Exponential Growth and Decay

Exponential Growth and Decay exercise

• Define exponential growth and decay.

  Hints

  Just look at the basis of the power:

  • $1+r>1$
  • $1-r<1$

  If you raise a number greater than $1$ by a positive power you get a value greater than $1$ as well.

  Look at the example, consider $x_t=10\times 1.5^{t}$.

  Here we have

  • $x_1=15$
  • $x_2=22.5$
  • ...

  Solution

  Here you see the formula for an exponential process. We distinguish between exponential growth and exponential decay.

  Let's have a look at the basis of the power, $(1\pm r)$. What does this mean?

  For $(1+r)$ we have a basis greater than $1$ and thus a power which is greater than $1$ as well. So $x_t$ is bigger than $x_0$. We can then conclude that we have an exponential growth. $(1+r)$ is called the growth factor.

  By the same argument we can determine an exponential decay for $(1-r)$. Here $(1-r)$ is called the decay factor.

  What's the meaning of the other terms? The other terms are independent of exponential growth or decay.

  • $\frac tn$ is the growing (or decaying) time in days by one growth (or decay) factor.
  • $x_t$ is the size after $t$ days.
  • $x_0$ is the original size, the starting size.

• Determine the growth of Bevo's horns.

  Hints

  Just put the given value for $t$ in the growth formula.

  To round a decimal number, like $3.1415$ for example, proceed as follows:

  • Look at the third decimal position; e.g. $1$ for $3.1415$.
  • If this is less than $5$ you have to round down.
  • Otherwise you have to round up.

  This gives us $3.1415\approx 3.14$.

  For $t=100$, we have

  $x_{100}=4\times 1.25^{\large \left(\frac {100}{10}\right)}=37.2529...\approx 37.25$.

  Solution

  We can see all solutions in the table to the right.

  How do we get these solutions?

  We put each $t$-value in the formula $x_t=4\times 1.25^{\large \left(\frac t{10}\right)}$:

  • $t=10$ gives us $x_{10}=4\times 1.25^{\large \left(\frac {10}{10}\right)}=5$.
  • $t=20$ gives us $x_{20}=4\times 1.25^{\large \left(\frac {20}{10}\right)}=6.25$.
  • $t=30$ gives us $x_{30}=4\times 1.25^{\large \left(\frac {30}{10}\right)}=7.8125\approx 7.81$.
  • $t=40$ gives us $x_{40}=4\times 1.25^{\large \left(\frac {40}{10}\right)}=9.765625\approx 9.77$.
  • $t=50$ gives us $x_{50}=4\times 1.25^{\large \left(\frac {50}{10}\right)}=12.2070...\approx 12.21$.

• Calculate the size of Bevo's horns.

  Hints

  Put the given value for $t$ in the formula

  $x_t=4\times (1+2.5)^{\large \left(\frac t{10}\right)}$.

  Have a look at the following example for rounding $2.71828$:

  • Look at the third decimal position; e.g. $8$.
  • If this is less than $5$ you have to round down.
  • Otherwise you have to round up.

  So $2.71828\approx 2.72$.

  For $t=60$:

  $x_{60}=4\times 3.5^{\large \left(\frac {60}{10}\right)}=7353.0625...\approx 7353.06$.

  Solution

  To get the values for $x_t$, put each $t$ into the formula $x_t=4\times 3.5^{\left(\frac t{10}\right)}$:

  • $t=10$ gives us $x_{10}=4\times 3.5^{\large \left(\frac {10}{10}\right)}=14$.
  • $t=20$ gives us $x_{20}=4\times 3.5^{\large \left(\frac {20}{10}\right)}=49$.
  • $t=30$ gives us $x_{30}=4\times 3.5^{\large \left(\frac {30}{10}\right)}=171.5$.
  • $t=40$ gives us $x_{40}=4\times 3.5^{\large \left(\frac {40}{10}\right)}=600.25$.
  • $t=50$ gives us $x_{50}=4\times 3.5^{\large \left(\frac {50}{10}\right)}=2100.875\approx 2100.88$.

• Decide if either exponential growth or decay is given.

  Hints

  The factor for exponential decay (or growth) is less than (or greater than) $1$.

  If we have a growth of $30\%$, we get the following growth factor: $1+30\%=1+0.3=1.3$.

  Solution

  For all the examples above we have to determine the growth, or decay, factor as well as the original size $x_0$.

  Water lilies - an exponential growing process:

  • $1+r=1+0.2=1.2$
  • $x_0=2$
  • $x_t=2\times 1.2^t$, where $t$ stands for the time in months.

  Trees - an exponential decaying process:

  • $1-r=1-0.15=0.85$
  • $x_0=2$
  • $x_t=2\times 0.85^t$, where $t$ stands for the time in years.

  Radiation - an exponential decaying process:

  • $1-r=1-0.2=0.8$
  • $x_0=2$
  • $x_t=2\times 0.8^t$, where $t$ stands for the time in $10$ years.

  Cellular - an exponential growing process:

  • $1+r=1+1=2$
  • $x_0=2$
  • $x_t=2\times 2^t$, where $t$ stands for the time in periods.

• Establish the formula for Bevo's shrinking horns.

  Hints

  What's the meaning of the terms in the formula?

  • $\frac tn$ is the decaying time by one decay factor in days.
  • $x_t$ is the size after $t$ days.
  • $x_0$ is the original size, the starting size.

  Keep in mind that you have to subtract the decay rate from $1$.

  Distinguish between the original size, that's the size at the beginning, and the size at $t$, which must be smaller than the original size.

  Solution

  We already know the formula for exponential decay, $x_t=x_0\times (1-r)^{\left(\frac t{10}\right)}$, so all we need to do is put in our known values.

  The original size of the horns is 49 feet, so $x_0=49$.

  The size at $t$ is unknown.

  To get the decay factor we have to subtract the rate $r=0.37$ from $1$ to get $1-0.37=0.63$.

  Putting all these values into the exponential decay equation, we get:

  $x_t=49\times (1-0.37)^{\left(\frac t{10}\right)}=49\times 0.63^{\left(\frac t{10}\right)}$.

• Decide the corresponding exponential function.

  Hints

  Put in the corresponding $x$-value in the respective function equation.

  You can find the starting value for each function; i.e. $f(0)$, $g(0)$, and $h(0)$.

  Check each point $(x,y)$. It must satisfy either $y=f(x)$, $y=g(x)$, or $y=h(x)$.

  Solution

  To check if any point belongs to a given function, put the $x$ value of the point into the function.

  Let's start with $x=0$:

  • $f(0)=3\times 1.2^0=3$ $\rightarrow$ $(0,3)$
  • $g(0)=4\times 0.9^0=4$ $\rightarrow$ $(0,4)$
  • $h(0)=5\times 0.5^0=5$ $\rightarrow$ $(0,5)$

  Similarly, you can check all the other points. For example, with $x=4$:

  • $f(4)=3\times 1.2^4\approx 6.2$ $\rightarrow$ $(4,6.2)$
  • $g(4)=4\times 0.9^4\approx 2.6$ $\rightarrow$ $(4,2.6)$
  • $h(4)=5\times 0.5^4\approx 0.3$ $\rightarrow$ $(4,0.3)$

  The given points for $x=4$ are $(4,1.5)$ and $(4,7.5)$. So, both points don't belong to any of the three functions.