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Math Help - exponential functions and depreciation

  1. #1
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    Exclamation exponential functions and depreciation

    A car depreciates at a rate of 30% per year. How much will a $20,000 car be worth after 5 years?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by math123456 View Post
    A car depreciates at a rate of 30% per year. How much will a $20,000 car be worth after 5 years?
    In the problem you are given the rate of decrease.

    Since the decay equation takes on the form P=P_0e^{-kt}, we can take note that k=.3, P_0=20000 and t=5. Plugging this into the equation, we have P=20000e^{-.3(5)}\approx 4462.60

    So after 5 years, the car will be worth about $4,462.60

    Does this make sense?
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    I would use the model

    V=A.P^t

    where V = current value, A = initial value, P = percentage change and t = time.

    Therefore

    V=20000\times (1-.3)^t

    V=20000\times .7^t

    then make t = 5

    V=20000\times .7^5= 3361.40
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    In the problem you are given the rate of decrease.

    Since the decay equation takes on the form P=P_0e^{-kt}, we can take note that k=.3, P_0=20000 and t=5. Plugging this into the equation, we have P=20000e^{-.3(5)}\approx 4462.60

    So after 5 years, the car will be worth about $4,462.60

    Does this make sense?
    I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%.

    With the given data you know that the car's value after 1 year is $14.000. If you now calculate the constant k you'll get: k\approx -0.356675

    Therefore your result will be too large.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by earboth View Post
    I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%.

    With the given data you know that the car's value after 1 year is $14.000. If you now calculate the constant k you'll get: k\approx -0.356675

    Therefore your result will be too large.
    Thank you for pointing that out to me. I didn't notice that.
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    Quote Originally Posted by earboth View Post
    I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%.
    Earboth, what is the reason for this?
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    Quote Originally Posted by pickslides View Post
    Earboth, what is the reason for this?
    If you use Chris L T 521's formula:

    A = A_0 \cdot e^{k \cdot t}

    then k = \ln(1-p) where p is the percentage of decrease.

    Now compare:

    decrease by 2% ====> ln(0.98) = -0.020202...
    decrease by 5% ====> ln(0.95) = -0.051293...
    decrease by 10% ====> ln(0.9) = -0.10536...
    ...
    decrease by 20% ====> ln(0.8) = -0.22314...

    You see that the approximation of k = -p is only sufficient for p < 10%.
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  8. #8
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    Hello, math123456!

    A car depreciates at a rate of 30% per year.
    How much will a $20,000 car be worth after 5 years?

    Each year the car is worth only 70% of its value during the previous year.

    . . Its value after n years is: . V \:=\:20,\!000(0.7)^n


    In 5 years, its value is: . V \;=\;20,\!000(0.7)^5 \;=\;\$3,\!361.40

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    Seeing the way that Chris approached the problem,
    I am obligated to ask: Is someone teaching that the e-formula
    . . is the only way to handle depreciation?
    If so, they are doing a great disservice.

    While P \:=\:P_oe^{-kt} is highly applicable to decay and population problems,
    I feel that it requires more work when applied to depreciation problem,
    . . especially when they give us the depreciation rate.


    Consider the original problem . . .
    . . Initial value: $20,000
    . . Depreciation rate: 30% (annually)


    We start with: . P \:=\:P_oe^{-kt}

    When t = 0,\:P = 20,\!000\!:\quad20,\!000 \:=\:P_oe^0 \quad\Rightarrow\quad P_o = 20,\!000

    . . Hence, the function (so far) is: . P \;=\;20,\!000\,e^{-kt}


    When t = 1,\;P = 14,\!000\!:\quad 14,\!000 \:=\:20,\!000e^{-k} \quad\Rightarrow\quad e^{-k} \:=\:0.7

    . . -k \:=\:\ln(0.7) \quad\Rightarrow\quad k \:=\:-\ln(0.7) \:=\:-0.356674944

    Hence, the function is: . \boxed{P \;=\;20,\!000\,e^{-0.356674944t}}

    This equation is correct . . . I have no complaint.



    But simplify the equation: . P \;=\;20,\!000\left(e^{-0.356674944}\right)^t

    Since e^{-0.356674944} \:=\:0.7, we have: . \boxed{P \;=\;20,\!000\,(0.7)^t}


    We had this equation way back then, just "eyeballing" the problem.

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  10. #10
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    Quote Originally Posted by earboth View Post
    If you use Chris L T 521's formula:

    A = A_0 \cdot e^{k \cdot t}

    then k = \ln(1-p) where p is the percentage of decrease.

    Now compare:

    decrease by 2% ====> ln(0.98) = -0.020202...
    decrease by 5% ====> ln(0.95) = -0.051293...
    decrease by 10% ====> ln(0.9) = -0.10536...
    ...
    decrease by 20% ====> ln(0.8) = -0.22314...

    You see that the approximation of k = -p is only sufficient for p < 10%.
    I see the error is becomes quite significant, so why isnít my solution (when given the depreciation level) always considered the better?


    V=20000\times (1-.3)^t

    V=20000\times .7^t

    then make t = 5

    V=20000\times .7^5= 3361.40
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  11. #11
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    Quote Originally Posted by pickslides View Post
    I see the error is becomes quite significant, so why isnít my solution (when given the depreciation level) always considered the better?


    V=20000\times (1-.3)^t

    V=20000\times .7^t

    then make t = 5

    V=20000\times .7^5= 3361.40
    The difficulty lies in the word "always" ...

    With the given problem the variable t is a discret number, that means actually you are dealing with a geometric sequence: a_n=20000 \cdot \left(\dfrac7{10} \right)^n

    If a value is increasing or decreasing continuously your method can't be applied, then you have to use the equation which was posted by ChrisLT521
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