Problem 35 - Calculus Early Transcendentals

Question

The temperature at a point \((x, y)\) is \(T(x, y),\) measured in degrees Celsius. A bug crawls so that its position after \(t\) seconds is given by \(X=\sqrt{1+t}, y=2+\frac{1}{3} t,\) where \(x\) and \(y\) are measured in centimeters. The temperature function satisfies \(T_{x}(2,3)=4\) and \(T_{y}(2,3)=3 .\) How fast is the temperature rising on the bug's path after 3 seconds?

Step-by-Step Solution

Answer

The temperature rises at 2°C per second when t=3.

Step 1: Understand the Problem

We need to find how fast the temperature is rising along the bug's path as it moves according to its position given by functions of time.

Step 2: Parametrize the Temperature Change

To determine how fast the temperature changes, we'll use the concept of the total derivative: \(\frac{dT}{dt} = T_x \frac{dx}{dt} + T_y \frac{dy}{dt}\). This expression considers the partial rates of change of temperature with respect to \(x\) and \(y\) alongside how \(x\) and \(y\) change with respect to time \(t\).

Step 3: Compute Derivatives of Position

Find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). We have \(x=\sqrt{1+t}\) and \(y=2+\frac{1}{3}t\). These give:\[\frac{dx}{dt} = \frac{1}{2\sqrt{1+t}}\]\[\frac{dy}{dt} = \frac{1}{3}\]

Step 4: Evaluate Derivatives at t=3

Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) at \(t=3\):\[\frac{dx}{dt}\Bigg|_{t=3} = \frac{1}{2\sqrt{1+3}} = \frac{1}{4}\]\[\frac{dy}{dt}\Bigg|_{t=3} = \frac{1}{3}\]

Step 5: Evaluate Position at t=3

Calculate the bug's position at \(t=3\):\[x(3) = \sqrt{1+3} = 2\]\[y(3) = 2 + \frac{1}{3} \cdot 3 = 3\]

Step 6: Apply Total Derivative Formula

Substitute the computed rates and known partial derivatives into the total derivative formula:\[\frac{dT}{dt} = T_x(2,3) \cdot \frac{dx}{dt} + T_y(2,3) \cdot \frac{dy}{dt}\]Substitute values: \[\frac{dT}{dt} = 4 \times \frac{1}{4} + 3 \times \frac{1}{3}\]Compute:\[\frac{dT}{dt} = 1 + 1 = 2 \]

Step 7: Conclusion

The rate at which the temperature is rising along the bug's path at \(t = 3\) seconds is \(2\) degrees Celsius per second.

Key Concepts

Partial Derivatives

Partial derivatives are a key concept in calculus that involves taking the derivative of a function with respect to one variable while holding other variables constant. This concept is crucial when dealing with functions of multiple variables, like our temperature function here, which depends on both coordinates \(x\) and \(y\).

In the original exercise, the temperature function \(T(x, y)\) has partial derivatives \(T_x = \frac{\partial T}{\partial x}\) and \(T_y = \frac{\partial T}{\partial y}\). These represent the rate of change of temperature with respect to \(x\) and \(y\), respectively, at a fixed point. For the point \((2,3)\), we know:

• \(T_x(2, 3) = 4\)
• \(T_y(2, 3) = 3\)

These values tell us how the temperature changes in small increments along the \(x\) and \(y\) directions at that specific point.

Grasping partial derivatives is essential for understanding how functions behave in space. They're like the pieces of a puzzle, helping us see how changes in each individual variable contribute to the overall change in a function's value.

Total Derivative

The total derivative is used to find the overall change of a function as its variables change, especially when these variables are functions of other variables, like time. In our exercise, \(T(x, y)\) changes with respect to time because both \(x\) and \(y\) depend on \(t\).

To compute this, we use the formula:\[\frac{dT}{dt} = T_x \frac{dx}{dt} + T_y \frac{dy}{dt}\]This formula is derived from the chain rule. It allows us to see how the temperature changes as both \(x\) and \(y\) change with time. It takes into account the rate at which \(x\) and \(y\) themselves are changing (\(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)) and how those changes affect \(T\).

In the step-by-step solution, we noted that the total derivative also required calculating these specific rates of change for \(x\) and \(y\) at \(t=3\). By plugging all the values into the formula, we determined how fast the temperature is rising at that instant. This illustration is central for understanding dynamic systems where multiple variables interact over time.

Parametrization

Parametrization is another vital mathematical tool used to express variables in terms of a parameter, often time \(t\), to simplify the analysis of complex systems.

In the problem at hand, the positions \(x\) and \(y\) of the bug are parametrized by time \(t\), given by the functions:

• \(x = \sqrt{1+t}\)
• \(y = 2 + \frac{1}{3}t\)

This framework provides a clear view of how \(x\) and \(y\) evolve over time, transforming the problem of tracking the bug's path into manageable mathematical functions.

Understanding parametrization is helpful not only in tracking movement but also in fields like computer graphics and physics. It's a strategy for transforming complicated expressions involving multiple variables into simpler, single-variable functions, easing computation and analysis.