Question

m sec sec Q4- Given A(0,0,0), B(1,-1,1), C(2,1,-2) and D(-1,2,-1) are vertices of tetrahedron. If the rate of increase in side AB = 0.5 m, BC = 0.3 and ACE 0.4 Find the change in altitude of tetrahedron ABCD to get the change in volume m3 0.1 m sec sec

Answer 1

D (21,2,-1) Sola c (2,1, - 2) B (1,-1,1) (0,0,0) eq? of plana (ABC) (2-0) (y-o) :) (1-0) o (1-0) (-1.0) (-2-0) (ico) (2-0) 2 y = > 2 0 n (2-1) - 9(+2=2) of 2 (1+2) nauy d 32 = Dist of altitude (d) lie, dist from point o to (ABC) Do (-1,2,-1) da (1)x11) + (4)*(2) + (3)*(-1)) 2 + 1492 + 3)? da O 4 526 1 جاهله :1.732 3 (4-0)2 + (a120)2 + (1-0)? Be an [22.02 + (1 - (-1)2 + (-2-1) 2 ib. ba (2.0)24 (1-0)2+(-2-02 Sty: 3.741 Ас 3

Given dCAB) d (c) 0.5 m/s at d (BC) . dla) 0.3 m/s dd ou m/s dow ) at d (A) ad d (volume) m/s and Now dt now volume of a detrahedron - - x ar (ABC) xd (ABC) (1-0) + (-1-0) Ŝ + (ir og ÍR . 19 & I + (-2.0) f To 2 (2-0) î+ (1-0) Ali AB XÃ 2 ch - Î auð + 3 2 34 | ABX All A?! - 31332+(4927(3)? (A AB) 34 als (0.3+0.440.5) ar ceasy) s(s-a) (5-6)(5-6) S 9464 2584. d for cameo) = S(S-a), al 45-6)(s-e)} + el {s(5-09] (5-1)(s-e) dt dc 7 dd ds at db da de u ) +(s.c) lds dt s(sc3460) ( +(5-6)(3-2) ds di Sandfar} da) + (s. a) as s(s-as{{$._1360.6 -0.5) +(5-0)(0.6 -0,4)} + (5-6) (s-e){$(0.6 -0.3) (s-a). G at

di 236 Sat Sud 3 S: ad bet ) Soy dav). S(S-a)20.35 - 016-0.2 } 4 (8-4)315-030.95=6.4 finale 4,236 (4.236 - 3.7413 (4,236-384.236 0.384.236 -0,1%3-0.2% 1.732 - By d (ar) dt 741 SX = 2.096 x 0.62 uy t 1.236x2.504x 1.567 6.160 dlar) 1.056 m/s dd sor from I (ar) x d vol. 3 a dlar) lay) dd) dt at d(rol.) at 3. 1.056] [using o 534 d (d) + 526 dt Jзч 4x 1.056 d (d) at 3526 34 a(d) 0.176 dt d(d) -0.181 mls dt