Calculus 1 Derivatives: Rules, Trig, and Inverse Functions

This comprehensive segment covers three classic related rates problems demonstrating how calculus models changing quantities in real-world scenarios. The first problem examines an oil spill spreading in a circular pattern, showing how the area increases as radius grows at 30 m/hour. The second problem tracks two planes approaching an airport from perpendicular directions, using the Pythagorean theorem to determine how their separation distance changes over time. Both problems follow a systematic approach: sketch the situation, identify variables, establish mathematical relationships, differentiate with respect to time, organize known values in tables, and substitute to solve for unknown rates. These examples illustrate the fundamental principle that related rates problems require understanding how different quantities are mathematically connected and how their rates of change relate through differentiation.

Related rates are important because they have many real-world applications such as snow melting, falling objects, boats moving away from each other at right angles, and temperature variation. The objectives of related rates are: (1) to use knowledge of derivatives to solve related rates problems; (2) to find the rate of change of one quantity knowing the rate of change of another quantity; (3) to find the rate of change of two or more variables that are changing with respect to time.

Related rates problems model how quantities change together through mathematical relationships. The strategy involves: (1) identifying given rates and desired rates; (2) writing relevant formulas connecting variables; (3) differentiating implicitly with respect to time; (4) substituting known values including signs (negative for decreasing quantities); (5) solving for the unknown rate. Common applications include geometry (spheres, circles) and motion problems. The key insight is that related quantities share common variables whose rates are connected through differentiation.

Related rates problems model real-world scenarios where multiple quantities change simultaneously, identified by multiple time derivatives (dx/dt, dy/dt). Common contexts include geometric relationships (Pythagorean theorem for right triangles, SOHCAHTOA for angles), geometric quantities (lengths, areas, volumes), and physical situations. The systematic solution process involves: (1) defining all time-dependent variables, (2) establishing relationships between them, (3) differentiating implicitly with respect to time, and (4) substituting known values to solve for unknown rates. The ladder problem exemplifies this: with ladder length extending at 8 ft/min and base moving away at 5 ft/min, Pythagorean theorem relates height, distance, and ladder length. Differentiating yields db/dt = 25/4 ft/min, showing the ladder's top moves upward at 6.25 ft/min.

This section covers related rates problems. Related rates problems involve finding how one quantity changes with respect to time when it is related to other quantities that are also changing. The key is to express the relationship between quantities, then differentiate both sides with respect to time. For example, if the radius of a sphere is increasing at a rate of 2 cm/min, and you need to find how fast the volume is increasing, use V = (4/3)πr³, differentiate with respect to t: dV/dt = 4πr²(dr/dt). If dr/dt = 2 and r = 3, then dV/dt = 4π(3)²(2) = 72π cm³/min.

Derivatives are fundamental tools for optimization in both economics and physics. In economics, a gas station selling 4800 liters daily at 50 cents per liter can maximize revenue by finding the optimal price increase. Taking the derivative of the revenue function and setting it to zero yields x = 5 cents, maximizing revenue to 242,000 cents. In physics, the electric field strength from a charged ring depends on distance from the ring's center. By differentiating the electric field function and setting the derivative to zero, we find the maximum occurs at x = r√2. The negative second derivative confirms these are maxima, demonstrating how calculus provides systematic methods for finding optimal solutions in practical scenarios.

This lesson covers two major applications of derivatives: (1) Economic optimization - finding production quantity that minimizes average cost by dividing total cost by x, differentiating, and setting derivative to zero; (2) Physics optimization - finding maximum height of projectile motion by differentiating height function to get velocity, setting velocity to zero, solving for time, then substituting back to find maximum height. Both applications follow the same three-step optimization process: obtain the function, differentiate, and set derivative to zero to find the optimal point.

Optimization problems apply derivatives to find maximum or minimum values in real-world scenarios. The derivative represents the slope of the tangent line, and when this slope equals zero, we identify critical points that may represent maxima or minima. The general procedure involves: (1) assigning variables to unknown quantities, (2) using constraints to express one variable in terms of another, (3) writing the objective function in terms of a single variable, (4) taking the derivative and setting it to zero, (5) solving for critical points, and (6) verifying the solution using number line analysis or derivative sign changes. Examples include finding two positive numbers whose product is 1000 and whose sum is minimized (yielding 10√10), maximizing the volume of an open-top rectangular prism with surface area 2700 cm² (yielding 30,500 cm³), and finding the optimal cut size for a box made from a 3x3 square (yielding 0.5 cm). Some problems involve letters instead of numbers, but the same procedure applies.

Derivatives are fundamental tools for finding maximum and minimum values in various fields: in physics, they determine velocity (first derivative of displacement) and acceleration (second derivative of displacement) to analyze motion; in economics and administration, they optimize costs, profits, and productivity by identifying critical points where the derivative equals zero, enabling businesses to minimize expenses and maximize gains through mathematical optimization.

Derivatives solve real-world optimization problems. For maximizing park visitors V = -T² + 50T, find dV/dT = -2T + 50, set to zero: T = 25°C. Second derivative -2 < 0 confirms maximum. For minimizing profit L = 2D² - 40D + 200, dL/dD = 4D - 40 = 0 gives D = 10. Second derivative 4 > 0 confirms minimum. These methods apply to any function where you can compute derivatives.

Curve sketching involves analyzing a function's first and second derivatives to determine its shape: the first derivative indicates increasing/decreasing behavior (positive means increasing, negative means decreasing), while the second derivative indicates concavity (negative means concave down, positive means concave up). By finding critical points (where first derivative is zero) and inflection points (where second derivative is zero), creating sign charts for both derivatives, and combining them into a single chart, you can identify local maxima, minima, and changes in concavity to accurately sketch the graph of a polynomial function.

To sketch a curve using first and second derivatives, analyze the sign of the first derivative to determine where the function is increasing (positive) or decreasing (negative), and use the sign of the second derivative to identify concavity (positive indicates concave up, negative indicates concave down); combine these analyses to identify critical points (where first derivative is zero or undefined) and inflection points (where second derivative is zero or undefined), then determine the overall shape of the graph by examining how the function behaves in each interval defined by these key points.

The fifth step uses the first derivative to determine where the function is increasing or decreasing. Differentiate using the quotient rule: f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]². Set f'(x) = 0 to find critical points. Test intervals: if f'(x) > 0, the function is increasing (تزايد); if f'(x) < 0, the function is decreasing (تناقص). For f(x) = x²/(x² + 1), f'(x) = 2x/(x² + 1)², giving critical point x=0 where the function changes from decreasing to increasing. The sixth step uses the second derivative to determine concavity. Differentiate the first derivative to get f''(x), then set f''(x) = 0 to find inflection points. For f(x) = x²/(x² + 1), f''(x) = 2(1 - 3x²)/(x² + 1)³, giving inflection points at x = ±1/√3.

To determine concavity in different intervals, create a number line with inflection points marked. Choose test values in each interval and evaluate the sign of f''(x). If f''(x) > 0 in an interval, the curve is concave up in that region. If f''(x) < 0, the curve is concave down. This information, combined with the first derivative test results, provides complete information about the curve's shape and behavior.

In calculus, the first derivative determines whether a function is increasing (positive) or decreasing (negative), with critical points occurring where the first derivative equals zero or is undefined; the second derivative indicates concavity (positive means concave up, negative means concave down), and inflection points occur where the second derivative equals zero and the concavity changes sign. By analyzing these derivatives, we can determine intervals of increase/decrease, locate local extrema using the first derivative test, and identify concave up/down regions along with inflection points.

Integration is defined as the inverse operation of differentiation. If differentiation takes a function y = f(x) and produces its derivative dy/dx, then integration takes the derivative dy/dx and reconstructs the original function y. For example, if dy/dx = 3x² - 6x + 4, then integrating this expression gives y = x³ - 3x² + 4x (plus a constant). This relationship allows us to work backwards from derivatives to find original functions.

Integration is defined as the inverse operation to differentiation (derivación). Just as factorization is the inverse of multiplication, integration reverses the process of finding derivatives. If differentiating a function f(x) yields f'(x), then integrating f'(x) should return us to the original function f(x) plus a constant. For example, if the derivative of 3x² is 6x, then the integral of 6x is 3x² + C, where C represents an arbitrary constant of integration.

Integration is the inverse operation of differentiation. An antiderivative (primitive) of a function f(x) is a function F(x) such that F'(x) = f(x). When finding antiderivatives, we always add a constant 'C' because the derivative of any constant is zero. For example, the antiderivative of 5x is (5/2)x² + C, not just (5/2)x². The constant C is essential because it represents the family of all possible antiderivatives.

For f(x) = (1/2)sin(2x) and g(x) = cos(2x), differentiating f(x) gives cos(2x), confirming the antiderivative relationship. The derivative of sin(2x) is 2cos(2x) using the chain rule, and the derivative of (1/2)sin(2x) is cos(2x). Integration is the inverse operation of differentiation. If f(x) is the antiderivative of g(x), then differentiating f(x) gives g(x), and integrating g(x) gives f(x). When asked to integrate g(x) and you know f(x) is its antiderivative, the definite integral from a to b equals f(b) - f(a). This inverse relationship is the foundation of the Fundamental Theorem of Calculus.

This comprehensive segment introduces integration as the inverse operation of differentiation. The instructor explains that integration involves finding the antiderivative of a function, also called the indefinite integral. The content establishes that if the derivative of F(x) equals f(x), then the integral of f(x) equals F(x) plus a constant C. The segment covers the power rule for integration, which states that the integral of x^n with respect to x equals (x^(n+1))/(n+1) + C, where C is the constant of integration. The instructor emphasizes that this formula applies when n is not equal to -1, as the case of n = -1 requires logarithmic integration. This foundational understanding sets the stage for learning specific integration techniques, emphasizing that integration reverses the differentiation process.