Question

For the function P=20e^-0.18t....

What is the initial quantity?

What is the growth rate in both decimal and percentage form?

Is this a growth or decay?

***not sure on this one since it's a natural log right? :/ thanks :)

Answer 1

What is the initial quantity?

Answer 2

umm 20e? :/

not sure if the e is included?

Answer 3

when t = 0, P = ???

Answer 4

umm i'm not very sure... 20?

Answer 5

e is just a number, it turns out that

e = 2.71828

It's like pi in a way

Answer 6

so if that 'e' bugs you, then replace it with 2.71828

Answer 7

ohh okay.. so would the initial quantity be 54.3656.. so 54.37 ?

Answer 8

plug in t = 0 and evaluate P = 20*e^(-0.18t)

Answer 9

P = 20*e^(-0.18t) is the same as P = 20*2.71828^(-0.18t)

since e = 2.71828 roughly

Answer 10

ohh oops... i multiplied the wrong thing

haha one sec...

20*0.835^7 ?

Answer 11

sorry i meant 20*0.835^t

Answer 12

P = 20*2.71828^(-0.18t)

P = 20*2.71828^(-0.18*0)

P = 20*2.71828^(0)

P = ???

Answer 13

P=20?

Answer 14

because anything to the 0 power is 1 right? so 20x1=20?

Answer 15

good

Answer 16

so 20 is the initial quantity?

Answer 17

yes it is

Answer 18

okay :)

and i'm a bit confused on the growth rate part... does that include the t? and i'm a bit lost on how to find the actual rate :/

Answer 19

Hint: solve for r

(1+r)^t = e^(-0.18t)

(1+r)^t = e^(-0.18) ^ t

1+r = e^(-0.18)

r = ??

Answer 20

-0.16 ?

Answer 21

good

Answer 22

so that's the growth rate?

-0.16 decimal form

-16% percentage form?

so that makes this decay?

Answer 23

yes it is

Answer 24

okay awesome! :) this one was confusing haha :P thank youuu!! :D

Answer 25

np

Answer 26

oh sorry one more thing... what's the difference between the continuous growth rate and the growth rate? :/ my question asks for continuous... but are they the same thing? :/

Answer 27

@jim_thompson5910 :)

Answer 28

This is continuous decay since we're involving the 'e' term

if we had something in the form P = a(1+r)^t then it would be either a normal growth problem or a normal decay problem

Answer 29

ohh okay so -0.16 and 16% are still the continuous decay rate?

Answer 30

yes that is correct

Answer 31

oh okay awesome!! thank you!! :D

Answer 32

sure thing