if we walk south x is negative direction vector =<-1,0>

rate =<-0.6,-0.8>.<-1,0>

=(-0.6*-1)+(-0.8*0)

=0.6 vertical meters per horizontal meter

ascend

b)

if we walk north west

direction vector =<-1/2,1/2>

rate =<-0.6,-0.8>.<-1/2,1/2>

=(-0.6*-1/2)+(-0.8*1/2)

=-0.2/2

=-0.1414

descend

rate=0.14vertical meters per horizontal meter

c) slope largest in direction of z =<-0.6,-0.8>

rate of ascent =[(-0.6)2+(-0.8)2]

rate of ascent =1 vertical meters perhorizontal meter

tan x =1

x =tan-1(1)

x =45o

angle above horizontal =45o

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