Finding the Rate of Change in a Biology Experiment

In a metabolic experiment, the mass \(M\) of glucose decreases according to the function \[ M(t) = 4.5-0.03t^{2} \]

where \(M\) is measured in grams (g) and \(t\) is the time in hours (h). Find the reaction rate \(M^\prime(t) at t=1h\).

Solution The reaction rate at \(t=1\) is \(M^\prime\) (1). \[ \begin{eqnarray*} M^\prime (1) &=&\lim\limits_{t\rightarrow 1}\dfrac{M( t) -M(1) }{t-1}=\lim\limits_{t\rightarrow 1}\dfrac{( 4.5-0.03t^{2}) -( 4.5-0.03) }{t-1}\\[5pt] &=&\lim\limits_{t\rightarrow 1}\dfrac{-0.03t^{2}+0.03}{t-1} =\lim\limits_{t\rightarrow 1}\dfrac{( -0.03) ( t^{2}-1 ) }{t-1}=\lim\limits_{t\rightarrow 1}\dfrac{( -0.03) ( t-1) ( t+1) }{t-1}\\[5pt] &=&( -0.03) (2) =-0.06\\[-20pt] \end{eqnarray*} \]

The reaction rate at \(t=1\) h is \(-0.06\)g/h. That is, the mass \(M\) of glucose at \(t=1\) h is decreasing at the rate of 0.06g/h.