Question

derive an expression to show law of conservation of energy

Answer 1

1. P{max(X,Y,Z) ≤ t} = P{X ≤ t and Y ≤ t and Z ≤ t} = P{X ≤ t}3 by independence. Thus the distribution function of the maximum is (t 6)3 = t 18, and the density is 18t 17, 0 ≤ t ≤ 1. 2. See Figure S1.1. We have P{Z ≤ z} = y≤zx fXY (x, y) dx dy = ∞ x=0 zx y=0 e−xe−y dy dx FZ(z) = ∞ 0 e−x(1 − e−zx) dx = 1 − 1 1 + z , z ≥ 0 fZ(z) = 1 (z + 1)2 , z ≥ 0 FZ(z) = fZ(z) = 0 for z < 0. 3. P{Y = y} = P{g(X) = y} = P{X ∈ g−1(y)}, which is the number of xi’s that map to y, divided by n. In particular, if g is one-to-one, then pY (g(xi)) = 1/n for i = 1,... ,n. 4. Since the area under the density function must be 1, we have ab3/3 = 1. Then (see Figure S1.2) fY (y) = fX(y1/3)/|dy/dx| with y = x3, dy/dx = 3x2. In dy/dx we substitute x = y1/3 to get fY (y) = fX(y1/3) 3y2/3 = 3 b3 y2/3 3y2/3 = 1 b3

Answer 2

m = mass of body

g = Acceleration due to gravity

Suppose a body is at point A initially. Then it falls freely, during the motion mechanical energy of the body is conserve.

At point A,

AC = height of object from ground

= h

Initial speed u = 0

So, P.E. = mgh



TE = P.E + K.E = mgh + 0

(TE)A = mgh ... (i)

At point B,

PE = mg (BC)

= mg (h – x) = mgh – mgx

Use kinematics equation



(TE)B = P.E. + K.E = mgh – mgx + mgx

= mgh ...(ii)

At point C,

P.E. = 0



Use kinematics equation



⇒ v 2 = 2gh

So,

(T.E.)C = K.E + P.E. = mgh +0 = mgh ...(iii)

(TE)A =(TE)B=(TE)C

Hence, the total energy of the body is conserve during free fall.