Find exponential decay equation for tiger population model

Question

I've forgotten how to do it first it starts..

In 1900, there were 100,000 wild tigers worldwide; in 2010 the number was 3200.

(a) Assuming that the tiger population has decreased exponentially, find a formula for f(t) , the number of wild tigers t years since 1900.

(b) Between 2000 and 2010, the number of wild tigers decreased by 40% . Is this percentage larger or smaller than the decrease in the tiger population predicted by your answer to part (a)?

For (a) I assumed

f(t) = 100'000 *N^t 
f(110) = 100'000 *N^110 = 3200
                      N = (32/1000)^(1/110)    
                        = 0.9691....  

Therefore...

f(t) = 100'000*0.9691^t

Which I know is wrong.

Answer 1

Your formula $$100000*0.9691^t$$ is actually an increasing function.

Try $$f(t)=100000\times N^{-t}$$ Spoiler:Answer:

$$f(t)=100000\times(1.0318)^{-t}$$

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From the formula get no. of tigers in 2000 and 2010, then find decrease percent.

Answer 2

$100000(0.9691)^t \; \text{is} \;\underline{\textbf{not}}\; \text{an increasing function.}$

$y=b^t \; \text{is an increasing function when }b>1, \; \textrm{but when }0<b<1,\; y=b^t \;\textrm{is a decreasing function} .$

$ \text{Suppose you let t = the number of }\; \underline{\text{decades}}\;\text{ since 1900. So here t = 11 decades.}$

$ \text {Then}\;\; f(11) = 3200=100000*N^{11} $

$ \text {Thus}\;\; N=\sqrt[11]{\frac{32}{1000}}=\sqrt[11]{0.032}\approx0.7313151 $

$ \text {So the function is}\;\; f(t) = 100000(0.7313151)^{t} ,\; \text{where }t\; \text{is measured in} \;\underline{\textbf {decades}}\;\text{after 1900.} $

$\text{In part (b) when it says "the number of wild tigers decreased by 40%", it means }$ $\displaystyle f(11)=f(10)-40\%*f(10)=f(10)-0.4*f(10)= f(10)(1-0.4)=0.6*f(10) $

$\displaystyle \text{In general, percent increase/decrease is defined as} \;\;\frac{\text{original number}\; \pm \; \text{new number}}{\text{original number}} $

$\text{Because}\;f(10)\;\text{comes before}\;f(11) ,\; \text{percent decrease means }$

$\displaystyle\frac{f(10)-f(11)}{f(10)}=\frac{f(10)}{f(10)}-\frac{f(11)}{f(10)}=1-N=1-\sqrt[11]{0.032}\approx1-0.7313151=.2686849$ $$ \approx 27\%$$

$\text{Or in your case,}$

$\displaystyle\frac{f(100)-f(110)}{f(100)}=\frac{f(100)}{f(100)}-\frac{f(110)}{f(100)}=1-N^{110-100}=1-N^{10}=1-\left(\;\sqrt[110]{0.032}\;\right)^{10}$ $$\approx1-0.7313151=.2686849 \approx 27\%$$

$\text{Therefore between 2000 to 2010, the number of wild tigers decreased by approximately 27%.}$