Problem 71 - Applied Calculus

Question

An experimental drug lowers a patient's blood serum cholesterol at the rate of \(t \sqrt{25}-t^{2}\) units per day, where \(t\) is the number of days since the drug was administered \((0 \leq t \leq 5)\). Find the total change during the first 3 days.

Step-by-Step Solution

Answer

The total change in cholesterol over the first 3 days is 13.5 units.

Step 1: Understand the Problem

We are given a rate function \(R(t) = t \sqrt{25} - t^2\) that represents the rate of change of cholesterol over time. We need to find the total change in cholesterol from day 0 to day 3. This involves integrating the rate function from \(t = 0\) to \(t = 3\).

Step 2: Set Up the Integral

To find the total change in cholesterol, we set up the definite integral of the rate function \(R(t)\) over the interval from 0 to 3. The integral is given by:\[\int_{0}^{3} (t \sqrt{25} - t^2)\, dt = \int_{0}^{3} (5t - t^2)\, dt.\]

Step 3: Compute the Integral

Calculate the integral:1. First, integrate each term separately: - \(\int 5t \, dt = \frac{5t^2}{2} + C\) - \(\int t^2 \, dt = \frac{t^3}{3} + C\)2. Put them together in the definite integral:\[\int_{0}^{3} (5t - t^2)\, dt = \left[\frac{5t^2}{2} - \frac{t^3}{3}\right]_{0}^{3}.\]

Step 4: Evaluate the Integral

Evaluate the integrated expression at \(t = 3\) and \(t = 0\), then subtract:- When \(t = 3\): \( \left[\frac{5(3)^2}{2} - \frac{(3)^3}{3}\right] = \left[\frac{45}{2} - 9\right] = 22.5 - 9 = 13.5 \)- When \(t = 0\): \( \left[\frac{5(0)^2}{2} - \frac{(0)^3}{3}\right] = 0. \)Subtract the two results:\(13.5 - 0 = 13.5\).

Step 5: State the Result

The total change in cholesterol levels over the first 3 days is 13.5 units.

Key Concepts

Definite Integral

In calculus, the definite integral is a mathematical concept that computes the accumulation of quantities, which in this context is used to find the total change over time. It is expressed as \[ \int_{a}^{b} f(t) \, dt \] where \(a\) and \(b\) are the bounds of integration, and \(f(t)\) is the function representing the rate of change.

• Bounds of Integration: The limits, 0 to 3 in our problem, tell us that we are interested in the change from day 0 to day 3.
• Function within the Integral: The rate function, \(t \sqrt{25} - t^2\), represents how the cholesterol level changes each day.

By evaluating a definite integral, we accumulate small changes over a specified interval, which gives us a total change. In our example, we calculated this total change in cholesterol over three days, yielding 13.5 units.

Rate of Change

The rate of change is a mathematical construct that describes how a certain quantity changes relative to another variable, often time. In this exercise, the rate of change is expressed by the function:\[ R(t) = t \sqrt{25} - t^2 \]

• This function tells us the change in cholesterol levels per day once the drug is administered.
• It varies with \(t\), which is the number of days since the drug was given.

Understanding Rate of Change:
The rate of change is dynamic – it varies every day. Different days can have different rates. Our aim here is to aggregate these daily changes over a specified period using integration.
In the context of our exercise, understanding this varying rate allows us to calculate the overall change more comprehensively. This emphasizes the importance of integration as a tool to consolidate rates spread across an interval.

Antiderivative

The antiderivative, or indefinite integral, is the reverse process of differentiation. It helps in finding a function whose derivative matches the given rate function. Example: For the rate function \[ R(t) = 5t - t^2 \] - The antiderivative is determined by integrating term by term:

• \( \int 5t \, dt = \frac{5t^2}{2} + C \)
• \( \int t^2 \, dt = \frac{t^3}{3} + C \)

This provides us with a new function, which describes accumulated change over time. In the context of definite integrals, we are often interested in the difference in values of this antiderivative function across a chosen interval.Purpose:
Understanding antiderivatives allows us to bridge rate functions and accumulated changes. Applying these concepts to our exercise, we begin by finding the antiderivative to calculate the total change in cholesterol, confirming the significance of the integration process.