Population of canada was 24 million in 1981 and the exponetial growth rate was 1.2% per year.?

Question

A.write a expo function describing pop of canada

B. When will pop b 30 million

C. what will the pop be in 1998? and 2010?

D. what is the doubling time ?

Answer 1

Hi :) Hope this helps!

For this one just use a simple exponential growth equation.

P(t) = P(0)e^(rt)

Where P(0) = 24 million Canadians, t = t − 1981, and r = 1.2% per year. And P(t) is an integer. We'll also use the flooring function ⌊x⌋ or floor(x) to round down to the nearest whole person. http://wikipedia.org/wiki/Floor_and_ceil... You can skip flooring but it make more sense to floor logically. Make sure that you have math encodings available to read this right; if you have little skinny rectangles, it's not right.

P(t) = ⌊24 million Canadians∙e^(1.2%∙(t − 1981))⌋

// Simplify.

Answer A.) P(t) = ⌊24 million Canadians∙e^(0.012t − 23.772)⌋

For B. set the population to 30 million Canadians and solve for t.

P(t) = 30 million Canadians // Plugin P(t)

⌊24 million Canadians∙e^(0.012t − 23.772)⌋ = 30 million Canadians // Factor out 6 million Canadians, also 30 million is integer, so forget flooring.

4e^(0.012t − 23.772) = 5 // Divide by 4.

e^(0.012t − 23.772) = 1.25 // Take the natural log of both sides.

0.012t − 23.772 = ln 1.25 // Factor out 0.004

3t − 5,943 = 250 ln 1.25 // Divide by 3

t − 1981 = 83.(3) ln 1.25 // Add 1981

t = 1981 + 83.(3) ln 1.25 // Parameter.

t(30 million Canadians) = 1981 + 83.(3) ln 1.25

We need to approximate. This is rounded to:

1999.5952959428508129805245908592 (all following numbers are rounded too.)

Find the month by subtracting the number rounded down to the nearest integer (1999) and multiplying by 12, then adding 1.

8.1435513142097557662950903098345

Find the month by subtracting the number rounded down to the nearest integer (8) and multiplying by 31 (31 days in August), then adding 1.

5.4500907405024287551477996048696

Find the hour by subtracting the number rounded down to the nearest integer (5) and multiplying by 24.

10.802177772058290123547190516871

Find the minute by subtracting the number rounded down to the nearest integer (10) and multiplying by 60.

48.130666323497407412831431012231

Do the same for seconds.

7.8399794098444447698858607338322.

Round the seconds. I prefered to the nearest hundredth.

7.84

Time zone is unknown :(

In ISO 8601 time that is: 1999-08-05T10:48:07.84, a fair representation.

Answer B.) t(30 million Canadians) = 1981 + 83.(3) ln 1.25 ≈ 1999-08-05T10:48:07.84, where percision, ɛ, is to the nearest hundredth of a second.

Check:

P(1981 + 83.(3) ln 1.25) = 30 million Canadians ≈ P(1999-08-05T10:48:07.84) // Use P(t).

⌊24 million Canadians∙e^ln 1.25⌋ = 30 million Canadians ≈ ⌊24 million Canadians∙e^(0.223 143 551 314 217 443 249 701 314 217 4)⌋ // Take the natural logarithm.

⌊24 million Canadians∙1.25⌋ = 30 million Canadians ≈ ⌊24 million Canadians∙1.250 000 000 000 009 609 354 257 779 921 4⌋ // Multiply and floor.

30 million Canadians = 30 million Canadians ≈ 30 million Canadians // Evaluate.

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Note: 0.223 143 551 314 217 443 249 701 314 217 4 and 1.250 000 000 000 009 609 354 257 779 921 4 are both approximations to the nearest ten-quintillionth (10^31).

For C:

P(t) = ⌊24 million Canadians∙e^(0.012t − 23.772)⌋ // Plug in the matrix t = [1998 2010] http://wikipedia.org/wiki/Matrix_%28math... Leave plenty of space between numbers in a matrix, I'm using an em space " ".

P([1998 2010]) = ⌊24 million Canadians∙e^[0.204 0.348]⌋

Remember flooring, round down to the nearest whole person. Use thousand Canadians as the unit since million Canadians gets a bit messier in terms of digit grouping.

Answer C.) P([1998 2010]) = [29,431.155 thousand Canadians 33,989.573 thousand Canadians].

D. Doubling time you get by setting P(t) to 2P(0) where P(t) = P(0)e^(rt).

P(t) = 2P(0)

P(0)e^(rt) = 2P(0) // Cancel out P(0).

e^(rt) = 2 // Take the natural log.

rt = ln 2 // Divide by r.

t = (ln 2)/r // Parameter, in this case a subscript.

t[doubling] = (ln 2)/r

Using our equation, P(t) = ⌊24 million Canadians∙e^(0.012t − 23.772)⌋, we get:

t[doubling] − 1981 = (ln 2)/0.012

t[doubling] = 1981 + 83.(3) ln 2

Finding the ISO 8601 time again, we get 2038-10-05T13:30:08.41.

Answer D.) t[doubling] = 1981 + 83.(3) ln 2 ≈ 2038-10-05T13:30:08.41, where percision, ɛ, is to the nearest hundredth of a second.

Check:

P(1981 + 83.(3) ln 2) = 2P(0) ≈ P(2038-10-05T13:30:08.41) // Use P(t).

⌊24 million Canadians∙e^ln 2⌋ = 2∙24 million Canadians ≈ ⌊24 million Canadians∙e^0.694 147 180 559 (289 127 837 514 934)⌋ // Exponentiate, multiply, and floor.

48 million Canadians = 48 million Canadians ≈ 48,048.024 thousand Canadians. // Evaluate.

true

Q.E.D.