document.write( "Question 1033098: T =a+b e^-kt (Equation 2)
\n" ); document.write( "where T is the temperature at time t, and a, b and k are constants with
\n" ); document.write( "values a = 20 C, b = 80 C and k = 2.6 × 10^-4 ^-s1.\r
\n" ); document.write( "\n" ); document.write( "Differentiate Equation 2 with respect to t and thus give values (with
\n" ); document.write( "appropriate units and to two significant figures) for the rate of change of
\n" ); document.write( "temperature with time at times of 1.0 × 10^3 s and 5.0 × 10^3 s.\r
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Algebra.Com's Answer #647997 by KMST(5328)\"\" \"About 
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Your post displayed with strange character codes,
\n" ); document.write( "but maybe your equation was \"T=a%2Bb%2Ae%5E%28-kt%29\" , with
\n" ); document.write( "\"a=20%5Eo\"\"C\" , \"b=80%5Eo\"\"C\" , and \"k=2.6%2A10%5E%28-4%29\"\"s%5E%28-1%29\" .
\n" ); document.write( "In that case, differentiating with respect to \"t\" , we get
\n" ); document.write( "\"dT%2Fdt=-bk%2Ae%5E%28-kt%29\" for the rate of change of temperature with time.
\n" ); document.write( "So for time \"t=1.0%2A10%5E%28-3%29\"\"s\" , we have
\n" ); document.write( "\"k%2At=%22%28%22\"\"2.6%2A10%5E%28-4%29\"\"s%5E%28-1%29\"\"%22%29%22\"\"%22%28%22\"\"t=1.0%2A10%5E%28-3%29\"\"s\"\"%22%29%22\"\"%22=+0.26%22\" .
\n" ); document.write( "and that rate in degrees Celsius per second (rounded to 2 significant figures) would be
\n" ); document.write( " .
\n" ); document.write( "For time \"t=5.0%2A10%5E%28-3%29\"\"s\" , we have
\n" ); document.write( "\"k%2At=1.3\" and the rate of temperature increase is
\n" ); document.write( "
\n" ); document.write( "(in degrees Celsius per second, and rounded to 2 significant figures, of course).
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\n" ); document.write( "NOTES:
\n" ); document.write( "Since the numerical values for \"a\", \"b\" , \"k\" and \"t\" were all given with 2 significant figures, it makes sense to report results with 2 significant figures.
\n" ); document.write( "I was not sure that I could get the symbols for \"degrees Celsius per second\" to display properly, so I spelled it. \n" ); document.write( "
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