Question

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

Answer 1

a. you're given the initial price, \(P_0\) = $20
a = 0.05

Just plug in the values into your formula

Answer 2

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

Answer 3

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

Answer 4

You are multiplying t

Answer 5

sorry, its actually raised to the t power

Answer 6

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

Answer 7

Oh. Then that changes the function completely!

Answer 8

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

Answer 9

yeah, i typed that in but i got it wrong

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Fo a.

Answer 15

wait sorry one second

Answer 16

I'm trying to get a screenshot of it

Answer 17

substitute a with the appropriate variables.

Answer 18

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

Answer 19

yeah it gave that as wrong, not sure why

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

25.53

Answer 24

Calculator gives me 25.52563
and i tried 25.52 and 25.53

Answer 25

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.

Answer 26

yea i tried 25.53 and 25.52 and both are wrong

Answer 27

That is very strange.

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

Why would it be the derivative?

Answer 32

yeah nevermind, i don't know :(

Answer 33

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

Answer 34

try 5.53

Answer 35

Actually,you would have to take the derivative.

Answer 36

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

Answer 37

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

Answer 38

yes please

Answer 39

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

Answer 40

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

Answer 41

no your fine man, i appreciate the help

Answer 42

\[P(t) = 20\cdot 1.05^t\]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.

Answer 43

who me?

Answer 44

to be honest i don't understand the 1.05 either

Answer 45

@benitob

Answer 46

Okay.

Answer 47

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

Answer 48

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... \(P(1) = \$20 + \left(\frac{0.05}{100}\right)\cdot \$20\)

Answer 49

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

Answer 50

ahh... im off on my decimal place.

Answer 51

yea that would be .0005 lol

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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.

Answer 56

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

Answer 57

Now moving on to part b.

Answer 58

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, \(\dfrac{dP}{dt}\)

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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) :)

Answer 63

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.

Answer 64

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

Answer 65

No problem :)

Answer 66

Does it make sense to you too, @Venomblast ?

Answer 67

Haha yea. Sorry I went to sleep