Problem 257 - Calculus 3

Question

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

Step-by-Step Solution

Answer

The temperature increases at a rate of 2°C per second after 3 seconds.

Step 1: Determine the Fly's Position at t = 3

To find the fly's position after 3 seconds, substitute \( t = 3 \) into the given path equations: \( x = \sqrt{1 + t} \) and \( y = 2 + \frac{1}{3}t \).\[ x = \sqrt{1 + 3} = \sqrt{4} = 2 \] \[ y = 2 + \frac{1}{3}\cdot 3 = 2 + 1 = 3 \] Therefore, the fly's position at \( t = 3 \) seconds is \( (2, 3) \).

Step 2: Write the Temperature at a Given Position

The temperature at a point \((x, y)\) is given by the function \( T(x, y) \). The rate at which the temperature changes at a point \((x, y)\) can be found using partial derivatives: \( T_x \) and \( T_y \).

Step 3: Calculate the Rate of Temperature Change Using the Chain Rule

The rate of temperature change with respect to time, \( \frac{dT}{dt} \), can be calculated using the chain rule:\[ \frac{dT}{dt} = T_x \frac{dx}{dt} + T_y \frac{dy}{dt} \]

Step 4: Find dx/dt and dy/dt at t = 3

Differentiate the path equations with respect to \( t \):\[ \frac{dx}{dt} = \frac{1}{2\sqrt{1+t}} \] \[ \frac{dy}{dt} = \frac{1}{3} \] Evaluate at \( t = 3 \):\[ \frac{dx}{dt} = \frac{1}{2\sqrt{4}} = \frac{1}{4} \] \[ \frac{dy}{dt} = \frac{1}{3} \]

Step 5: Calculate dT/dt at t = 3

Use the values \( T_x(2,3) = 4 \), \( T_y(2,3) = 3 \), \( \frac{dx}{dt} = \frac{1}{4} \), and \( \frac{dy}{dt} = \frac{1}{3} \) into the chain rule formula:\[ \frac{dT}{dt} = 4 \times \frac{1}{4} + 3 \times \frac{1}{3} = 1 + 1 = 2 \]

Key Concepts

Partial Derivatives

In calculus, partial derivatives are a way to measure how a function changes as its variables change. Unlike regular derivatives which deal with functions of one variable, partial derivatives involve functions with multiple variables. It's like looking at the slope of a hill if you only care about walking on one axis, say north-south, or east-west.

• Consider a function of two variables, like temperature depending on position \(T(x, y)\).
• To find how the temperature changes as you move along the x-direction, while keeping y constant, you use the partial derivative with respect to x, written as \(T_x\).
• Similarly, the partial derivative with respect to y, \(T_y\), tells you how temperature changes when moving in the y-direction.

In our problem, we have that \(T_x(2,3) = 4\), which means at point (2,3), a small step in the x-direction changes the temperature by 4 degrees Celsius per centimeter.
Similarly, \(T_y(2,3) = 3\) means a step in the y-direction results in a 3 degrees change per centimeter. These are neither total changes nor over time, but rather instantaneous rates at that point along their respective axes.

Calculus Problem Solving

Calculus problems, like the one involving the fly and temperature change, often require combining multiple calculus concepts. Here, the chain rule is particularly useful. The chain rule lets you find the rate of change of a function relative to a variable that indirectly causes the change. This problem specifically illustrates such an application.

• Identify the direct functions of time, here it's \(x = \sqrt{1+t}\) and \(y = 2 + \frac{1}{3}t\).
• Using these, find how fast x and y change over time, giving \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
• The chain rule: \[ \frac{dT}{dt} = T_x \frac{dx}{dt} + T_y \frac{dy}{dt} \] gives the overall temperature change rate.

Thus, coupling the partial derivatives of temperature with the rates of \(x\) and \(y\) over time, yields the desired rate long the path of the fly's movement, showing just how fundamental calculus techniques like derivatives and the chain rule are to problem-solving.

Temperature Change Rate

In physics and math, the rate of change of temperature often connects fluidly through real-world situations. For this example, the question specifically investigates how quickly a fly experiences a change in temperature along its path. This rate is vital for understanding dynamic systems where variables interact and change together.

• We begin by confirming the fly's position at any time, here \(t=3\) seconds, using its path formulas \(x = \sqrt{1+t}\) and \(y = 2 + \frac{1}{3}t\).
• Knowing the position allows for calculating the specific partial derivatives at that point, providing a contextual rate of change \(T_x(2,3) = 4\) and \(T_y(2,3) = 3\).
• Finally, the temperature change rate there is \( \frac{dT}{dt} = 2\) Celsius per second, meaning at \(t = 3\), the temperature increases by 2 degrees Celsius each second along the fly's path.

Capturing how quickly conditions like temperature change is essential in many fields, from meteorology to cooking, demonstrating the wide applications of these calculus strategies beyond theoretical exercises.