Calculus serves as a pivotal tool in comprehending and modeling the dynamics of change across various contexts. This segment of study notes delves into the essence of calculus, highlighting its role in generalizing knowledge about motion and addressing diverse problems of change through the concept of limits. By engaging with these notes, students will gain insights into utilizing limits to conceptualize and solve problems involving dynamic change, a foundational aspect of calculus critical for further studies in this field.

Introduction to Limits

Limits are fundamental in calculus, allowing us to explore values that functions approach as the input reaches a certain point.

Definition of a Limit: Let $f(x)$ be a function defined on an interval containing $c$ (except possibly at $c$) and let $L$ be a real number. The statement

$\lim _{x \rightarrow c} f(x) = L$

means that for every \epsilon > 0 there exists a \delta > 0 such that if 0 < |x - c| < \delta, then |f(x) - L| < \epsilon.

Intuitive Understanding: As $x$ approaches $c$, the value of $f(x)$ gets arbitrarily close to $L$.

The Role of Limits in Understanding Dynamic Change

Limits provide the framework to quantify changes occurring in an instant, which are pivotal in understanding and modeling dynamic systems.

Example 1: Calculating Instantaneous Speed

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Scenario: A car travels in a straight line and its position $s$ at time $t$ is given by $s(t) = t^2 + 2t$, where $s$ is in meters and $t$ is in seconds. Find the car's instantaneous speed at $t = 3$ seconds.

Solution:

$\begin{aligned} \text{Instantaneous speed at } t = 3 & = \lim {h \rightarrow 0} \frac{s(3+h) - s(3)}{h} \\ & = \lim {h \rightarrow 0} \frac{[(3+h)^2 + 2(3+h)] - (3^2 + 2\cdot3)}{h} \\ & = \lim {h \rightarrow 0} \frac{9 + 6h + h^2 + 6 + 2h - 9 - 6}{h} \ & \\ & = \lim {h \rightarrow 0} \frac{h^2 + 8h}{h} \\ & = \lim _{h \rightarrow 0} (h + 8) \\ & = 8. \end{aligned}$

Thus, the instantaneous speed of the car at $t = 3$ seconds is $8$ meters per second.

Example 2: Understanding Change in Temperature

Scenario: The temperature $T$ in degrees Celsius at a specific location as a function of time $t$ in hours is given by $T(t) = 3t^2 - 2t + 1$. Calculate the rate of temperature change at $t = 4$ hours.

Solution:

$\begin{aligned} \text{Rate of temperature change at } t = 4 & = \lim _{h \rightarrow 0} \frac{T(4+h) - T(4)}{h} \\ & = \lim _{h \rightarrow 0} \frac{[3(4+h)^2 - 2(4+h) + 1] - (3\cdot4^2 - 2\cdot4 + 1)}{h} \\ & = \lim _{h \rightarrow 0} \frac{3(16 + 8h + h^2) - 2(4 + h) + 1 - (48 - 8 + 1)}{h} \\ & = \lim _{h \rightarrow 0} \frac{48 + 24h + 3h^2 - 8 - 2h + 1 - 41}{h} \\ & = \lim _{h \rightarrow 0} \frac{3h^2 + 22h}{h} \\ & = \lim _{h \rightarrow 0} (3h + 22) \\ & = 22. \end{aligned}$

The rate of temperature change at $t = 4$ hours is $22$ degrees Celsius per hour.

Application of Limits in Calculus

The application of limits extends beyond physical motion and temperature changes, encompassing economic models, biological growth patterns, and chemical reaction rates.

Example 3: Economic Model - Profit Maximization

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Scenario: A company's profit (P) in thousands of dollars is modeled by the function $P(x) = -2x^2 + 12x - 20$, where $x$ represents the number of units produced and sold in thousands. Determine the number of units that must be produced and sold to maximize profit.

Solution:

First, we find the derivative of $P(x)$ to identify the rate of change of profit with respect to the number of units. The maximum profit occurs where this derivative is zero.

$\begin{aligned} P'(x) & = \frac{d}{dx}(-2x^2 + 12x - 20) \\ & = -4x + 12. \end{aligned}$

Setting $P'(x) = 0$ to find the critical point:

$-4x + 12 = 0 \\ 4x = 12 \\ x = 3.$

Conclusion: To maximize profit, the company must produce and sell $3,000$ units.

Application in Biology: Population Growth Rate

Consider a population of bacteria that grows according to the function $P(t) = 100e^{0.5t}$, where $P$ is the population size and $t$ is time in hours.

Scenario: Determine the rate of population growth at $t = 2$ hours.

Solution:

The growth rate is the derivative of $P(t)$ with respect to $t$:

$\begin{aligned} \frac{dP}{dt} & = \frac{d}{dt}(100e^{0.5t}) \\ & = 100 \cdot 0.5e^{0.5t} \\ & = 50e^{0.5t}. \end{aligned}$

Substitute $t = 2$ to find the rate at that time:

$\begin{aligned} \frac{dP}{dt} \bigg|_{t=2} & = 50e^{0.5\cdot2} \\ & = 50e^1 \\ & = 50e \\ & \approx 135.9. \end{aligned}$

Conclusion: The rate of population growth at $t = 2$ hours is approximately $135.9$ bacteria per hour.

Chemical Kinetics: Rate of Reaction

In a chemical reaction, the concentration of a reactant $A$ decreases over time according to the function $A(t) = 50e^{-0.3t}$, where $A$ is the concentration in moles per liter and $t$ is time in seconds.

Scenario: Find the rate of decrease of $A$ at $t = 5$ seconds.

Solution:

The rate of decrease is the negative derivative of $A(t)$ with respect to $t$:

$\begin{aligned} -\frac{dA}{dt} & = -\frac{d}{dt}(50e^{-0.3t}) \\ & = 50 \cdot 0.3e^{-0.3t} \\ & = 15e^{-0.3t}. \end{aligned}$

Substituting $t = 5$ gives:

$\begin{aligned} -\frac{dA}{dt} \bigg|_{t=5} & = 15e^{-0.3\cdot5} \\ & = 15e^{-1.5} \\ & \approx 15 \cdot 0.2231 \\ & \approx 3.347. \end{aligned}$

Conclusion: The rate of decrease in the concentration of $A$ at $t = 5$ seconds is approximately $3.347$ moles per liter per second.

Practice Questions

Question 1

Evaluate the limit $\lim_{x \rightarrow 2} \frac{x^2 - 4}{x - 2}$.

Question 2

A box with a square base and no top is to be made from a square piece of cardboard by cutting squares from each corner and folding up the sides. If the piece of cardboard is 12 inches on each side, what size square should be cut from each corner to maximize the volume of the box?

Solutions to Practice Questions

Solution to Question 1

1. Identify the Problem: Substituting $x = 2$ directly into the function yields the indeterminate form $0/0$, suggesting the need for simplification.

2. Simplify the Expression: Factor the numerator:

$\frac{x^2 - 4}{x - 2} = \frac{(x + 2)(x - 2)}{x - 2}.$

Cancel out the $(x - 2)$ terms:

$\frac{(x + 2)(x - 2)}{x - 2} = x + 2.$

3. Evaluate the Limit: With the expression simplified, substitute $x = 2$:

$\lim_{x \rightarrow 2} (x + 2) = 4.$

Conclusion: The limit $\lim_{x \rightarrow 2} \frac{x^2 - 4}{x - 2} = 4$.

Solution to Question 2

1. Formulate the Problem: Define $x$ as the length of the side of the square cut from each corner. The dimensions of the box will be $12 - 2x$ by $12 - 2x$ by $x$.

2. Volume Function: The volume $V$ is given by:

$V = x(12 - 2x)^2.$

Simplify this to:

$V = 4x^3 - 48x^2 + 144x.$

3. Find the Critical Points: Derive $V$ with respect to $x$ and set to 0 to find critical points:

$\frac{dV}{dx} = 12x^2 - 96x + 144.$

Simplify and solve the quadratic equation:

$12x^2 - 96x + 144 = 0 \Rightarrow x^2 - 8x + 12 = 0.$

Factoring:

$(x - 6)(x - 2) = 0,$

yields $x = 6$ or $x = 2$.

4. Determine the Maximum: By logical constraints, $x = 2$ is the viable solution as $x = 6$ would leave no material for the box sides.

5. Evaluate the Volume at $x = 2$: Substitute $x = 2$ into the volume equation:

$V = 4(2)^3 - 48(2)^2 + 144(2) = 128 \text{ cubic inches}.$

6. Conclusion: To maximize the volume, squares of 2 inches should be cut from each corner, yielding a box with a volume of 128 cubic inches.