Problem 86 - Brief Applied Calculus

Question

It is well known that the hearts of smaller animals beat faster than the hearts of larger animals. The actual relationship is approximately $$ \text { (Heart rate) }=250 \text { (Weight) }^{-1 / 4} $$ where the heart rate is in beats per minute and the weight is in pounds. Use this relationship to estimate the heart rate of: A 625 -pound grizzly bear.

Step-by-Step Solution

Answer

The heart rate of a 625-pound grizzly bear is 50 beats per minute.

Step 1: Identify the Formula

We are given a formula that relates heart rate to weight: \( \text{Heart rate} = 250 \times (\text{Weight})^{-1/4} \). We will use this formula to find the heart rate of the bear.

Step 2: Substitute the Bear's Weight

The weight of the grizzly bear is given as 625 pounds. Substitute this value into the formula: \( \text{Heart rate} = 250 \times (625)^{-1/4} \).

Step 3: Calculate the Expression \((625)^{-1/4}\)

To solve \((625)^{-1/4}\), note that the fourth root of 625 is 5, so \(625^{1/4} = 5\). Therefore, \((625)^{-1/4} = 1/5\).

Step 4: Evaluate the Heart Rate

Now, substitute \((625)^{-1/4} = 1/5\) back into the equation: \( \text{Heart rate} = 250 \times 1/5 \). This results in \( \text{Heart rate} = 250/5 = 50 \).

Step 5: Conclude the Heart Rate

The heart rate of a 625-pound grizzly bear is calculated to be 50 beats per minute.

Key Concepts

Heart Rate Calculation

Heart rate calculation using mathematical models is a fascinating intersection between biology and mathematics. This particular model is intriguing because it involves using an animal's weight to estimate its heart rate. In the given exercise, we deal with the formula: \[ \text{Heart rate} = 250 \times (\text{Weight})^{-1/4} \]This formula tells us that the heart rate is a function of the weight raised to the negative quarter power. To understand how to use this, we'll explore the given example of a 625-pound grizzly bear.
We first substitute the bear's weight into the equation and transform the mathematical expression to calculate the heart rate:

• Substitution: Plug in 625 pounds for the weight.
• Computation: Find \((625)^{-1/4} \). The fourth root of 625 is 5, hence \((625)^{-1/4} = 1/5 \).
• Final Calculation: Multiply 250 by the computed value to get 50 beats per minute.

Understanding these computations is essential to solving similar problems and appreciating how mathematics models biological phenomena.

Mathematical Modeling

Mathematical modeling is a powerful tool that lets us use equations to represent real-world phenomena, like calculating the heart rate from the size of an animal in this exercise. Through modeling, we take variables like weight and express their interplay using mathematical functions. In our formula, the relationship between heart rate and weight is shown using a specific power function: \[ \text{Heart rate} = 250 \times (\text{Weight})^{-1/4} \] This model illustrates how smaller animals typically have faster heart rates by inversely relating the weight. Such models help us predict outcomes with reasonable accuracy. Additionally, they form the basis for further exploration and understanding of complex systems. By using this model, we are not just finding answers to isolated problems but are tapping into a system of equations that can describe a wider range of scenarios.
As students explore these formulas, they develop skills that are crucial for sciences like biology and bioengineering, where modeling plays a pivotal role in research and applications.

Power Functions

Power functions are integral in mathematical modeling due to their capability to describe various phenomena, including scaling relationships like heart rate to weight. A power function typically takes the form \( f(x) = ax^n \), where \( a \) and \( n \) are constants. In our exercise, the function:
\[ \text{Heart rate} = 250 \times (\text{Weight})^{-1/4} \]has the weight raised to the negative quarter power, which reflects the specific biological scaling of heart rates.
This means that the heart rate decreases as the weight increases, following an inverse power law. Such relationships are common when dealing with physiology across various species.

• The constant 250 in the function acts as a scaling factor and adjusts the output to match observed data.
• The exponent \( -1/4 \) determines how sensitively the heart rate responds to changes in weight.

Power functions like these are invaluable for highlighting underlying patterns and establishing empirical laws in nature. As students work with these functions in calculus, they gain insights into how mathematical principles elucidate real-world relationships.