# Depreciation/Rate of Change (Calculus I)

• Nov 19th 2008, 08:39 PM
DirtMcGirt
Depreciation/Rate of Change (Calculus I)
Problem: The value of a piece of machinery depreciates exponentially. When purchased new, it was worth $70,000. After 4 years, it was worth$55,200. Find the rate of change of the value of the equipment after 4 years.

I started with:

4r = ln (55,200/70,000)
so..
4r = -0.2375322...
divide by 4 on both sides..
r = -.05938...
multiply by initial value (70000)
I got:
$4,156.60 which was not an answer choice, lol I'm obviously setting this up wrong, any help is appreciated :D • Nov 19th 2008, 09:28 PM sobadin Well if it is depreciation in business it is the same every year so 70,000-55200= 14800 so divide by 4 14800/4=$3700
• Nov 19th 2008, 09:38 PM
Soroban
Hello, DirtMcGirt!

Quote:

The value of a piece of machinery depreciates exponentially.
When purchased new, it was worth $70,000. After 4 years, it was worth$55,200.
Find the rate of change of the value of the equipment after 4 years.

We have: . $V \;=\;70,\!000\;\!e^{-kt}$

When $t = 4,\;V = 55,\!200$

Then: . $55,\!200 \:=\:70,\!000\;\!e^{-4k} \quad\Rightarrow\quad e^{-4k} \:=\:\tfrac{55,\!200}{70,\!000} \:=\:\tfrac{138}{175}$

Hence: . $-4k \:=\:\ln\left(\tfrac{138}{175}\right) \quad\Rightarrow\quad t \:=\:-\tfrac{1}{4}\ln\left(\tfrac{138}{175}\right) \;\approx\; 0.059$

. . The function is: . $V \;=\;70,\!000\;\!e^{-0.059t}$

The rate of change is: . $\frac{dV}{dt} \;=\;70,\!000(-0.059)\;\!e^{-0.059t} \;=\;-4130\;\!e^{-0.059t}$

When $t = 4\!:\;\;\frac{dV}{dt} \;=\;-4130\;\!e^{-0.059(4)} \;\approx\;\boxed{-3261.79}$

At the end of 4 years, it is depreciating at the rate of \$3261.79 per year.