Worksheet: Application of Derivatives

The derivative of a function at a point measures the rate at which the function value changes as its input changes. Mathematically defined as f'(x) = lim(h→0) [f(x+h) - f(x)]/h, it indicates how f(x) changes concerning x. In real-world applications, derivatives are used in physics to find velocity (rate of change of distance with respect to time), in economics to evaluate marginal cost or revenue, and in biology to model population growth rates. An example would include calculating how quickly a car accelerates (change in speed over time).

To find the equation of the tangent line to the curve y = f(x) at the point (a, f(a)), follow these steps: 1. Determine f'(a), the derivative at x = a, to find the slope of the tangent. 2. Use the point-slope form of a line: y - f(a) = f'(a)(x - a). For example, if f(x) = x^2 at the point x = 3, f(3) = 9 and f'(x) = 2x implies f'(3) = 6. Therefore, the tangent line equation is y - 9 = 6(x - 3).

Derivatives help find local maxima and minima by identifying critical points where f'(x) = 0 or f'(x) does not exist. To classify these points, use the first derivative test: if f' changes from positive to negative at a critical point, it's a local maximum; if it changes from negative to positive, it's a local minimum. For example, consider f(x) = -x^2 + 4x. The critical point occurs at x = 2. Analyzing f'(x) shows it changes signs around this point, confirming a maximum. Calculate f(2) = 8 to find the maximum value.

The second derivative test evaluates concavity; if f''(x) > 0, the function is concave up, and if f''(x) < 0, it is concave down. This test helps identify inflection points. For example, consider f(x) = x^3. We find f'(x) = 3x^2 and f''(x) = 6x. At x = 0, f''(0) = 0 signals a potential inflection point. Analyzing f'' around this point shows a change in concavity, confirming its nature. Thus, the second derivative provides deeper insights into the function's behavior.

Optimization problems involve finding maximum or minimum values of functions. To set up, identify the function to optimize, usually subject to constraints, express it in terms of one variable, and find critical points by taking derivatives. For example, to maximize the area A of a rectangle with a fixed perimeter P = 20, express A = lw with l + w = 10. Substitute w = 10 - l into A to get A(l) = l(10 - l) = 10l - l^2. The critical point occurs when A' = 10 - 2l = 0, giving l = 5 for a maximum area of 25.

In physics, velocity is the rate of change of position with respect to time, defined by the first derivative of displacement s(t). Mathematically, v(t) = ds/dt. Acceleration is the rate of change of velocity, represented by the second derivative, a(t) = dv/dt = d^2s/dt^2. For example, if s(t) = t^3 - 6t^2 + 9t, then v(t) = 3t^2 - 12t + 9 and a(t) = 6t - 12. By evaluating these derivatives at specific time values, one finds instantaneous velocity and acceleration.

Implicit differentiation is used when y is not isolated in an equation involving x and y. By differentiating both sides concerning x and applying the chain rule, dy/dx can be extracted. For instance, in the equation x^2 + y^2 = 1, differentiate to get 2x + 2y(dy/dx) = 0. Solving for dy/dx yields dy/dx = -x/y. This method is vital for curves not easily manipulated into y = f(x) form.

The application of derivatives helps classify functions as increasing or decreasing by analyzing f'(x). If f'(x) > 0 for an interval, the function is increasing there; if f'(x) < 0, it is decreasing. This behavior is crucial in real-world contexts, informing us about trends such as profit maximization in economics or speed in mechanics. For example, if sales increase with time can be modeled by a function where its derivative is positive in that interval, indicating a growing market.

Let the width be y meters. Then, xy = 240 implies y = 240/x. The perimeter P = 2(x + y) = 2(x + 240/x). To minimize P, find P'(x) and set it to zero: P'(x) = 2(1 - 240/x²). Setting P'(x) = 0 gives x = √(240) = 15.49 m. Thus y = 240/15.49 = 15.49 m. The dimensions that minimize the perimeter are approximately 15.49 m by 15.49 m, forming a square.

Let r be the radius and h the height of the cylinder. Since the volume V = πr²h = 128π, we have r²(2r) = 128, giving r^3 = 64, thus r = 4 cm and h = 8 cm. The maximum volume of the cone can then be calculated as V_cone = (1/3)πr²h_cone = (1/3)π(4²)(8) = 128π/3 cm³ when the cone height equals the cylinder's height.

Let x be the width and y be the length of the field. Then, y = (200 - x). The area A = xy = x(200 - x) = 200x - x². To maximize A, find A'(x) = 200 - 2x and set it to zero, giving x = 100 m and y = 200 m. Therefore, the dimensions that maximize the area are 100 m by 200 m.

The volume of a sphere is given by V = (4/3)πr³. The derivative is dV/dt = 4πr²(dr/dt). When r = 5 cm and dr/dt = 0.1 cm/s, dV/dt = 4π(5)²(0.1) = 50π cm³/s.

Let x be the distance from the wall to the foot of the ladder, y be the height above the ground. We can use the Pythagorean theorem: x² + y² = 10². Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. When x = 6, y = √(10² - 6²) = 8. Thus, 2(6)(1) + 2(8)(dy/dt) = 0 ⇒ dy/dt = -3/4 ft/s (the negative sign indicates the height is decreasing).

Given y = x² - 4x + 6, find dy/dx = 2x - 4. Setting dy/dx = 2 gives 2x - 4 = 1, yielding x = 2.5. Therefore, y = (2.5)² - 4(2.5) + 6 = 8.25. The required point is (2.5, 8.25).

The volume of the cylinder is V = πr²h = π(3²)h = 9πh. The rate of change of volume is dV/dt = 9π(dh/dt). Thus, setting 10 = 9π(dh/dt) gives dh/dt = 10/(9π) m/hr at h = 5m.

Marginal revenue R'(x) = dR/dx = 8x + 12. Evaluating at x = 6, R'(6) = 8(6) + 12 = 60.

Area A = 0.5 * base * height = 0.5 * b * h. Substituting values for b and h, A = 0.5 * 6 * 4 = 12 m². The maximum area is 12 m².

Analyze the use of critical points within a contextual framework, focusing on how neglecting concavity can mislead conclusions.

Use the derivative to find critical points of the area function, applying the second derivative test for confirmation.

Discuss how the velocity function's derivative indicates changes in height, including real-life applications in projectiles.

Present the revenue function, find its derivative for maximum profit under given constraints, and verify your solution.

Create a hypothetical stock price function and analyze its maxima, minima, and inflection points during critical market events.

Define the cost function, utilize differentiation to find marginal cost, and relate these to output decisions.

Apply the derivative to find inflection points that indicate maximum sustainable population levels.

Analyze the profit function through calculus, employing both first and second derivative tests for critical points.

Illustrate with examples how different concavity behaviors affect perceived value and adjustment of prices.

Utilize constraints effectively, with careful consideration of the resource functions and their derivatives.