The value of a particular item can be modeled by
P(t) = P0(a)t
where P is in dollars and t is the number of years since the item was purchased. Suppose the value of the item increases 5% each year and the item was purchased for $20.
(a) Write a formula for P(t) according to the model.
(b) How fast is the value of the item increasing when t = 5 years? Round your answer to two decimal places

- anonymous

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- Jhannybean

a. you're given the initial price, \(P_0\) = $20
a = 0.05
Just plug in the values into your formula

- Jhannybean

b. t = 5 years, a = 0.05, \(P_0\)= 20

- anonymous

for
a. i put P(t) = 20(.05)^t but i got it wrong

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## More answers

- Venomblast

You are multiplying t

- anonymous

sorry, its actually raised to the t power

- Venomblast

So you would get, 20 * .05 * 5. To make this easier do 20*5 then multiply by .05
Hint: .05 = \[\frac{ 1 }{20 }\]

- Venomblast

Oh. Then that changes the function completely!

- Venomblast

So I assume your equation would be \[20 \times(.05)^{t}\]

- anonymous

yeah, i typed that in but i got it wrong

- Venomblast

This formula is similar to the compound formula. Put in your answer and tell me if it is correct?

- anonymous

sorry, i am confused, for what letter a or b?

- Venomblast

I gave you (a) already. Now plug in 5 for t

- Jhannybean

Then you are multiplying by t and it is not raised to a power, perhaps?

- Jhannybean

Fo a.

- anonymous

wait sorry one second

- anonymous

I'm trying to get a screenshot of it

- Venomblast

substitute a with the appropriate variables.

- Venomblast

Try this\[20(.05)^{t}\]

- anonymous

##### 1 Attachment

- anonymous

yeah it gave that as wrong, not sure why

- Venomblast

Try adding the times in. Could be a error. Maybe it s compounding formula. Try this \[20\times(1.05)^{t}\]

- anonymous

ok that was right for the equation, but for the part b
i plug in chug but it gives the wrong answer?

- anonymous

i get 20(1.05)^5 = 25.52 rounding to two decimal places, says it is wrong tho

- Venomblast

25.53

- anonymous

Calculator gives me 25.52563
and i tried 25.52 and 25.53

- Venomblast

when it says round to the nearest cent, It really telling you to round to the nears hundredth. Thus you look at the hundredth place which is a 2. look next to it to the right. if the number is 5 or above, you round up. The 2 becomes a 3.

- anonymous

yea i tried 25.53 and 25.52 and both are wrong

- Venomblast

That is very strange.

- anonymous

##### 1 Attachment

- anonymous

would it maybe have to do with it asking for dollar/year?

- Venomblast

dollar per year? That a strange way of saying it.

- anonymous

wait wouldn't it be the derivative????

- Venomblast

Why would it be the derivative?

- anonymous

yeah nevermind, i don't know :(

- Venomblast

If it's a physic problem, then yes I would take the derivative.

- Venomblast

try 5.53

- Jhannybean

Actually,you would have to take the derivative.

- anonymous

yeah, my time limit, expired :( oh well got a 80% on it :)

- Jhannybean

Do you still want to know how to figure it out? Or nah.

- anonymous

yes please

- Venomblast

Sorry man. It was confusing. It seems like an interest problem but then a physic problem?

- Venomblast

If it is a physic problem, take the derivative (chain rule).

- anonymous

no your fine man, i appreciate the help

- Jhannybean

\[P(t) = 20\cdot 1.05^t\]Just to be clear... can you explain to me how you got \(a=1.05\) ? I know how this works, but I want you to explain it t me.

- Venomblast

who me?

- anonymous

to be honest i don't understand the 1.05 either

- Jhannybean

@benitob

- Jhannybean

Okay.

- Venomblast

I can tell you why. When money is growing over certain amount of year (or days or month). You use that formula. P(1+i)^t. Where i is the interest rate, P is your initial amount and t is the amount of years/month/days

- Jhannybean

So we have out function \(P(t) = P_0 \cdot a^t\).
Were given that \(P_0 = 20\) an the growth factor = \(5\%\).
Therefore for every year it is increasing... \(P(1) = \$20 + \left(\frac{0.05}{100}\right)\cdot \$20\)

- Venomblast

Yes correct. You would get the constant rate of interest i if you are headed for the right direction

- Jhannybean

ahh... im off on my decimal place.

- Venomblast

yea that would be .0005 lol

- Jhannybean

uld have said \[P(1) = \$20 + \left(\frac{5}{100}\right)\cdot \$20\]

- Venomblast

I can prove it if you want me to. How i derive the formula.

- anonymous

i understand the formula now, but what i don't understand is the part for b?

- Jhannybean

That gives us, \(P(1) = \$20 + \$1 = \$21\)
And now we want to find the growth factor, \(a\) to find the generalization of the growth function of \(t\) years.

- Jhannybean

So we have \(P(1) = \$21,\qquad ~P_0 = \$20\qquad ~,~\qquad t=1\)
\[P(t) = P_0 \cdot a^t\]\[P(1) = 20 \cdot a^1\]\[\$21= \$20 \cdot a^1\]\[a= \frac{21}{20} = 1.05\]\[\implies P(t) = \$20 \cdot 1.05^t\]

- Jhannybean

Now moving on to part b.

- Jhannybean

We want to find `how fast` the value is increasing with respect to time, therefore we're finding the rate of change of the growth function, \(\dfrac{dP}{dt}\)

- Jhannybean

\[P(t)= 20 \cdot 1.05^t\]\[\frac{d}{dt} (P(t) = 20 \cdot 1.05^t) \]\[P'(t) = 20\frac{d}{dx}( 1.05^t)\qquad \qquad \frac{d}{dx}(a^x) = a^x \cdot \ln(a)\]\[P'(t) = 20[1.05^t \cdot \ln(1.05)]\]\[\color{red}{P'(t) \approx 0.98 \cdot 1.05^t}\]

- Jhannybean

Sorry, I'm extremely lagged out, took a while to type up.

- anonymous

no your fine, i see the answer now, i should've taken the derivative :/

- Jhannybean

yeah, when a question asks you `how fast` something is changing within a given time, it's asking you basically for a certain value at a point. (your slope) :)

- Jhannybean

I was working it out on a separate piece of paper while you guys were discussing the question earlier and i had to reread the question a few times to notice it was a rate of change question.

- anonymous

thank you so much for explaining the question to me, i understand it well now, thank you!!!

- Jhannybean

No problem :)

- Jhannybean

Does it make sense to you too, @Venomblast ?

- Venomblast

Haha yea. Sorry I went to sleep

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