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Plotting a 3d Direction of maximum incre...

peter ursa 5
3d plotting multivariate-calculus
April 14 2018
2 2

Probably I am overthinking this but in any case I have hit a road block. If anyone can point me in the right direction I woould greatly appreciate it! Peter_Ursa.mw
 

restart

with(Student[MultivariateCalculus])

with(VectorCalculus) 

with(plots)

2. The surface of a mountain is modeled by the equation "h(x,y)=400-2 x^(2)-y^(2)."  A mountain climber is at the point 10, 10.  

"h(x,y):=400-2 x^(2)-y^(2)"

proc (x, y) options operator, arrow; VectorCalculus:-`+`(VectorCalculus:-`+`(400, VectorCalculus:-`-`(VectorCalculus:-`*`(2, x^2))), VectorCalculus:-`-`(y^2)) end proc

 (1)

p := `<,>`(10, 10)

Vector[column](%id = 18446746427608139406)

 (2)

a) In what direction should the climber move in order to ascend at the greatest rate?

 

 

H := Gradient(h(x, y), [x, y])

Vector[column](%id = 18446746427608141086)

(1.1)

eval(H, [x = 10, y = 10])

Vector[column](%id = 18446746427608133630)

(1.2)

``NULL

 

 

b) Determine a rectangular equation for the path of the mountain climber.

 

 

restart

  `<,>`(10, 10)+`<,>`(-40*t, -20*t)

Vector[column](%id = 18446746427574846638)

(2.1)

 

 

sys := {10-40*t = x, 10-20*t = y}

{10-40*t = x, 10-20*t = y}

(2.2)

solve(sys, {x, y})

{x = 10-40*t, y = 10-20*t}

(2.3)

t := solve({x = 10-40*t, y = 10-20*t}[1], t)

1/4-(1/40)*x

(2.4)

10-20*t

5+(1/2)*x

(2.5)

````

``

NULL

c) Plot the path of the climber along a 2-D contour diagram of the surface of the mountain.

``

 

 

restart

with(plots):

p1 := contourplot(-2*x^2-y^2+400, x = -15 .. 15, y = -15 .. 15)

p2 := plot(5+(1/2)*x, x = 0 .. 10, y = 5 .. .10)

display(p1, p2)

 

``

d) Show a 3-D view of the path of the climber along the surface of the mountain.

p3 := plot3d(-2*x^2-y^2+400, x = -15 .. 15, y = -15 .. 15)

p4 := plot3d(5+(1/2)*x)

display(p3, p4)

  

e) Animate the path along the surface of the mountain.NULL

``


 

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