document.write( "Question 953768: Hi, I'd like some explanation this question please:\r
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document.write( "3. A population of rats grows at a rate which is proportional to the number of rats, y, existing at the end of month t. Initially there are 50 rats, and after 2 months there are 100 rats.\r
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document.write( "(i) Find a relationship between y and t.
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document.write( "(given answer --> y = 50(sqrt(2))^t)\r
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document.write( "(ii) Find the number of rats present after 1 year. (given answer --> 3200)\r
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document.write( "My answer to (i) is different but still gives the correct values for 2 months and 12 months:\r
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document.write( "y= exp ( (ln2)/2 * t + ln50)\r
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document.write( "Obviously mine is much less elegant, but I can't see how the given answer is arrived at.\r
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document.write( "My equation is derived from:\r
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document.write( "dy/dt = ky\r
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document.write( "Thanks, \r
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document.write( "Robert \n" );
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Algebra.Com's Answer #582511 by rothauserc(4718)![]() ![]() You can put this solution on YOUR website! We can solve this using the constant of proportionality, k \n" ); document.write( "we are given, \n" ); document.write( "y = 50*k^t and y = 50 for t = 0, hence t is the exponent of k \n" ); document.write( "100 = 50*k^2 and y = 100 for t = 2, then \n" ); document.write( "k^2 = 2 \n" ); document.write( "k = square root (2) \n" ); document.write( "therefore \n" ); document.write( "y = 50 *(square root(2)^t) \n" ); document.write( " \n" ); document.write( " |