# Thread: Depreciation/Rate of Change (Calculus I)

1. ## 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 2. 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

3. Hello, DirtMcGirt!

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

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

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

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

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

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

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

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

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### एक मशीन संतुलन को कम करने के मूल लागत रु 10000 और उसके ulɵmate स्क्रैप मूल्य रु था 3750 था मशीन के effecɵve जीवन को ख

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