Related Rates - Volume of a tetrahedron

[shaitan82]

Hey i've been working on this question and i keep on getting stuck on somehow trying to find a way to write height in terms of the side length of the base triangle. Any ideas or solutions would be greatly appreciated!

A crystal is growing in the shape of a tetrahedron in such a way that its base is an equilateral triangle and its slant faces are isosceles triangles. The growth rate of each base edge is 1mm/week and the growth rate of the vertical height is 1.5 mm/week.

Find the grwoth rate of the volume at the instant when the area of the triangular base is 100srqrt(3) mm^2.




Thanks!

 

[soroban]

Hello, shaitan82!

A crystal is growing in the shape of a tetrahedron in such a way that
its base is an equilateral triangle and its slant faces are isosceles triangles.
The growth rate of each base edge is 1mm/week and the growth rate of the vertical height is 1.5 mm/week.

Find the growth rate of the volume at the instant when the area of the triangular base is 100sqrt(3) mm^2.

I made an assumption . . .
The crystal begins at "size zero".
. . The base-triangle has side 0 and its height is 0.
. . After that the side (x) grows at 1 mm/wk and the height (h) at 1.5 mm/wk.
. . That is, at <u>any</u> time, h will be 1.5 times x.

The area of an equilateral triangle with side x is given by: . B .= .(sqrt{3})/4)x²

The volume of a pyramid is: . V .= .(1/3)Bh
. . . where B = area of base, h = height

The volume of this tetrahedron is: . V .= .(1/3)[(sqrt{3})/4]x²h .= .(\sqrt{3})/12)x²h

Differentiate with respect to time ... (and use the Product Rule):
. . . dV/dt . = . (sqrt{3}/12)[x²(dh/dt) + 2xh(dx/dt)]

The left side is what we want (the growth rate of the volume).

Do we know everything on the right side?
. . . We know that: . dx/dt = 1 mm/wk, . dh/dt = 1/5 mm/wk
. . . Do we know x and h? . . . well, not yet.
. . . We told that B = 100sqrt{3}.
. . . . . So: . (sqrt{3}{4})x² = 100sqrt{3} . ---> . x = 20
. . . . . Then: . h .= . (1.5)(x) .= .30

Now we're all set . . . just plug those values in!

 

[shaitan82]

Slant height/slant length?

thanks for the reply, but hmm... i don't think you're allowed making that assumption (that at any time, the height'll be 1.5 times the side length). hmm


is there anyway to relate the slant height (height of the isoceles triangle) or the slant length (length of one a side of the isoceles triangle) to the side length of the base? that's the problem i keep running into...

any other ideas out there?


thanks!