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Questions about gradient

  • Thread starter don_anon25
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If someone could gtive me a general idea about how to approach these problems, I would be very grateful! Our class time was devoted to derivation rather than application.

1) Find the angle between the surfaces defined by r^2=9 and x+y+z^2=1 at the point (2,-2,1).

2) The height of a hill is given by z = 2xy - 3x^2 - 4y^2 - 18x +28y +12. x is the distance east and y is the distance north of the origin. i) Where is the top of the hill and how high is it? ii) What is the angle between a vector perpendicular to the hill and the z axis? I really have no idea where to start with this one!!!
 

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  • #2
HallsofIvy
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don_anon25 said:
If someone could gtive me a general idea about how to approach these problems, I would be very grateful! Our class time was devoted to derivation rather than application.

1) Find the angle between the surfaces defined by r^2=9 and x+y+z^2=1 at the point (2,-2,1).
Find the angle between the normal vectors. Since you labled this "gradients" I presume you know how to find those normal vectors!

2) The height of a hill is given by z = 2xy - 3x^2 - 4y^2 - 18x +28y +12. x is the distance east and y is the distance north of the origin. i) Where is the top of the hill and how high is it? ii) What is the angle between a vector perpendicular to the hill and the z axis? I really have no idea where to start with this one!!!
i) Do you know how to find the maximum of a function of two variables? Do you remember how to find the maximum of y= f(x) from Calculus I? (Find the derivative and set it equal to 0. Same here!)
ii) A vector along the z-axis is 0i+ 0j+ 0k. Do you know how to find a vector perpendicular (normal) to a surface? (Think "gradient vector". {3 dimensional, not 2!})
 
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I understand 2a now! But could you elaborate a little more on the first problem? How do I start?
 
  • #4
HallsofIvy
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1) Find the angle between the surfaces defined by r^2=9 and x+y+z^2=1 at the point (2,-2,1).
r2= 9? Is that in polar coordinates? The sphere of radius 3? Must be since (2, -2, 1)satisfy that. In that case, a normal vector is easy! Any radius is perpendicular to a sphere so 2i- 2j+ k is normal to the sphere at (2, -2, 1).

To find a normal vector to x+y-z2= 1, think of it as a level surface of the function F(x,y,z)= x+ y- z2. The gradient of F, i+ j- 2zk, is normal to that suface at each point. In particular, taking z= 1, i+ j- 2k is normal to that surface at (2, -2, 1). Now, what is the angle between the vectors 2i- 2j+ k and i+ j- k?
 

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