Suppose you are climbing a hill whose shape is given by the equation z-1200-0.00
ID: 2895130 • Letter: S
Question
Suppose you are climbing a hill whose shape is given by the equation z-1200-0.005x2-001r, where x, y, and z are measured in meters, and you are standing at a point with coordinates (120, 80, 1064). The positive x-axis points east and the positive y-axis points north (a) If you walk due south, will you start to ascend or descend? ascend descend At what rate? vertical meters per horizontal meter (b) If you walk northwest, will you start to ascend or descend? ascend O descend At what rate? (Round your answer to two decimal places.) vertical meters per horizontal meter (c) 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
given z=f(x,y)=1200-0.005x2-0.01y2
f=<-0.01x ,-0.02y>
at (x,y)=(120,80)
f=<-0.01*120 ,-0.02*80>
f=<-1.2 ,-1.6>
(a)
unit direction vector along south ,u=<0,-1>
directional derivative =f.u
directional derivative =<-1.2 ,-1.6>.<0,-1>
directional derivative =(-1.2*0)+(-1.6*-1)
directional derivative =1.6 >0
ascend
rate =1.6 vertical meters per horizontal meter
(b)
unit direction vector along northwest ,u=<-1/2,1/2>
directional derivative =f.u
directional derivative =<-1.2 ,-1.6>.<-1/2,1/2>
directional derivative =(-1.2*(-1/2))+(-1.6*(1/2))
directional derivative =-0.4/2
directional derivative =-0.28<0
descend
rate =0.28 vertical meters per horizontal meter
(c)
slope is lagest in direction of f
direction of lagest slope =<-1.2 ,-1.6>
rate of ascent =[(-1.2)2+(-1.6)2]
rate of ascent =2 vertical meters per horizontal meter
angle =tan-1(2)
angle=63.43o