Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose \(m\) is a positive integer and \(u(t)=t^{-m}=1 / t^{m}\) for \(t \neq 0 .\) Show that \(u^{\prime}(t)=-m t^{-m-1}\), thus proving the \(t^{n}\) rule for negative integers. Begin your argument with $$\begin{aligned}u^{\prime}(t) &=\lim _{b \rightarrow t} \frac{\frac{1}{b^{m}}-\frac{1}{t^{m}}}{b-t} \\ &=\lim _{b \rightarrow t} \frac{t^{m}-b^{m}}{b^{m} t^{m}} \times \frac{1}{b-t}\end{aligned}$$

Step-by-Step Solution

Answer

The derivative of \( t^{-m} \) is \( -mt^{-m-1} \).

Step 1: Apply the Difference of Powers Formula

Recall the difference of powers formula: \( b^{m} - a^{m} = (b-a)(b^{m-1} + b^{m-2}a + \ldots + ba^{m-2} + a^{m-1}) \). Applying this to \( t^{m} - b^{m} \), we have: \( t^{m} - b^{m} = (t-b)(t^{m-1} + t^{m-2}b + \ldots + tb^{m-2} + b^{m-1}) \).

Step 2: Substitute and Simplify the Limit Expression

Substitute the expression from Step 1 into the limit. Thus, \[ \lim _{b \rightarrow t} \frac{t^{m}-b^{m}}{b^{m} t^{m}} \times \frac{1}{b-t} = \lim _{b \rightarrow t} \frac{(t-b)(t^{m-1} + t^{m-2}b + \ldots + tb^{m-2} + b^{m-1})}{b^{m} t^{m} (b-t)} \].

Step 3: Cancel Out Common Factors

The factor \((t-b)\) is present in both the numerator and the denominator. Cancel them to simplify the expression: \[ \lim _{b \rightarrow t} \frac{t^{m-1} + t^{m-2}b + \ldots + tb^{m-2} + b^{m-1}}{b^{m} t^{m}} \].

Step 4: Evaluate the Limit as b Approaches t

As \( b \rightarrow t \), every term \( t^{m-k}b^{k-1} \) in \( t^{m-1} + t^{m-2}b + \ldots + tb^{m-2} + b^{m-1} \) tends to \( t^{m-1} \). Hence, we have \[ \lim _{b \rightarrow t} \frac{mt^{m-1} }{b^{m} t^{m}} = \frac{mt^{m-1}}{t^{2m}}= \frac{m}{t^{m+1}} \].

Step 5: Conclude the Derivative

Since \(u'(t)\) equals \( \frac{d}{dt}t^{-m} \), we find that \[ u^{\prime}(t) = -m t^{-m-1} \]. This proves the derivative of \( t^{n} \) for negative integers is \(-m t^{-m-1} \).

Key Concepts

Understanding Difference of Powers

The difference of powers formula is a useful algebraic tool in calculus. It allows us to rewrite an expression involving powers, such as \( b^{m} - a^{m} \), as the product of a factor \( (b-a) \) and a sum of other terms: \( b^{m-1} + b^{m-2}a + \ldots + ba^{m-2} + a^{m-1} \).
The importance of this formula lies in its ability to simplify problems involving limits, derivatives, and algebraic expressions.
In our exercise, we apply this formula to express \( t^{m} - b^{m} \) as \((t-b)(t^{m-1} + t^{m-2}b + \ldots + tb^{m-2} + b^{m-1})\).
This transformation helps us to later simplify and evaluate the limit. Such simplifications are pivotal when computing derivatives, especially for expressions involving powers.

The Role of Derivatives

In calculus, the derivative is a measure of how a function changes as its input changes. It gives the slope of the function at any given point.
For the function \( u(t) = t^{-m} = 1/t^{m} \), we need to find \( u'(t) \), its derivative. Using the definition of a derivative, we express it as a limit:

• \( u'(t) = \lim_{b \to t} \left( \frac{u(b) - u(t)}{b-t} \right) \).

This involves calculating the difference quotient, which we examined using the difference of powers.
Finding the derivative of a negative exponent function like \( t^{-m} \) demonstrates how derivatives handle power rules.

Mastering Limit Evaluation

Limit evaluation is one of the fundamental concepts in calculus. It helps us find the behavior of a function as a variable approaches a particular value.

• In the context of derivatives, calculating limits allows us to determine the rate of change precisely.
• In our example, we simplify the expression \( \lim _{b \rightarrow t} \frac{t^{m}-b^{m}}{b^{m} t^{m}} \times \frac{1}{b-t} \).

Once simplified, the method of evaluating limits lets us determine such expressions' behavior as \( b \) approaches \( t \).
This step-by-step assessment shows moments in calculus when algebraic manipulations achieve the intended simplification. This process lets us isolate terms depending only on \( t \), simplifying limit evaluation.

Understanding Negative Exponents

Negative exponents are a critical concept in algebra and calculus. They represent the reciprocal of the positive exponents:

• \( t^{-m} = 1/t^{m} \).
• This means that to raise a number to a negative exponent, we take the inverse of the number raised to the positive equivalent exponent.

In differentiating functions with negative exponents, the process involves recognizing the negative impact on rate changes.
Derivatives of functions like \( t^{-m} \) follow patterns similar to positive exponents, where the power rule applies:
\( \frac{d}{dt}t^{-m} = -mt^{-(m+1)} \), which makes sense given how negative exponents are reciprocal in nature.
Understanding how negative exponents influence derivatives broadens students' capability to handle a range of mathematical problems.