The number $N$ of E. coli bacteria is modelled as growing exponentially so that, at time t minutes, the number of bacteria present is given by $N=10000e^{kt}$, where $k$ is a constant. After $10$ minutes there are $15000$ bacteria present.
What is the rate of growth at $10$ minutes? Give your answer to 2 significant figures.
I am a GCSE student attempting this question. So far it has stumped me and I have gained many answers all of which have been rejected by the automatic answer checker and I have no idea how to proceed. Any help would be appreciated.
I have already worked out that $k= 0.04054651081$ $e^{10k}$ is also equal to $1.5$
What you want to do is find the slope at $t=10$, which is equal to rate of growth in $\text{bacteria}/\text{min}$.
You can do this several ways.
You can do this using differentiation to obtain the exact rate. You have $N=10000e^{kt}$, which you can differentiate with respect to $t$ to obtain $\frac{dN}{dt}=10000ke^{kt}$ from the chain rule. Substituting for $t=10$ gives you the rate $\frac{dN}{dt}$ at that time.
A way to do this without differentiation (approximate though) is by selecting values very close to $t=10$ and using the fact that the slope is given by $$m=\frac{\Delta N}{\Delta t}=\frac{N_2-N_1}{t_2-t_1} \tag{1}$$
For example, you can select $t_1=9.999$ and $t_2=10.001$ and evaluate the corresponding values of $N_1$ and $N_2$.
Another way to do this (Even more approximate) is by simply drawing the graph and draw a tangent line to the graph at $t=10$. You can evaluate its slope by using the same as equation $(1)$.