Integral Problem

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April 3rd 2011, 03:07 PM
joatmon

Integral Problem

I seriously have no idea how to approach this, even after staring at it for an hour.

A high-tech company purchases a new computing system whose intitial value is V. The system will depreciate at the rate f = f(t) where t is the time measured in months. Suppose that

$f(t) = \frac{V}{15} - \frac{V}{450}t$ if $0<t\leq{30}$ . If (note: if t>30, f(t) = 0)

Determine the length of time T for the total depreciation $D(t) = \int^t_0 f(s)ds$ to equal the initial value V.

Can somebody please help me get this set up? I'm totally frustrated.

Thanks.(Headbang)
April 3rd 2011, 03:36 PM
skeeter

Quote:

Originally Posted by joatmon

I seriously have no idea how to approach this, even after staring at it for an hour.

A high-tech company purchases a new computing system whose intitial value is V. The system will depreciate at the rate f = f(t) where t is the time measured in months. Suppose that

$f(t) = \frac{V}{15} - \frac{V}{450}t$ if $0<t\leq{30}$ . If (note: if t>30, f(t) = 0)

Determine the length of time T for the total depreciation $D(t) = \int^t_0 f(s)ds$ to equal the initial value V.

Can somebody please help me get this set up? I'm totally frustrated.

Thanks.(Headbang)

initial value - total depreciation = 0

$\displaystyle V - \int_0^t \frac{V}{15} - \frac{V}{450} \cdot s \, ds = 0$

solve for $t$
April 3rd 2011, 05:45 PM
joatmon

Thanks, Skeeter. I think that I got this, but I spent so much time spinning my wheels that it would be great if somebody would please look at the math to make sure that I didn't make a mistake.

$V - \int_0^t (\frac{V}{15} - \frac{V}{450} \cdot s) \, ds = 0$

$V - \int_0^t \frac{V}{15}ds + \int_0^t \frac{V\cdot s}{450} ds = 0$

$V - \frac{V \cdot t}{15} \, + \frac{V}{450} \int_0^t s\, ds = 0$

$V - \frac{V \cdot t}{15} \, + \frac{V}{450} \cdot \frac{t^2}{2} = 0$

$\frac{900V - 60Vt + Vt^2}{900} = 0$

$\frac{V(t^2 - 60t +900)}{900} = 0$

$\frac{V(t-30)^2}{900} = 0$

$t = 30$
April 3rd 2011, 07:11 PM
skeeter

looks fine to me