Question

**Will Give Medals** A piece of machinery is worth $1600. The rate of depreciation of the value V (in dollars) at time t (in years) is proportional to the square root of its value V . The piece of machinery will be worth $900 two years later. When will the piece of machinery be worth $625?

Answer 1

i need to make a differential equation which will lead into integrals

Answer 2

dv/dt = sqrt(v)

dv/sqrt(v) = dt

1/2 sqrt(v) = t + C

sqrt(v) = 2t + C

v = (2t + C)^2
.............................

v' = 2(2t + C) = 2 sqrt(v)

just taking a stab at it

Answer 3

C = 40 since 40^2 = 1600

Answer 4

yeah, this function doesnt decrease over time does it

Answer 5

and your answer key gives us? if you have one ...

Answer 6

@amistre64: dv/dt = k*sqrt(v) (where k is some arb. const) rather than dv/dt = sqrt(v) (we can't assume that k=1 in this context).

Answer 7

@bpaquin007 & @amistre64: So the equation written by amistre64 becomes modified to read as follows:

dv/dt = k*(√v)

∫ dv/k*(√v) = ∫ dt

(2/k)*(√v) = t + C

V = [k*(t+C)/2]^2 ---------- (1)

Given that when t=0, V=1600 & when t=2, V=900, we have:

kC/2 = 40 ----------- (2)
k*(2+C)/2 = 30 ------- (3)

From (2), we have:
kC = 80 ------------- (4)

From (3), we have:
2k + kC = 60 --------- (5)

Substituting (4) into (5), we have:
2k = -20
k = -10

Substituting k = -10 into (4), we have:
C = -8

Substituting these values of k & C into (1), we have:
V = [(-5)*(t-8)]^2 ---------- (6)

Given that V = 625, we now have from (6):
625 = [(-5)*(t-8)]^2
25 = (-5)*(t-8)
t-8 = -5
t = 3

Hence, the machinery will be valued at $625 after 3 years from its initial valuation.

Answer 8

thats good, i was thinking the constant of proportionality, k, was going to be taken care of in the arbitrary constant .... its good to know better now :)