Question

At \(0^{\circ}\) Celsius, the heat loss \(H\) (in kilocalories per square meter per hour) from a person's body can be modeled by \(H=33(10 \sqrt{v}-v+10.45)\) where \(v\) is the wind speed (in meters per second). (a) Find \(\frac{d H}{d v}\) and interpret its meaning in this situation. (b) Find the rates of change of \(H\) when \(v=2\) and when \(v=5\)

Short answer

The derivative, \(\frac{d H}{d v}= 33 [5v^{-1/2} - 1]\), represents the rate of change of heat loss with respect to wind speed. When \(v=2\), \(\frac{d H}{d v} = 49.5\) indicating an increase in heat loss, whereas at \(v=5\), \(\frac{d H}{d v} = 0\) indicating no change in heat loss with the wind speed.

Step-by-step solution

Find the derivative

We are supposed to find \(\frac{d H}{d v}\), the derivative of heat loss with respect to wind speed. The given function is \(H=33(10\sqrt{v} - v+10.45)\). Applying the product and chain rules, we obtain: \(\frac{d H}{d v}= 33 [5v^{-1/2} - 1]\)

Interpretation of the derivative

\(\frac{d H}{d v}\) represents the rate at which the heat loss changes with respect to the wind speed. If \(\frac{d H}{d v}\) is positive, the heat loss increases with increasing wind speed. If \(\frac{d H}{d v}\) is negative, the heat loss decreases with increasing wind speed. The magnitude of \(\frac{d H}{d v}\) gives the speed at which this change happens

Rates of change at \(v=2\) and \(v=5\)

To determine the rate of change at specific wind speeds, substitute \(v=2\) and \(v=5\) into the derivative to get the slopes at those points. Thus, when \(v=2\) we get \(\frac{d H}{d v} = 33 [5(2)^{-1/2} - 1] = 33[5/2 - 1] = 33[1.5] = 49.5\). And when \(v=5\), \(\frac{d H}{d v} = 33 [5(5)^{-1/2} - 1] = 33[5/5 - 1] = 33[0] = 0\). This shows that at \(v=2\) the heat loss is increasing at a rate of 49.5 kilocalories per square meter per hour for each metre per second increase in wind speed. At \(v=5\) the heat loss does not change with the wind speed.

Key concepts

Derivative Application

Understanding the application of derivatives in physical contexts is a powerful tool in modeling and predicting phenomena. The process to find the derivative of a function, like the heat loss as a function of wind speed in the given exercise, is grounded in the principles of calculus. By taking the derivative of the heat loss function with respect to wind speed, we are able to calculate how sensitive heat loss is to changes in wind speed.

Essentially, the derivative, commonly denoted as \(\frac{dH}{dv}\), tells us the rate at which one quantity changes with respect to another. Performing this calculation allows us to make predictions and understand the relationship between these two quantities—the heat loss from a person's body and the wind speed, in this exercise. Through the derivative application, one becomes equipped to handle a multitude of real-world problems beyond textbook examples, such as estimating energy consumption or optimizing engineering designs.

Rate of Change

The concept of rate of change is fundamental in describing how a system evolves over time or in response to different conditions. In our scenario, it translates to how quickly heat loss from a person's body occurs as the wind speed changes. When we compute the derivative of the heat loss function with respect to wind speed and find it to be positive, as in the case when the wind speed is 2 meters per second, this indicates that heat loss is increasing as the wind gets faster.

A rate of change can be constant, or it can be variable. In the exercise, we witness a variable rate of change, which depends on the wind speed. Key takeaway: The value obtained from the derivative gives us the specific rate of heat loss at a particular wind speed. Examiner learners can use this understanding to interpret the behavior of differentiable functions in various scientific fields.

Wind Speed Effect on Heat Loss

Wind speed has a profound effect on heat loss from the body, a concept well known as wind chill. With our exercise, we are given a mathematical model that describes how heat loss is affected by wind speed. Knowing that wind can enhance the rate at which a body loses heat, the derivative helps us quantify that effect.

When the derivative \(\frac{dH}{dv}\) equals zero, as it does when the wind speed is 5 meters per second, it signifies a point of equilibrium where heat loss does not increase or decrease with a change in wind speed. This could mean that other factors have a more significant effect on heat loss at this point, or that we have reached a maximum or minimum rate of heat loss with respect to wind speed. Insights such as these are vital in fields such as meteorology, climatology, and even in designing clothing for extreme weather conditions.