Respuesta :
Answer:
a
The movement is ascension and the rate is 0.8
b
The movement is descending and the rate is 0.141
c
The direction of the largest slope is [tex]\Delta z (60,40)= (-0.6, -0.8)[/tex]
The rate of the rate of ascent is [tex]|\Delta z| = 1[/tex]
The angle above the horizontal is [tex]\theta = 45^o[/tex]
Step-by-step explanation:
From the question we are given an equation
[tex]z = 1000 - 0.005x^2 - 0.01y^2[/tex]
The directional derivative for the for walk along the y-axis (North and south ) is mathematically evaluated as
[tex]\frac{\delta z}{\delta y} = -0.02y[/tex]
The directional derivative for the for walk along the x-axis (East and west ) is mathematically evaluated as
[tex]\frac{\delta z}{\delta x} = -0.01x[/tex]
When the walk is due south directional directional derivative would be
[tex]D_{-j}z = - j[{\frac{\delta x}{\delta x} ,\frac{\delta z}{\delta y} ][/tex]
Note : the negative j signifies that the walk is due south (i.e negative y direction )
Since the movement we are considering is in the the y direction then the equation becomes
[tex]D_{-j}z = - \frac{\delta z}{\delta y}[/tex]
At point (60,40,966) =(x,y,z) the rate of the walk due south is
[tex]-[\frac{\delta z}{\delta y} ]_{[60,40,966]} = 0.02(40)[/tex]
[tex]= 0.8[/tex]
Since the values is positive then the movement is ascension and the is 0.8
The unit vector for northwest is obtained as
Considering the north with respect to the west
[tex]\r r = \frac{\= r}{|r|}[/tex]
[tex]\= r = I * j* k[/tex]
Here i = 1
j = 1
k = 0
[tex]\= r = 1[/tex]
[tex]|r | =\sqrt{1^2 + 1^2 + 0^2}[/tex]
= [tex]\sqrt{2}[/tex]
[tex]\r r = \frac{1}{\sqrt{2} }[/tex]
Considering the west with respect to the north
[tex]\r r = \frac{\= r}{|r|}[/tex]
[tex]\= r = I * j* k[/tex]
Here i = - 1
j = 1
k = 0
[tex]\= r = 1[/tex]
[tex]|r | =\sqrt{1^2 + 1^2 + 0^2}[/tex]
= [tex]\sqrt{2}[/tex]
[tex]\r r = - \frac{1}{\sqrt{2} }[/tex]
Thus the unit vector for north is
[tex](i, j , k ) =\r a= (-\frac{1}{\sqrt{2} } , \frac{1}{\sqrt{2} } ,0 )[/tex]
The directional derivative is given as
[tex]D_{\r a} z = [\frac{\delta z}{\delta x} , \frac{\delta z}{\delta y} ] \cdot [-\frac{1}{\sqrt{2} } , \frac{1}{\sqrt{2} } ][/tex]
Note the equation above is different from the first one because in the first one we were only considering the south direction but here we are considering the north and the west direct so x and y are involved
[tex]= [-0.01x, 0.02y] \cdot [-\frac{1}{\sqrt{2} } , \frac{1}{\sqrt{2} } ][/tex]
[tex]= [\frac{0.01}{\sqrt{2} }x , - \frac{0.02}{\sqrt{2} } y][/tex]
Now at the point (60,40,966)
[tex]D_{\r a} z (60,40,966) = [\frac{0.01}{\sqrt{2} } (60) , -\frac{0.02}{\sqrt{2} } (40) ][/tex]
[tex]= 0.3\sqrt{2} - 0.4 \sqrt{2}[/tex]
[tex]= -0.1\sqrt{2}[/tex]
[tex]= -0.141[/tex]
Since the value is negative then the walk is descending and the rate is 0.141
The gradient vector at the point (60,40,966). is mathematically evaluated as
[tex]\Delta z (60,40)= [(-0.01 (60) , -0.02(40))][/tex]
[tex]\Delta z (60,40)= (-0.6, -0.8)[/tex]
The rate of ascent is mathematically evaluated as
[tex]|\Delta z| = \sqrt{(-0.6)^2 + (-0.8)^2}[/tex]
[tex]= \sqrt{0.36 + 0.64}[/tex]
[tex]= \sqrt{1}[/tex]
[tex]= 1[/tex]
Therefore the direction in which the slope is largest is evaluated as
[tex]tan \theta = 1[/tex]
[tex]\theta = tan ^{-1} [1][/tex]
[tex]\theta = 45^o[/tex]
