differentiation

a) as PQ=x & PS=y , the area will be (xy)
now we know raten change of PQ is 0.2 so dx/dt = 0.2
PS is 0.3 dy/dt = 0.3

differentiate area w.r.t. time gives d/dt(xy) = (dx/dt)y + (dy/dt)x (by product rule)
then sub. x=20 , y=15 gives --- (0.2)(15)+(0.3)(20) = 9 (m^2/s)

b) diagonal PR = (x^2+y^2)^(1/2) by Pythagoras theorem
then differential w.r.t time again gives

d/dt[(x^2+y^2)^(1/2)] = 0.5*(x^2+y^2)^(-1/2) * [(2x)(dx/dt) + (2y)(dy/dt)]
by product & chain rule. then sub x=20, y=15 gives

0.5*(20^2+15^2)^(-1/2) * [(2*20)(0.2) + (2*15)(0.3)] = 17/50 m/s or 0.34m/s

2007-10-27 06:03:09 補充：
sorry .... mis-read the question... dy/dt should be (-0.3)a) (0.2)(15) (-0.3)(20) = -3 (m^2/s)b) 0.5*(20^2 15^2)^(-1/2) * [(2*20)(0.2) (2*15)(-0.3)] = -0.02 m/s

(a) Let area of PQRS be A.
Then A = xy.
rate of change of A, dA/dt = xdy/dt + ydx/dt
Since dx/dt = 0.2 and dy/dt = -0.3,
rate of change of A when x =20 and y = 15 = 20(-0.3) + 15(0.2) = -3.

(b) let s be the diagonal of PQRS.

Then s = (x^2 + y^2)^(1/2)
rate of change of s, ds/dt = 1/2(x^2 + y^2)^(-1/2) x (2xdx/dt + 2ydy/dt)
Therefore rate of change of s when x = 20 and y = 15
= 1/2(20^2 + 15^2)^(-1/2) x [2(20)(0.2) + 2(15)(-0.3)]
= -0.02