Suppose you are climbing a hill whose shape is given by the equation z = 1400 -

Question

Suppose you are climbing a hill whose shape is given by the equation z = 1400 - 0.005x^2 - 0.01y^2, where x, y, and z are measured in meters, and you are standing at a point with coordinates (60, 40, 1366). 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 start to ascend or descend? At what rate? (Round your answer to two decimal places.) vertical meters per horizontal meter In which direction is the slope largest? What is the rate of ascent in that direction? 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

z=f(x,y)=1400-0.005x2-0.01y2

f =<-0.01x,-0.02y>

f at (60,40,1366)=<-0.01*60,-0.02*40>

f at (60,40,1366)=<-0.6,-0.8>

a)unit vector along south=<0,-1>

Du=<-0.6,-0.8>.<0,-1> =0+0.8=0.8 >0

==>ascend

rate =0.8

b)unit vector along nothwest=<-1/sqrt, 1/sqrt2>

Du=<-0.6,-0.8><-1/sqrt, 1/sqrt2>=0.6/sqrt2 -0.8/sqrt2=-0.2/sqrt2 =-0.1414

==>decend

rate=0.1414

c)solpe larget in direction of f=<-0.6,-0.8>

rate =|f|=sqrt[(-0.6)2+(-0.8)2]

rate =sqrt[(-0.6)2+(-0.8)2] =1

rate =tan x

tanx =1

==>x=45o

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