Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

If the temperature of a plate at the point \((x, y)\) is \(T(x, y)=10+x^{2}-y^{2}\), find the path a heat-seeking particle (which always moves in the direction of greatest increase in temperature) would follow if it starts at \((-2,1)\). Hint: The particle moves in the direction of the gradient $$ \nabla T=2 x \mathbf{i}-2 y \mathbf{j} $$ We may write the path in parametric form as $$ \mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j} $$ and we want \(x(0)=-2\) and \(y(0)=1\). To move in the required direction means that \(\mathbf{r}^{\prime}(t)\) should be parallel to \(\nabla T\). This will be satisfied if $$ \frac{x^{\prime}(t)}{2 x(t)}=-\frac{y^{\prime}(t)}{2 y(t)} $$ together with the conditions \(x(0)=-2\) and \(y(0)=1\). Now solve this differential equation and evaluate the arbitrary constant of integration.

Step-by-Step Solution

Answer

The path is parametrically defined by \( x(t) = -2 e^{2kt} \) and \( y(t) = e^{-2kt} \).

Step 1: Write Down the Gradient

First, let's write down the expression for the gradient of the temperature function given by \(T(x, y)=10+x^{2}-y^{2}\). We are already provided the gradient: \( abla T = 2x\mathbf{i} - 2y\mathbf{j} \).

Step 2: Express the Path Parametrically

The path of the particle can be described parametrically as \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \). Our initial condition is \( x(0) = -2 \) and \( y(0) = 1 \).

Step 3: Set Up the System of Differential Equations

The direction \( \mathbf{r}^{\prime}(t) \) is parallel to \( abla T \), meaning: \( \frac{x^{\prime}(t)}{2x(t)} = -\frac{y^{\prime}(t)}{2y(t)} \). Let's denote \( x'(t) = 2x(t)k \) and \( y'(t) = -2y(t)k \) for some scalar \( k \), which implies proportionality.

Step 4: Solve the Differential Equations

From the relations \( x'(t) = 2x(t)k \) and \( y'(t) = -2y(t)k \), divide through by \( x(t) \) and \( y(t) \) respectively to integrate:\[ \frac{x'(t)}{x(t)} = 2k, \quad \int \frac{x'(t)}{x(t)} dt = \int 2k dt \]This gives \( \ln |x(t)| = 2kt + C_1 \).For \( y(t) \):\[ \frac{y'(t)}{y(t)} = -2k, \quad \int \frac{y'(t)}{y(t)} dt = \int -2k dt \]Which results in \( \ln |y(t)| = -2kt + C_2 \).

Step 5: Exponentiate and Use Initial Conditions

Exponentiate the above equations to solve for \( x(t) \) and \( y(t) \):\[ x(t) = e^{2kt + C_1} = e^{C_1} e^{2kt} \]\[ y(t) = e^{-2kt + C_2} = e^{C_2} e^{-2kt} \]Using initial conditions, where \( x(0) = -2 \) and \( y(0) = 1 \), we find:\[ -2 = e^{C_1} \quad \Rightarrow \quad e^{C_1} = -2 \]\[ 1 = e^{C_2} \quad \Rightarrow \quad e^{C_2} = 1 \].

Step 6: Write the Final Parametric Equations

Incorporating the constants into the equations for \( x(t) \) and \( y(t) \), we obtain:\[ x(t) = -2 e^{2kt} \]\[ y(t) = e^{-2kt} \].This describes the path of our heat-seeking particle over time, with \( k \) being a parameter related to its speed.

Key Concepts

Parametric Equations

Parametric equations allow us to describe a path or a curve in terms of a parameter, often denoted as \( t \). In our exercise, the position of a heat-seeking particle on a plane is represented by the parametric equations for \( x(t) \) and \( y(t) \). This means instead of simply using \( x \) and \( y \) directly, we define each coordinate as a function of \( t \), such as \( x(t) = -2e^{2kt} \) and \( y(t) = e^{-2kt} \).

This approach is helpful because it provides a way to track how the particle's path evolves over time. Parametric equations can handle more complex paths than traditional Cartesian coordinates, as we can control direction and speed through the parameter \( t \). Utilizing initial conditions, such as \( x(0) = -2 \) and \( y(0) = 1 \), allows us to specify where the particle starts, thereby fully defining its trajectory from that point.

Differential Equations

A differential equation relates functions with their derivatives, providing a language for describing the behavior of dynamic systems. In our case, we're given a system that dictates the movement of the particle in relation to the temperature gradient: \( \frac{x'(t)}{2x(t)} = -\frac{y'(t)}{2y(t)} \).

This equation implies that the rate of change of \( x(t) \) and \( y(t) \) are proportional to each other but in opposite directions. Solving these equations involves integrating to find functions for \( x(t) \) and \( y(t) \).

The solution reveals how each component of the particle's path changes over time, which is essential to determining the particle's trajectory. Differential equations are crucial in fields like physics and engineering, where they model how changes in one variable affect another over time.

Temperature Gradient

The temperature gradient is a vector that points in the direction of the greatest rate of increase of a function, in this case, temperature. It is derived from the function \( T(x, y) = 10 + x^2 - y^2 \), resulting in the gradient vector \( abla T = 2x\mathbf{i} - 2y\mathbf{j} \).

This vector field represents the direction and rate at which temperature changes the fastest. For the heat-seeking particle, this gradient acts as a compass, guiding it towards hotter regions of the plate. The direction is crucial because it indicates where the particle should move to achieve the fastest increase in temperature.

The magnitude of the gradient can also give the rate of temperature rise. In practical applications, understanding gradients allows us to interpret phenomena such as heat flow, which is essential in thermodynamics and climate science.

Initial Conditions

Initial conditions specify the state of a system at the start of an observation or measurement. For the heat-seeking particle, the initial conditions are given by \( x(0) = -2 \) and \( y(0) = 1 \). These conditions anchor the solution of differential equations, ensuring the resulting curve passes through a specific point at \( t = 0 \).

Using these values, we calculate the constants of integration in the parametric equations \( x(t) \) and \( y(t) \). Solving for these constants provides a unique solution that describes the particular behavior of our particle from its starting position.

Initial conditions are vital because they ensure the mathematical model reflects the real-world scenario accurately. They allow for personalized solutions that accommodate specific situations, making them indispensable in simulations and predictive models across various disciplines such as meteorology and mechanical systems design.