Question

Tortoise growth In a study conducted at University of New Mexico, it was found that the mass (weight) of juvenile desert tortoises exhibited exponential growth after a diet switch. One of these tortoises had a mass of about \(64 \mathrm{g}\) at the time of the diet switch, and 33 days later the mass was 73 g. How many days after the diet switch did the tortoise have a mass of \(100 \mathrm{g} ?\)

Short answer

Answer: Approximately 55 days.

Step-by-step solution

Exponential Growth Equation

An exponential growth equation typically follows the form: \(y(t) = y_0 \cdot (1 + r)^t\), where \(y(t)\) represents the mass at a certain time, \(y_0\) represents the initial mass, \(r\) represents the rate of growth, and \(t\) represents the time (in this case, days).

Plugging in Known Values

We know the initial mass (\(y_0 = 64g\)), the mass after 33 days (\(y(33) = 73g\)), and the time it took to reach that mass (\(t = 33\) days). We can plug these values into the exponential growth equation: \(73 = 64 \cdot (1 + r)^{33}\).

Solving for the Rate of Growth (r)

Now, we need to solve for the rate of growth (\(r\)) in the equation: \(73 = 64 \cdot (1 + r)^{33}\). To do this, follow these steps: 1. Divide both sides of the equation by \(64\) to isolate the exponential term: \(\frac{73}{64} = (1 + r)^{33}\). 2. Take the 33rd root of both sides to remove the exponent: \(\sqrt[33]{\frac{73}{64}} = 1 + r\). 3. Subtract 1 from both sides to isolate \(r\): \(r = \sqrt[33]{\frac{73}{64}} - 1 \approx 0.01024\). So, the rate of growth (\(r\)) is approximately \(1.024\%\).

Finding the Time to Reach a Mass of \(100g\)

We want to find the number of days (\(t\)) it takes for the tortoise's mass to reach \(100g\). We use the exponential growth equation with the known rate of growth (\(r = 0.01024\)) and the target mass (\(y(t) = 100g\)): \(100 = 64 \cdot (1 + 0.01024)^t\). Follow these steps to solve for \(t\): 1. Divide both sides of the equation by \(64\): \(\frac{100}{64} = (1 + 0.01024)^t\). 2. Take the natural logarithm (ln) of both sides to remove the exponent: \(\ln(\frac{100}{64}) = t\cdot \ln(1 + 0.01024)\). 3. Divide by the natural logarithm of \((1 + 0.01024)\) to isolate \(t\): \(t = \frac{\ln(\frac{100}{64})}{\ln(1 + 0.01024)} \approx 54.75\). Since the time (\(t\)) must be an integer (as we are counting the number of days), we can round up to the nearest whole number: \(t \approx 55\) days.

Final Answer

The tortoise will have a mass of \(100g\) approximately 55 days after the diet switch.

Key concepts

Rate of Growth

The rate of growth is a crucial concept when we talk about how quickly something like an animal, plant, or even money increases over time. In the context of the tortoise problem, the rate of growth is denoted by the variable \( r \) and represents how fast the tortoise's mass grows each day after the diet switch. It is expressed as a percentage.To determine the rate of growth, we use an equation derived from the exponential growth formula. We first isolate the exponential term by dividing the mass after 33 days by the initial mass:\[ \frac{73}{64} = (1 + r)^{33} \]. By taking the 33rd root of both sides, we eliminate the exponent; this is a common technique to solve exponential equations. Once isolated, \( r \) can be found by subtracting 1 from the result. In this example, \( r \) was calculated to be approximately 0.01024, or 1.024%. This means that the tortoise's mass increases by about 1.024% per day.Knowing the rate of growth helps us understand the pace at which changes occur and allows us to predict when the tortoise will reach a certain mass, given enough time.

Exponential Growth Equation

Exponential growth is a pattern of increase where the amount grows by a constant percentage over equal time intervals. The exponential growth equation is given by:

• \(y(t) = y_0 \cdot (1 + r)^t\)

Here, \(y(t)\) is the value at time \(t\), \(y_0\) is the initial value, \(r\) is the rate of growth, and \(t\) is the time.In many everyday problems, including the growth of the tortoises, this equation helps describe the situation where an initial quantity increases at a rate proportional to its current size. It is particularly useful because it accounts for continuous growth instead of linear growth.By applying this formula to our problem, we can predict future size based on current growth trends. For example, given an initial mass of \(64 \text{g}\) and a growth rate \(r\), we can use the equation to determine when the tortoise will reach a mass of \(100 \text{g}\). Such information is valuable not just corporately, but in a broad range of scientific studies exploring population growth, finance, and biological processes.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is an essential mathematical concept that plays a significant role in solving exponential growth problems. It is the inverse of the exponential function, meaning if you apply the natural logarithm to an exponential expression, it helps you solve for the exponent.In the tortoise growth problem, once the equation \( \frac{100}{64} = (1 + 0.01024)^t \) was set up to find the time when the mass reaches \(100\text{g}\), taking the natural logarithm of both sides allows us to bring the exponent \(t\) down from the exponential function. This makes it easier to solve for \(t\):

• \( \ln\left(\frac{100}{64}\right) = t \cdot \ln(1 + 0.01024) \)

By isolating \(t\) using division, \( \ln \) gives us the ability to compute exact values that might otherwise be cumbersome.Understanding natural logarithms not only empowers you to handle exponential equations but extends to numerous applications such as calculating time for investments in finance or natural processes like cooling or heating.