GRE #34,53

Forum for the GRE subject test in mathematics.
aas56
Posts: 29
Joined: Sun Jun 28, 2009 3:42 pm

GRE #34,53

Postby aas56 » Wed Sep 23, 2009 10:25 am

Can any one give me hints for sloving these two questions.
n34
the minimal distance between any point on the sphere (x-2)^2+(y-1)^2+(z-3)^2=1 and any point on the sphere (x-3)^2+(y-2)^2+(z-4)^2=4 is
a)0 B)4 C)root(27) D)2(root(2)+1) E)3(root(3)-1)

answer is E

n53
what is the minimal value of the expression x+4z as a function defined on R^3 subject to the constraint x^2+y^2+z^2<= 2
A)0 B)-2 C)-root(34) D)-root(35) E)-5root(2)

answer is C

EHart
Posts: 6
Joined: Mon Sep 14, 2009 1:38 pm

Re: GRE #34,53

Postby EHart » Wed Sep 23, 2009 1:38 pm

For 34,

The center of the first sphere is (2,1,3) and the center of the second sphere is (3,2,4); find the distance between the two centers and subtract the sum of their radii to get the smallest distance between any two points on their respective surfaces. The distance between the centers can be found in a number of ways (I used right triangles, IIRC) and working from memory I believe the distance between their centers is 3*sqrt(3), and their radii are 2 and 1, so 3*sqrt(3)-(2+1) = 3(sqrt(3)-1).

prong
Posts: 24
Joined: Thu Sep 24, 2009 12:17 am

Re: GRE #34,53

Postby prong » Thu Sep 24, 2009 1:54 am

For 53, the expression doesn't depend on y at all; you maximize the possible configurations of x,z available to you when y = 0, so your condition can just be x^2 + z^2 <=2.

You're trying to find the minimum of f(x,y,z) = x + 4z. Extrema occur either where the derivative is 0 or on the boundary. In this case the derivative is grad(f) = (df/dx, df/dy, df/dz) [really those df/dx_i's are partial derivatives but I don't want to bother to pull out some LaTeX].

f is nice so grad(f) is nice, grad(f) = (1,0,4). this is never 0 so any extrema occur at the boundary of our region, i.e. x^2 + z^2 = 2. therefore x = -sqrt(2-z^2).

now you want to minimize x+4z = -sqrt(2-z^2) + 4z. this is the same as d/dz(-sqrt(2-z^2) + 4z) = 0, which is the same as -z/sqrt(2-z^2) + 4 = 0.

solving this, z/sqrt(2-z^2) = 4 so z^2/(2-z^2) = 16 and z^2 = 32-16z^2 gives us 17z^2 = 32. thus z = sqrt(32/17)

crap, did i screw up somewhere?

thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am

Re: GRE #34,53

Postby thmsrhn » Fri Mar 26, 2010 3:29 pm

Ehart,

The circles are centered at (2,1,3) and (-3,2,4), not (3,2,4).

Therefore the distance between the two centers is calculated using the distance formula viz., ((2+3)^2 + (1-2)^2 + (3-4)^2)^.5 which equates to the root of 26, which is roughly 5.

Now, the radii of the two spheres are 1 and 4 units, so the minimal distance should be (the root of 26 ) -(1+4) which is a very insignificant number.

The answer I am getting is A, not E.

jerseyj282
Posts: 4
Joined: Mon Apr 05, 2010 3:35 pm

Re: GRE #34,53

Postby jerseyj282 » Mon Apr 05, 2010 3:59 pm

thmsrhn wrote:Ehart,

The circles are centered at (2,1,3) and (-3,2,4), not (3,2,4).

Therefore the distance between the two centers is calculated using the distance formula viz., ((2+3)^2 + (1-2)^2 + (3-4)^2)^.5 which equates to the root of 26, which is roughly 5.

Now, the radii of the two spheres are 1 and 4 units, so the minimal distance should be (the root of 26 ) -(1+4) which is a very insignificant number.

The answer I am getting is A, not E.


It looks like you have the distance formula set up correctly. You should get root 27 when you evaluate that. The radii are 1 and 2 respectively (x^2+y^2+z^2=r^2, not r). The minimal distance is then 27^(1/2)-(1+2) which reduces to E

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

Re: GRE #34,53

Postby enork » Wed Apr 07, 2010 12:03 am

For 53, as prong said y=0. The min of x+4z is on the circle x^2 + z^2 = 2 since it clearly doesn't have a min inside. In fact it'll be the point with the farthest perpendicular distance from the line x+4z=0. Thinking about the geometry here, that point will be on the line 4x=z. Solving the two equations, x = -\sqrt{2/17} and the min is 17x = -\sqrt{34}.

prong's method works too and his/her math is correct despite his/her doubts.

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

Re: GRE #34,53

Postby enork » Wed Apr 07, 2010 8:28 pm

Oh, here is totally a better way to consider this problem:
Given the vector u = (1,0,4), find the minimum possible value of v.u where v is a vector satisfying |v|^2 <= 2.
The minimum of v.u is obtained when v points in the opposite direction from u and has the maximum possible magnitude. That value is -|v||u| = -sqrt(2)sqrt(17).

lecherme
Posts: 1
Joined: Sun Apr 18, 2010 4:15 am

Re: GRE #34,53

Postby lecherme » Sun Apr 18, 2010 10:02 pm

the centres of two spheres are (2,1,3)and(3,2,4),the distance between these centres are ((2-3)^2+(1-2)^2+(3-4)^2)^(1/2)=root(3)<the plus of radiis of the spheres=1+2=3,so the relationship of these two spheres are intersectant,that is,if keep the spheres in one coordinate system of 3 dimension,there must be at least a meeting point of the spheres.so the distance between the two spheres is 0.

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: GRE #34,53

Postby DDswife » Thu Aug 28, 2014 2:07 pm

lecherme wrote:the centres of two spheres are (2,1,3)and(3,2,4),the distance between these centres are ((2-3)^2+(1-2)^2+(3-4)^2)^(1/2)=root(3)<the plus of radiis of the spheres=1+2=3,so the relationship of these two spheres are intersectant,that is,if keep the spheres in one coordinate system of 3 dimension,there must be at least a meeting point of the spheres.so the distance between the two spheres is 0.


You are right, but the problem was miscopied.




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