## anonymous one year ago 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 • This Question is Closed 1. anonymous a. you're given the initial price, $$P_0$$ =$20 a = 0.05 Just plug in the values into your formula

2. anonymous

b. t = 5 years, a = 0.05, $$P_0$$= 20

3. anonymous

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

4. Venomblast

You are multiplying t

5. anonymous

sorry, its actually raised to the t power

6. Venomblast

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

7. Venomblast

Oh. Then that changes the function completely!

8. Venomblast

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

9. anonymous

yeah, i typed that in but i got it wrong

10. Venomblast

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

11. anonymous

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

12. Venomblast

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

13. anonymous

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

14. anonymous

Fo a.

15. anonymous

wait sorry one second

16. anonymous

I'm trying to get a screenshot of it

17. Venomblast

substitute a with the appropriate variables.

18. Venomblast

Try this$20(.05)^{t}$

19. anonymous

20. anonymous

yeah it gave that as wrong, not sure why

21. Venomblast

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

22. anonymous

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

23. anonymous

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

24. Venomblast

25.53

25. anonymous

Calculator gives me 25.52563 and i tried 25.52 and 25.53

26. 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.

27. anonymous

yea i tried 25.53 and 25.52 and both are wrong

28. Venomblast

That is very strange.

29. anonymous

30. anonymous

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

31. Venomblast

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

32. anonymous

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

33. Venomblast

Why would it be the derivative?

34. anonymous

yeah nevermind, i don't know :(

35. Venomblast

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

36. Venomblast

try 5.53

37. anonymous

Actually,you would have to take the derivative.

38. anonymous

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

39. anonymous

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

40. anonymous

41. Venomblast

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

42. Venomblast

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

43. anonymous

no your fine man, i appreciate the help

44. anonymous

$P(t) = 20\cdot 1.05^t$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.

45. Venomblast

who me?

46. anonymous

to be honest i don't understand the 1.05 either

47. anonymous

@benitob

48. anonymous

Okay.

49. 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

50. anonymous

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

51. Venomblast

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

52. anonymous

ahh... im off on my decimal place.

53. Venomblast

yea that would be .0005 lol

54. anonymous

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

55. Venomblast

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

56. anonymous

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

57. anonymous

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.

58. anonymous

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$

59. anonymous

Now moving on to part b.

60. anonymous

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, $$\dfrac{dP}{dt}$$

61. anonymous

$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}$

62. anonymous

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

63. anonymous

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

64. anonymous

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) :)

65. anonymous

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.

66. anonymous

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

67. anonymous

No problem :)

68. anonymous

Does it make sense to you too, @Venomblast ?

69. Venomblast

Haha yea. Sorry I went to sleep