Volume of tetrahedra formed from coordinate and tangent planes

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SUMMARY

The volume of the tetrahedron T formed by the coordinate planes and the tangent plane P to the surface defined by xyz=a³ at the point (r,s,t) is independent of the coordinates (r,s,t). The equation of the tangent plane is given by (x/r) + (y/s) + (z/t) = 3, which intersects the coordinate planes at x=3r, y=3s, and z=3t. The calculated volume of T is confirmed to be 9a³/2, demonstrating that it does not depend on the specific point (r,s,t) chosen on the surface.

PREREQUISITES
  • Understanding of tangent planes in multivariable calculus
  • Familiarity with the volume formula for tetrahedra
  • Knowledge of the surface equation xyz=a³
  • Basic skills in algebraic manipulation and geometry
NEXT STEPS
  • Explore the geometric properties of tangent planes in three-dimensional space
  • Study the derivation of the volume formula for tetrahedra
  • Investigate the implications of surface equations in multivariable calculus
  • Learn about the relationship between intercepts and volumes in geometric shapes
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Students and professionals in mathematics, particularly those studying calculus and geometry, as well as educators looking to enhance their understanding of geometric properties related to tangent planes and volumes.

Juggler123
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I have that P is the tangent plane to the surface xyz=a^{3} at the point (r,s,t). I need to show that the volume of the tetrahedron, T, formed by the coordinate planes and the tangent plane to P is indepedent of the point (r,s,t).

I have found that P is;

\frac{x}{r} + \frac{y}{s} + \frac{z}{t} = 3

A know that the volume of a tetrahedron is giving by 1/3(area of base \times height)

But I just can't picture what this looks like, as far as I can see the volume of T has to be dependent of the point (r,s,t).

Any help anyone could give would be great! Thanks.
 
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Juggler123 said:
I have that P is the tangent plane to the surface xyz=a^{3} at the point (r,s,t). I need to show that the volume of the tetrahedron, T, formed by the coordinate planes and the tangent plane to P is indepedent of the point (r,s,t).

I have found that P is;

\frac{x}{r} + \frac{y}{s} + \frac{z}{t} = 3

A know that the volume of a tetrahedron is giving by 1/3(area of base \times height)

But I just can't picture what this looks like, as far as I can see the volume of T has to be dependent of the point (r,s,t).

Any help anyone could give would be great! Thanks.

You would expect the volume to depend on r,s, and t. But if you work it out you will find that it doesn't. Just write the equation of the tangent plane, find its intercepts and the corresponding volume. Here is a picture with a = 1 of one of the tangent planes to help you visualize it:
pyramid.jpg
 
Thanks for that, think I might be starting to understand this a little bit more now.

Right I have that

\frac{x}{r} + \frac{y}{s} + \frac{z}{t} = 3

and so this plane intersects the coordinate planes at x=3r, y=3s and z=3t but all of these points you know that xyz=a^{3} so is right to then say that the intersects occur at x=3a^{3}, y=3a^{3} and z=3a^{3}.

Hence the volume of T is given by \frac{9a^{9}}{2}
 
Almost right. But check the plane intercepts again. For example, 3r doesn't equal 3a3.
 
Right think I've got it this time.

The intercepts are at x=3r, y=3s and z=3t.

Now 3r=\frac{3a^{3}}{st}, 3s=\frac{3a^{3}}{rt} and 3z=3r=\frac{3a^{3}}{rs}

Hence the volume of T is given by \frac{9a^{3}}{2}
 
Sorry that should say 3t=\frac{3a^{3}}{rs}
 
Juggler123 said:
Hence the volume of T is given by \frac{9a^{3}}{2}

Looks good.
 
I get the feeling that this can be proved with only geometric considerations with eyes closed...
 

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