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c3 differentiation problem

The population of Cambridge was 37000 in 1900 and was about 109000 in 2000. Find an equation of the form P = (Po)(K^t) to model this data, where t is measured in 1900. Evaluate Dp/dt in the year 2000. What does this value represent?

So I have got dp/dt as being 37000 x k^t x lnk

The calculation I am doing is 37000 x 109/37 x ln109/37

This gives 117766.8 which is one hundred times the answer which is 1178. Am I overlooking something?

Sorry this is a c4 problem not c3.
(edited 9 years ago)
Original post by readyman
The population of Cambridge was 37000 in 1900 and was about 109000 in 2000. Find an equation of the form P = (Po)(K^t) to model this data, where t is measured in 1900. Evaluate Dp/dt in the year 2000. What does this value represent?

So I have got dp/dt as being 37000 x k^t x lnk

The calculation I am doing is 37000 x 109/37 x ln109/37

This gives 117766.8 which is one hundred times the answer which is 1178. Am I overlooking something?

Sorry this is a c4 problem not c3.

Could you give your equation? I don't like the 109/37 - where did it come from?
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readyman
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Original post by Smaug123
Could you give your equation? I don't like the 109/37 - where did it come from?


When t=100 p =109000 (year is 2000)

so 109000 = 37000 x k^100

therefore K^100 = 109/37
Original post by readyman
When t=100 p =109000 (year is 2000)

so 109000 = 37000 x k^100

therefore K^100 = 109/37

Yes, but 37000×10937×log⁡(10937)37000 \times \frac{109}{37} \times \log(\frac{109}{37}) is not 37000klog⁡(k)37000 k \log(k) - it's 37000klog⁡(k100)37000 k \log(k^{100}). Sorry, I wasn't clear about which 109/37 I meant.
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Original post by Smaug123
Yes, but 37000×10937×log⁡(10937)37000 \times \frac{109}{37} \times \log(\frac{109}{37}) is not 37000klog⁡(k)37000 k \log(k) - it's 37000klog⁡(k100)37000 k \log(k^{100}). Sorry, I wasn't clear about which 109/37 I meant.


I found the solutions for the question here https://e1934e482e1b0d2c8580c3a986eb005fbbe9e596.googledrive.com/host/0B1ZiqBksUHNYcHVKUE1IOEZ3UTQ/CH4.pdf
It is exercise 4c q4

I see your point but I don't quite get what they have done with the 1/100x ln109/37 part
Original post by readyman
I found the solutions for the question here https://e1934e482e1b0d2c8580c3a986eb005fbbe9e596.googledrive.com/host/0B1ZiqBksUHNYcHVKUE1IOEZ3UTQ/CH4.pdf
It is exercise 4c q4

I see your point but I don't quite get what they have done with the 1/100x ln109/37 part

log⁡(k100)=100log⁡(k)\log(k^{100}) = 100 \log(k).
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Original post by Smaug123
log⁡(k100)=100log⁡(k)\log(k^{100}) = 100 \log(k).

why is it 1/100 then?
Original post by readyman
why is it 1/100 then?

You took kk inside the log to be 100 powers bigger than it actually is. The expression you need is 37000ktlog⁡(k)=37000×((10937)1/100)100×log⁡((10937)1/100)37000 k^t \log(k) = 37000 \times ((\dfrac{109}{37})^{1/100})^{100} \times \log((\frac{109}{37})^{1/100}).
Reply 8
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readyman
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Original post by Smaug123
You took kk inside the log to be 100 powers bigger than it actually is. The expression you need is 37000ktlog⁡(k)=37000×((10937)1/100)100×log⁡((10937)1/100)37000 k^t \log(k) = 37000 \times ((\dfrac{109}{37})^{1/100})^{100} \times \log((\frac{109}{37})^{1/100}).

Oh I see, thanks for your help!

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