anonymous
  • anonymous
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
Mathematics
katieb
  • katieb
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Jhannybean
  • Jhannybean
a. you're given the initial price, \(P_0\) = $20 a = 0.05 Just plug in the values into your formula
Jhannybean
  • Jhannybean
b. t = 5 years, a = 0.05, \(P_0\)= 20
anonymous
  • anonymous
for a. i put P(t) = 20(.05)^t but i got it wrong

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Venomblast
  • Venomblast
You are multiplying t
anonymous
  • anonymous
sorry, its actually raised to the t power
Venomblast
  • 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
  • Venomblast
Oh. Then that changes the function completely!
Venomblast
  • Venomblast
So I assume your equation would be \[20 \times(.05)^{t}\]
anonymous
  • anonymous
yeah, i typed that in but i got it wrong
Venomblast
  • Venomblast
This formula is similar to the compound formula. Put in your answer and tell me if it is correct?
anonymous
  • anonymous
sorry, i am confused, for what letter a or b?
Venomblast
  • Venomblast
I gave you (a) already. Now plug in 5 for t
Jhannybean
  • Jhannybean
Then you are multiplying by t and it is not raised to a power, perhaps?
Jhannybean
  • Jhannybean
Fo a.
anonymous
  • anonymous
wait sorry one second
anonymous
  • anonymous
I'm trying to get a screenshot of it
Venomblast
  • Venomblast
substitute a with the appropriate variables.
Venomblast
  • Venomblast
Try this\[20(.05)^{t}\]
anonymous
  • anonymous
anonymous
  • anonymous
yeah it gave that as wrong, not sure why
Venomblast
  • Venomblast
Try adding the times in. Could be a error. Maybe it s compounding formula. Try this \[20\times(1.05)^{t}\]
anonymous
  • anonymous
ok that was right for the equation, but for the part b i plug in chug but it gives the wrong answer?
anonymous
  • anonymous
i get 20(1.05)^5 = 25.52 rounding to two decimal places, says it is wrong tho
Venomblast
  • Venomblast
25.53
anonymous
  • anonymous
Calculator gives me 25.52563 and i tried 25.52 and 25.53
Venomblast
  • 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
  • anonymous
yea i tried 25.53 and 25.52 and both are wrong
Venomblast
  • Venomblast
That is very strange.
anonymous
  • anonymous
anonymous
  • anonymous
would it maybe have to do with it asking for dollar/year?
Venomblast
  • Venomblast
dollar per year? That a strange way of saying it.
anonymous
  • anonymous
wait wouldn't it be the derivative????
Venomblast
  • Venomblast
Why would it be the derivative?
anonymous
  • anonymous
yeah nevermind, i don't know :(
Venomblast
  • Venomblast
If it's a physic problem, then yes I would take the derivative.
Venomblast
  • Venomblast
try 5.53
Jhannybean
  • Jhannybean
Actually,you would have to take the derivative.
anonymous
  • anonymous
yeah, my time limit, expired :( oh well got a 80% on it :)
Jhannybean
  • Jhannybean
Do you still want to know how to figure it out? Or nah.
anonymous
  • anonymous
yes please
Venomblast
  • Venomblast
Sorry man. It was confusing. It seems like an interest problem but then a physic problem?
Venomblast
  • Venomblast
If it is a physic problem, take the derivative (chain rule).
anonymous
  • anonymous
no your fine man, i appreciate the help
Jhannybean
  • 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
  • Venomblast
who me?
anonymous
  • anonymous
to be honest i don't understand the 1.05 either
Jhannybean
  • Jhannybean
Jhannybean
  • Jhannybean
Okay.
Venomblast
  • 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
  • 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
  • Venomblast
Yes correct. You would get the constant rate of interest i if you are headed for the right direction
Jhannybean
  • Jhannybean
ahh... im off on my decimal place.
Venomblast
  • Venomblast
yea that would be .0005 lol
Jhannybean
  • Jhannybean
uld have said \[P(1) = \$20 + \left(\frac{5}{100}\right)\cdot \$20\]
Venomblast
  • Venomblast
I can prove it if you want me to. How i derive the formula.
anonymous
  • anonymous
i understand the formula now, but what i don't understand is the part for b?
Jhannybean
  • 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
  • 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
  • Jhannybean
Now moving on to part b.
Jhannybean
  • 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
  • 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
  • Jhannybean
Sorry, I'm extremely lagged out, took a while to type up.
anonymous
  • anonymous
no your fine, i see the answer now, i should've taken the derivative :/
Jhannybean
  • 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
  • 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
  • anonymous
thank you so much for explaining the question to me, i understand it well now, thank you!!!
Jhannybean
  • Jhannybean
No problem :)
Jhannybean
  • Jhannybean
Does it make sense to you too, @Venomblast ?
Venomblast
  • Venomblast
Haha yea. Sorry I went to sleep

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