document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197444>Question 1197444</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Climate Change   The most abundant greenhouse gas is carbon dioxide. According to figures from the Intergovernmental Panel on Climate Change (IPCC), the amount of carbon dioxide in the atmosphere (in parts of volume per million) can be approximated by<BR>\n" );
document.write( "C(t) ≈ 280e0.00127t parts per million<BR>\n" );
document.write( "where t is time in years since 1750.†<BR>\n" );
document.write( "(a)<BR>\n" );
document.write( "Use the model to estimate the amount of carbon dioxide in the atmosphere in <BR>\n" );
document.write( "1975, 2000, 2050, and 2100.<BR>\n" );
document.write( " (Round your answers to the nearest whole number.)<BR>\n" );
document.write( "year	1975	2000	2050	2100\r<BR>\n" );
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document.write( "C(t) parts per million	\r<BR>\n" );
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document.write( "(b)<BR>\n" );
document.write( "According to the model, in what year, to the nearest decade, will the level surpass 380 parts per million?\r<BR>\n" );
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document.write( "Part (a)\r<BR>\n" );
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document.write( "t is the number of years since 1750<BR>\n" );
document.write( "t = 0 represents 1750<BR>\n" );
document.write( "t = 1 represents 1751<BR>\n" );
document.write( "and so on\r<BR>\n" );
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document.write( "The year 1975 means t = 1975-1750 = 225<BR>\n" );
document.write( "The gap from 1750 to 1975 is 225 years<BR>\n" );
document.write( "Use this idea for the other years of 2000, 2050, and 2100.\r<BR>\n" );
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document.write( "Use spreadsheet software to quickly compute the outputs.<BR>\n" );
document.write( "You'll need to type in the <font color=blue>exp</font> function instead of something like e^x<BR>\n" );
document.write( "Example calculation:  <font color=blue>=280*EXP(0.00127*225)</font>  yields the approximate result of 372.612722644863 which rounds to 373.\r<BR>\n" );
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document.write( "This is what you should get when rounding each C(t) value to the nearest whole number.<BR>\n" );
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document.write( "<tr><TD >Year</TD><TD >t = Number of years since 1750</TD><TD >C(t) = Amount of CO2 (ppm)</TD></tr><tr><TD >1975</TD><TD >225</TD><TD >373</TD></tr><tr><TD >2000</TD><TD >250</TD><TD >385</TD></tr><tr><TD >2050</TD><TD >300</TD><TD >410</TD></tr><tr><TD >2100</TD><TD >350</TD><TD >437</TD></tr>\n" );
document.write( "</TABLE>ppm = parts per million\r<BR>\n" );
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document.write( "Part (b)\r<BR>\n" );
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document.write( "The table above shows <BR>\n" );
document.write( "C(225) = 373 .... year 1975<BR>\n" );
document.write( "C(250) = 385 .... year 2000\r<BR>\n" );
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document.write( "The level surpasses 380 ppm between the years 1975 and 2000.<BR>\n" );
document.write( "It's likely it's somewhere close to the year 2000 compared to the year 1975.\r<BR>\n" );
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document.write( "Use a natural logarithm to determine a more accurate time value.<BR>\n" );
document.write( "C(t) = 280*e^(0.00127t)<BR>\n" );
document.write( "380 = 280*e^(0.00127t)<BR>\n" );
document.write( "380/280 = e^(0.00127t)<BR>\n" );
document.write( "0.00127t = Ln(380/280)<BR>\n" );
document.write( "t = Ln(380/280)/0.00127<BR>\n" );
document.write( "t = 240.457991772584\r<BR>\n" );
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document.write( "Rounding to the nearest year gets us t = 240<BR>\n" );
document.write( "240 years after 1750 is 1750+240 = 1990 which also is the start of a decade, meaning we don't have to round to the nearest decade.\r<BR>\n" );
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document.write( "Answer: <font color=red>1990</font> <BR>\n" );
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