Question

Historical data shows that the amount of money sent out of Canada for interest and dividend payments during the period from 1967 to 1979 can be approximated by the model P = (5 x 10^8)e^0.20015t, where t is measured in years (t = 0 in 1967) and P is the total payment in Canadian dollars.

a.Determine and compare the rates of increase for the years 1968 and 1978.
b.Assuming this trend continues, compare the rate of increase for 1988 with the rate of increase for 1998.

Answer 1

so what is here the question ?

Answer 2

so the questions ar

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) @marcgoeL
so the questions ar
\(\color{#0cbb34}{\text{End of Quote}}\)
?

Answer 4

i edited the question

Answer 5

What do you think the first thing to do is?

Answer 6

can someone send the answer and how they solved it

Answer 7

We can't give direct answers, but we can help guide you and explain it so it makes sense

Answer 8

so first of all do you know the value of e from this given formula ?

Answer 9

you cant give the ans cause u also don't know how to do it

Answer 10

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
so first of all do you know the value of e from this given formula ?
\(\color{#0cbb34}{\text{End of Quote}}\)
no u don't know that's the challenge

Answer 11

the number of years so the value of t how you calcule - do you know the way ?

Answer 12

I think we have to do P=5×108×e^0.20015t and then \[dp/dt=?\]

Not sure though idk calc that much

Answer 13

\(\color{#0cbb34}{\text{Originally Posted by}}\) @supie
I think we have to do P=5×108×e^0.20015t and then \[dp/dt=?\]

Not sure though idk calc that much
\(\color{#0cbb34}{\text{End of Quote}}\)

Yep!

It's a simple rule:

\( \dfrac{d}{dt} \left( e^{f(t)} \right) = \frac{df}{dt} e^{f(t)} \)

And by linearity:

\( \dfrac{d}{dt} \left(\alpha e^{f(t)} \right) = \alpha \frac{df}{dt} e^{f(t)} \)


So for:

\(P(t) = (5 \times 10^8)e^{0.20015t}\)

You get:
\( \dfrac{dP}{dt} (t) = (0.20015 \times 5 \times 10^8)e^{0.20015t}\)