Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Let \(P(t)=\sqrt{t}\) and \(a\) be a positive number. Give reasons for the following steps to find \(P^{\prime}(a)\). $$ \begin{aligned} P^{\prime}(a) &=\lim _{b \rightarrow a} \frac{P(b)-P(a)}{b-a} \\ &=\lim _{b \rightarrow a} \frac{\sqrt{b}-\sqrt{a}}{b-a} \\ &=\lim _{b \rightarrow a} \frac{1}{\sqrt{b}+\sqrt{a}} \\ &=\frac{1}{\sqrt{a}+\sqrt{a}}=\frac{1}{2 \sqrt{a}} \end{aligned} $$

Step-by-Step Solution

Answer

\( P'(a) = \frac{1}{2\sqrt{a}} \) for \( P(t) = \sqrt{t} \).

Step 1: Understand Derivative Definition

We are tasked to find the derivative \( P'(a) \) using the limit definition of a derivative: \( P'(a) = \lim_{b \rightarrow a} \frac{P(b)-P(a)}{b-a} \). For the given function \( P(t) = \sqrt{t} \), substitute into the definition to express the derivative as a limit.

Step 2: Substitute Function into Derivative

Substitute \( P(t) = \sqrt{t} \) into the derivative limit expression. This gives \( P'(a) = \lim_{b \rightarrow a} \frac{\sqrt{b} - \sqrt{a}}{b - a} \). This expression represents the change in the function \( \sqrt{t} \) over the interval \( [a, b] \).

Step 3: Rationalize the Numerator

To simplify \( \frac{\sqrt{b} - \sqrt{a}}{b-a} \), we need to rationalize the numerator by multiplying the numerator and the denominator by the conjugate \( \sqrt{b} + \sqrt{a} \). This yields \( \frac{(\sqrt{b} - \sqrt{a})(\sqrt{b} + \sqrt{a})}{(b-a)(\sqrt{b} + \sqrt{a})} \).

Step 4: Simplify the Expression

The numerator becomes \( b - a \) after applying the difference of squares: \( (\sqrt{b})^2 - (\sqrt{a})^2 = b - a \). So the expression simplifies to \( \lim_{b \rightarrow a} \frac{b-a}{(b-a)(\sqrt{b} + \sqrt{a})} \).

Step 5: Cancel Terms and Evaluate Limit

Cancel \( b-a \) in the numerator and denominator, resulting in the limit \( \lim_{b \rightarrow a} \frac{1}{\sqrt{b} + \sqrt{a}} \). As \( b \) approaches \( a \), this limit simplifies to \( \frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}} \).

Step 6: Conclusion

The derivative \( P'(a) \) of the function \( P(t) = \sqrt{t} \) at any positive number \( a \) is \( \frac{1}{2\sqrt{a}} \). This was obtained through the limit process and algebraic simplification.

Key Concepts

Limit Definition of Derivative

The limit definition of a derivative is a fundamental concept in calculus. It allows us to find the slope of the tangent line to a curve at any given point. For the function given, \(P(t) = \sqrt{t}\), we are tasked with finding the derivative, \(P'(a)\), at some point \(a\). The derivative in terms of limits is defined as:
\[P'(a) = \lim_{b \rightarrow a} \frac{P(b) - P(a)}{b-a}\]Substituting the function into this definition involves replacing \(P(b)\) and \(P(a)\) with \(\sqrt{b}\) and \(\sqrt{a}\) respectively. This gives the new expression:
\[P'(a) = \lim_{b \rightarrow a} \frac{\sqrt{b} - \sqrt{a}}{b - a}\]This setup captures how the value of \(P(t)\) changes as \(b\) approaches \(a\). By understanding this limit, we can find the instantaneous rate of change, which is essentially the derivative.

Rationalization Technique

The rationalization technique is a useful algebraic method designed to eliminate radicals from the numerator. In our exercise, we use this technique to simplify the expression:
\[\frac{\sqrt{b} - \sqrt{a}}{b-a}\]We do this by multiplying both the numerator and the denominator by the conjugate of the numerator, \(\sqrt{b} + \sqrt{a}\). This approach takes advantage of the identity:
\[(a-b)(a+b) = a^2 - b^2\]Thus, the transformed expression becomes:
\[\frac{(\sqrt{b} - \sqrt{a})(\sqrt{b} + \sqrt{a})}{(b-a)(\sqrt{b} + \sqrt{a})}\]This step is essential to eliminate the radicals, specifically in the numerator, to make further simplifications possible.

Difference of Squares

This concept refers to the identity \(a^2 - b^2 = (a-b)(a+b)\), which is particularly useful for rationalization. Applying this identity reduces the expression:
\[(\sqrt{b} - \sqrt{a})(\sqrt{b} + \sqrt{a})\]to just \(b - a\). This remarkably simplifies our problem because the radicals disappear, and the numerator becomes simpler:
\[\frac{b-a}{(b-a)(\sqrt{b} + \sqrt{a})}\]With the numerator \(b-a\) and denominator containing \(b-a\), we can now cancel these terms. This simplification is pivotal as it prepares the expression for the next and final step.

Simplification of Expressions

Simplifying expressions is an important skill in calculus, making complex problems more manageable. After applying the difference of squares, our goal is to cancel out common terms. As seen in our formula:
\[\frac{b-a}{(b-a)(\sqrt{b} + \sqrt{a})}\]The \(b-a\) terms cancel each other, yielding:
\[\lim_{b \rightarrow a} \frac{1}{\sqrt{b} + \sqrt{a}}\]To evaluate this limit as \(b\) approaches \(a\), substitute \(b = a\) to obtain:
\[\frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}}\]By following these simplification steps, we arrive at the derivative \(P'(a) = \frac{1}{2\sqrt{a}}\). This shows how these techniques work together to find derivatives of functions involving radicals efficiently.