*Mathematics This engineering resource asks the question: how can you predict future power requirements?*

*Mathematics This engineering resource asks the question: how can you predict future power requirements?*

Solution: Here there is no direct mention of differential equations, but use of the buzz-phrase ‘growing exponentially’ must be taken as indicator that we are talking about the situation $$f(t)=ce^$$ where here $f(t)$ is the number of llamas at time $t$ and $c,k$ are constants to be determined from the information given in the problem.

And the use of language should probably be taken to mean that at time $t=0$ there are $1000$ llamas, and at time $t=4$ there are $2000$.

And, in case you were wondering where you get to take a derivative here, the answer is that you don't really: all the ‘calculus work’ was done at the point where we granted ourselves that all solutions to that differential equation are given in the form $f(t)=ce^$.

First to look at some general ideas about determining the constants before getting embroiled in story problems: One simple observation is that $$c=f(0)$$ that is, that the constant $c$ is the value of the function at time $t=0$.

This resource could be used as a prelude to the ' Power Demand' activity.

requires students to form equations given a set of cards and to determine, with examples, whether the equation is always, sometimes or never true and to attempt to say why.(This is easier than using the bulkier more general formula for finding $c$).And use the formula for $k$: $$k== == $$ Therefore, we have $$f(t)=10\cdot e^=10\cdot 200^$$ as the general formula.This little section is a tiny introduction to a very important subject and bunch of ideas: solving differential equations.We'll just look at the simplest possible example of this.But it's still not so hard to solve for $c,k$: dividing the first equation by the second and using properties of the exponential function, the $c$ on the right side cancels, and we get $$=e^$$ Taking a logarithm (base $e$, of course) we get $$\ln y_1-\ln y_2=k(t_1-t_2)$$ Dividing by $t_1-t_2$, this is $$k=$$ Substituting back in order to find $c$, we first have $$y_1=ce^$$ Taking the logarithm, we have $$\ln y_1=\ln c t_1$$ Rearranging, this is $$\ln c=\ln y_1-t_1= $$ Therefore, in summary, the two equations $$y_1=ce^$$ $$y_2=ce^$$ allow us to solve for $c,k$, giving $$k=$$ $$c=e^$$ A person might manage to remember such formulas, or it might be wiser to remember the way of deriving them.A herd of llamas has 00$ llamas in it, and the population is growing exponentially. Write a formula for the number of llamas at arbitrary time $t$.Then, either repeating the method above or plugging into the formula derived by the method, we find $$c=\hbox = 1000$$ $$k==$$ $$== = (\ln 2)/4$$ Therefore, $$f(t)=1000\;e^=1000\cdot 2^$$ This is the desired formula for the number of llamas at arbitrary time $t$. At time $t=0$ it has $ bacteria in it, and at time $t=4$ it has 00$. Solution: Even though it is not explicitly demanded, we need to find the general formula for the number $f(t)$ of bacteria at time $t$, set this expression equal to 0,000$, and solve for $t$.Again, we can take a little shortcut here since we know that $c=f(0)$ and we are given that $f(0)=10$.Such a relation between an unknown function and its derivative (or derivatives) is what is called a differential equation.Many basic ‘physical principles’ can be written in such terms, using ‘time’ $t$ as the independent variable.

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