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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 $1.5$

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    Do you know how to [differentiate](https://en.wikipedia.org/wiki/Derivative) functions? – 2017-02-28
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    Is it calculus because I have just begun learning that as a part of further maths gcse? – 2017-02-28
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    @Matt Yes, it is part of calculus. Did you get this question on further maths? – 2017-02-28
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    No it is a pre-workshop question for an a-level maths/physics workshop I am planning to go to. – 2017-02-28
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    So the answer would be 10000*kte^kt-1 – 2017-02-28
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    @Matt No it would not. Note that $\frac{d}{dx} x^n=nx^{n-1}$ only applies for polynomial functions. – 2017-02-28
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    So how do I differentiate an exponential function? – 2017-02-28
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    @Matt I will post an answer explaining this. – 2017-02-28
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    I am not quite sure why but brief research suggests it is 10000*k*e^(kt) – 2017-02-28
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    @Matt Yes, that's right! This is correct due to the [chain rule](https://en.wikipedia.org/wiki/Chain_rule). Does this work when substituting $k$ onto your automatic answer checker? – 2017-02-28
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    Is k what I need to find? – 2017-02-28
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    Also would you be able to explain why the exponential differentiation results in that equation? – 2017-02-28
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    I just did 10000*k*e^(kt) and got 608.xxx It wants it to 2 sig figs but told me 608 was wrong. – 2017-02-28
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    Does the fact that the unit is per minute (min^-1) change the answer in some way? – 2017-02-28
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    @Matt $608$ to 2 significant figures is $610$. Does this work? – 2017-02-28
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    I can't believe I didn't think of that! It works and I am feeling rather idiotic. Thanks for all your help. – 2017-02-28
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    @Matt No problem. I've posted an answer containing several methods you can use to solve this problem since differentiation may not be the method they are trying to get you to use. – 2017-02-28

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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 $(1)$.

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    I originally, about 1 hour ago, tried the question using 10.5 and 9.5 (one minute difference so like the per minute part of the question)and then subtracting and gained 608.xxx but once again forgot to turn that into 610. – 2017-02-28