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Chapter 5: Problem 71

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.

Short Answer

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

Step by step solution

01

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 \(t = 3\).
02

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:\[\int_{0}^{3} (t \sqrt{25} - t^2)\, dt = \int_{0}^{3} (5t - t^2)\, dt.\]
03

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:\[\int_{0}^{3} (5t - t^2)\, dt = \left[\frac{5t^2}{2} - \frac{t^3}{3}\right]_{0}^{3}.\]
04

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:\(13.5 - 0 = 13.5\).
05

State the Result

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 鈥� 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.

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Most popular questions from this chapter

Find the derivative of each function. $\ln \left(x^{2}+5 x\right)

Suppose that you have a positive, increasing, concave up function and you approximate the area under it by a Riemann sum with midpoint rectangles. Will the Riemann sum overestimate or underestimate the actual area? [Hint: Make a sketch.]

\(85-94 .\) The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int(x+1)(x-5)^{4} d x $$

A company expects profits of \(60 e^{0.02 t}\) thousand dollars per month, but predicts that if it builds a new and larger factory, its profits will be \(80 e^{0.04 t}\) thousand dollars per month, where \(t\) is the number of months from now. Find the extra profits resulting from the new factory during the first two years \((t=0\) to \(t=24\) ). If the new factory will cost \(\$ 1,000,000\), will this cost be paid off during the first two years?

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int x(x+4)^{7} d x $$
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