Problem 11 - Applied Calculus

Question

Ice is forming on a pond at a rate given by $$ \frac{d y}{d t}=\frac{\sqrt{t}}{2} \text { inches per hour, } $$ where \(y\) is the thickness of the ice in inches at time \(t\) measured in hours since the ice started forming. (a) Estimate the thickness of the ice after 8 hours. (b) At what rate is the thickness of the ice increasing after 8 hours?

Step-by-Step Solution

Answer

(a) The thickness is approximately 7.542 inches after 8 hours. (b) The rate is 1.414 inches per hour after 8 hours.

Step 1: Understand the Problem

You are given the rate of change of the ice thickness, \( \frac{dy}{dt} = \frac{\sqrt{t}}{2} \). You need to estimate the thickness at \( t = 8 \) hours and find the rate of thickness increase at that time.

Step 2: Integrate to Find Thickness Function

To find the thickness of the ice, \(y\), integrate the rate function. The rate function is given by \( \frac{dy}{dt} = \frac{\sqrt{t}}{2} \). Integrate the right side with respect to \( t \). \[ y(t) = \int \frac{\sqrt{t}}{2} \, dt \].

Step 3: Solve the Integral

The integral \( \int \frac{\sqrt{t}}{2} \, dt \) can be rewritten as \( \frac{1}{2} \int t^{1/2} \, dt \). Use the power rule for integration: \( \int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C \). Thus, \( \frac{1}{2} \times \frac{t^{3/2}}{3/2} + C = \frac{1}{3} t^{3/2} + C \), where \( C \) is the integration constant.

Step 4: Determine the Integration Constant

Assume at \( t = 0 \), the initial thickness \( y = 0 \), since no ice has formed yet. Substituting \( t = 0 \) and \( y = 0 \) into the equation \( y(t) = \frac{1}{3} t^{3/2} + C \), gives \( 0 = \frac{1}{3} (0)^{3/2} + C \). Thus, \( C = 0 \).

Step 5: Formulate the Thickness Function

With \( C = 0 \), the function simplifies to \( y(t) = \frac{1}{3} t^{3/2} \). This function represents the thickness of the ice at any time \( t \).

Step 6: Calculate Thickness After 8 Hours

Substitute \( t = 8 \) into \( y(t) = \frac{1}{3} t^{3/2} \). Compute \( y(8) = \frac{1}{3} (8)^{3/2} = \frac{1}{3} \times 22.627 = 7.542 \) inches, approximately.

Step 7: Calculate the Rate of Thickness Increase After 8 Hours

Use the original rate formula \( \frac{dy}{dt} = \frac{\sqrt{t}}{2} \) and substitute \( t = 8 \). Thus, \( \frac{dy}{dt} = \frac{\sqrt{8}}{2} = \frac{2.828}{2} = 1.414 \) inches per hour.

Key Concepts

Rate of Change

In calculus, the concept of "rate of change" is pivotal and allows us to understand how a quantity varies over time. In the context of this exercise, the rate of change represents how quickly the thickness of ice is growing on the pond. We are presented with the rate equation \( \frac{dy}{dt} = \frac{\sqrt{t}}{2} \), which indicates that the rate of change depends on the square root of time \( t \) divided by 2. This expression tells us that as time progresses, the rate at which ice accumulates initially increases because the square root of \( t \) grows with time.

The rate at any given time point, like at 8 hours in our exercise, can be found by substituting the time value into the rate equation. This provides us with a snapshot of how fast the quantity—in this case, ice thickness—is changing at that moment. This understanding is essential in predicting future states and making informed decisions based on dynamic conditions.

Integration

Integration is the mathematical process that allows us to determine the accumulated quantity, like the thickness of ice, from its rate of change. In our problem, we are given the rate of change, \( \frac{dy}{dt} \), and want to find the total thickness \( y(t) \) of the ice after a certain period. To do this, we integrate the rate equation \( \int \frac{\sqrt{t}}{2} \, dt \).

Using the power rule for integration, we derive that \( y(t) = \frac{1}{3} t^{3/2} + C \), where \( C \) is the constant of integration. The antiderivative tells us how the total quantity in question, here the ice thickness, is built up over time from its rate of change. Essentially, integration reverses differentiation, allowing us to uncover a function which represents the accumulated change over a given interval.

Integration is vital in applied calculus as it provides a tool for computing areas, volumes, and in this context, evolving quantities like the ice thickness.

Initial Condition

The initial condition in calculus problems, such as this, is often a starting point or boundary we use to solve for unknown constants like \( C \) in our integration problem. When we attempt to find the thickness function, it is crucial to recognize the situation at time \( t = 0 \). Here, the exercise tells us that there is no ice yet, so the thickness \( y = 0 \).

By substituting the initial condition \( y = 0 \) and \( t = 0 \) into the integrated function \( y(t) = \frac{1}{3} t^{3/2} + C \), we solve for the integration constant, finding \( C = 0 \). This eliminates unknowns and gives us the specific function \( y(t) = \frac{1}{3} t^{3/2} \), tailored to our particular problem starting from no initial ice.

Initial conditions like these are crucial as they ensure our mathematical models align with real-world scenarios and constraints, bringing precise and relevant solutions to practical problems.