Velocity and Acceleration
Velocity
- Definition: Velocity is defined as the rate of change of displacement with respect to time.
- Mathematical Expression: If s(t) represents the displacement function, then the velocity, v(t), is given by the first derivative of s(t) with respect to time t.v(t) = ds(t)/dt
Acceleration
- Definition: Acceleration is defined as the rate of change of velocity with respect to time.
- Mathematical Expression: If v(t) represents the velocity function, then the acceleration, a(t), is given by the first derivative of v(t) with respect to time t.a(t) = dv(t)/dt
Example
Consider a particle moving along a line with its displacement given by the function s(t) = 3t3 - 15t2 + 18t.
- Finding Velocity: The velocity function v(t) is the derivative of the displacement function s(t) with respect to time t. Thus,v(t) = 9t2 - 30t + 18
- Finding Acceleration: The acceleration function a(t) is the derivative of the velocity function v(t) with respect to time t. Calculating this gives us the acceleration function.a(t) = 18t - 30
Optimization
Optimization involves finding the maximum or minimum values of a function, which is crucial in various fields like economics, physics, and engineering.
Critical Points
- Definition: Critical points are values of the variable for which either the derivative is zero or undefined.
- Mathematical Expression: If f(x) is the function to be optimized, then the critical points, c, satisfy the condition:f'(c) = 0 or f'(c) is undefined
Maximum and Minimum Values
- Local Maximum: A function f(x) has a local maximum at c if f(c) is greater than or equal to all nearby function values.
- Local Minimum: A function f(x) has a local minimum at c if f(c) is less than or equal to all nearby function values.
The First Derivative Test
- Increasing/Decreasing Test: If f'(x) > 0 for all x in an interval, then f is increasing on that interval. If f'(x) < 0 for all x in an interval, then f is decreasing on that interval.
Example
Consider a company whose profit, P, in thousands of dollars, is given by the function P(x) = -2x2 + 28x - 50, where x is the number of units produced and sold in thousands.
- Finding Critical Points: To find the critical points, we find the derivative of P(x) and set it equal to zero.P'(x) = -4x + 28Solving P'(x) = 0 gives us the critical point x = 7.
- Determining Maximum Profit: Using the second derivative test or analyzing the sign change of the first derivative around the critical point, we can determine that x = 7 gives a maximum profit.P''(x) = -4 (which is negative, confirming a maximum point)
Thus, producing and selling 7000 units will maximize the company's profit.
Applications in Physics
In physics, differentiation is widely used to describe motion, force, and energy. For instance, the position function of an object can be differentiated to find velocity and acceleration, providing insights into the object’s motion.
Example
Given the position function s(t) = t3 - 6t2 + 9t, find the velocity and acceleration at time t = 3 seconds.
- Velocity: Differentiating the position function with respect to time gives the velocity function. Calculating this and substituting t = 3 will provide the velocity at that instant.v(t) = 3t2 - 12t + 9v(3) = 0
- Acceleration: Differentiating the velocity function with respect to time gives the acceleration function. Calculating this and substituting t = 3 will provide the acceleration at that instant.a(t) = 6t - 12a(3) = 6
Thus, at t = 3 seconds, the object has a velocity of 0 units/second and an acceleration of 6 units/second2.
These applications of differentiation in various contexts illustrate its utility in deciphering and solving real-world problems, especially in understanding and describing change, motion, and optimization scenarios
FAQ
In demographics, the rate of population growth is determined by how the population size changes with respect to time. Mathematically, this rate is the derivative of the population function with respect to time (dP/dt). If P(t) represents the population at time t, then dP/dt represents the rate of population growth at time t. This application is crucial in various fields, such as ecology and urban planning, to predict future population sizes, understand the factors influencing population growth, and plan resources and infrastructure accordingly.
The second derivative in economic graphs, like cost curves, is used to analyse concavity and determine whether a function is concave up or down. If the second derivative of a function is positive, the function is concave up, indicating that as production increases, the marginal cost is also increasing. If the second derivative is negative, the function is concave down, implying that as production increases, the marginal cost is decreasing. This analysis is vital in economics to understand cost structures and to find inflection points where the nature of the cost changes, which can influence pricing and production strategies.
In physics, particularly in studying motion under gravity, derivatives are used to find velocity and acceleration. If s(t) represents the position of an object at time t, then the first derivative ds/dt (or v(t)) gives its velocity, indicating how the position changes with time. The second derivative d2s/dt2 (or a(t)) gives its acceleration, showing how the velocity changes with time. In the context of gravity, this helps in understanding how objects move under gravitational pull, allowing physicists to predict future positions and velocities of objects and to understand the underlying forces acting upon them.
Elasticity in economics, specifically price elasticity of demand, measures how the quantity demanded responds to a change in price. It's calculated as the percentage change in quantity demanded divided by the percentage change in price. In calculus, this ratio is essentially a derivative: it measures how one variable changes in response to a small change in another. The derivative of the demand (or supply) curve with respect to price gives the rate of change of quantity with respect to price, which, when multiplied by the ratio of price to quantity, gives the price elasticity of demand (or supply).
Marginal cost refers to the cost added by producing one additional unit of a product or service. In calculus, particularly in derivatives, the marginal cost at a given level of production is the rate of change of the total cost with respect to the level of production. Mathematically, it's the derivative of the total cost function with respect to quantity (dC/dQ). This concept is crucial in business applications as it helps to determine the level of production that minimises or maximises cost, thereby aiding in optimising production levels for maximum profitability.
Practice Questions
To find the velocity, we differentiate the position function s(t) with respect to time t. So, v(t) = ds(t)/dt = 3t2 - 10t + 6. Substituting t = 4 into the velocity function gives us v(4) = 3(4)2 - 10(4) + 6 = 48 - 40 + 6 = 14. Since v(4) = 14 is positive, the particle is moving forwards at t = 4 seconds.
To find the number of items that maximize the profit, we need to find the critical point of the profit function, P(x). The critical point occurs where the derivative, P'(x), is zero. So, P'(x) = dP(x)/dx = -2x + 12. Setting P'(x) equal to zero and solving for x gives us -2x + 12 = 0, which simplifies to x = 6. Thus, to maximize its profit, the company should produce and sell 6000 items. The second derivative, P''(x) = -2, is negative, confirming that this is a maximum point.
