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Consider the differential equation
\begin{align*}
\frac{d x}{d t} &= 3 t^{2} + 4 t - 3\\
x(0) &= 4
\end{align*}
We start by finding the indefinite integral of $3 t^{2} + 4 t - 3$, since that is the family of all functions whose derivatives equal $3 t^{2} + 4 t - 3$.
$\displaystyle \int 3 t^{2} + 4 t - 3\, dt =$
To determine which one of these functions $x$ is, we plug in our initial condition (don't simplify, and leave terms in the same order as they were in the integral, with the constant added at the end):
$=$
Solving for $C$ yields $C=$
.
Then $x(t) =$
.
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The volume of a cell is increasing at a rate of $\frac{10}{2 t + 5}$ $\mu {\rm m}^3$ per minute. If the volume at $t=0$ is $500$ $\mu {\rm m}^3$, find a function describing the volume of the cell and determine the volume of the cell after $10$ minutes.
Letting $v(t)$ denote the volume after $t$ minutes, the differential equation describing the volume is
\begin{align*}
\frac{d v}{d t} &= \frac{10}{2 t + 5} \\
v(0) &= 500
\end{align*}
We solve this by first finding the indefinite integral
$\displaystyle \int \frac{10}{2 t + 5}\, dt =$
When $t=0$, $2t+5$ is
, so we can ignore the absolute values and just consider
$v(t)=$
.
We can now plug in our initial condition to find $C$ (note: unless otherwise specified, use exact values rather than rounding):
$=$
So $C=$
.
Then the function describing the cell volume is $v(t) =$
, and the population after $10$ minutes will be $v(10) = $
$\approx$
$\mu {\rm m}^3$ (enter the exact answer in the first blank, and then round to $4$ significant figures in the second).
Hint
Type
log(stuff) to get $\log (stuff)$.
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A population of bacteria is growing at a rate of $2 e^{\frac{t}{20}}$ bacteria per hour. If the population after two hours is $300$, what will the population be after twelve hours?
Letting $p(t)$ denote the population after $t$ hours, write down the differential equation describing the situation:
$\frac{d p}{d t}=$
, with $p($
$)=$
.
To find the function describing the bacteria population, we start with the indefinite integral:
$\displaystyle \int 2 e^{\frac{t}{20}}\, dt =$
Next, plug in the initial condition:
$=$
Solving for $C$ gives us $C=$
.
Then the function describing the bacteria population is $p(t) =$
, and the population after twelve hours will be $p(12) = $
$\approx$
bacteria (enter the exact answer in the first blank, and then the nearest whole number in the second).
Hint
Use
exp(stuff) or
e^(stuff) for $e^{stuff}$.
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