Antiderivative As Expressions Of Pj Problems
Peter Oye Sagay
Differential calculus and integral calculus are the two parts of the branch of mathematics known as calculus. The subject of antiderivatives is usually considered part of integral calculus. This categorization is justified because the integral sign is a major component of antiderivatives. However, the role of antiderivatives as the bridge between differential calculus and integral calculus is often lost in the general discussions of integral calculus. The desire to highlight this role is the reason for this separate presentation of antiderivatives .
Antiderivatives
In differential calculus, the instantaneous rate of change of a variable with respect to another variable is called the derivative of the function that relates both variables. For example, y = f(x) implies that the variable y is a function of the variable x (y is the dependent variable and x is the independent variable). The instantaneous rate of change of y with respect to x, that is, the derivative of f(x) = dy/dx = f'(x). This derivative is also the slope of f(x) at the point x.
The definition of the derivative implies that every derivative has a function from which it is derived. The function from which a derivative is derived is called the antiderivative of its derivative. For example, suppose F(x) is a function having the function f(x) as its derivative, that is, F'(x) = f(x), then F(x) is an antiderivative of f(x). The process of determining the antiderivative is called antidifferentiation. The following is an example of the concepts of the derivative and the antiderivative:
f(x) = dy/dx = x2, implies that there is a function F(x) for which F'(x) = x2.
The function F(x) = (1/3)x3 is one such function. In fact, so is the function F(x) = (1/3)x3 + C (where C is any constant) since the derivative of a constant is zero. Consequently, one can obtain all antiderivatives of the derivative f(x) by adding an infinite number of all possible constants C to F(x). Hence, the following important theorem:
Theorem I: If F1(x) and F2(x) are two antiderivatives of the same function f(x), then F2(x) = F1(x) + C. In other words, F1(x) and F2(x) differ by a constant. This theorem is based on the fact that the antiderivative F(x) = C (where C is some constant) implies F'(x) = 0 for all x. A formal proof of the theorem requires the mean value theorem.
The implication of theorem I is that one can find all the antiderivatives of a given function once one antiderivative is known. The standard mathematical representation of this fact is as follows:
The symbol ∫ is called an integral sign. f(x) is a function whose antiderivatives are F(x) + C. The entire notation ∫ f(x)dx is called an indefinite integral and it represents the antidiffrentiation of the function f(x). The variable of interest is always indicated by prefacing it with the letter d. For example, if t is the variable of interest, equation (1) in figure 1.1 will become ∫ f(t)dt = F(t) + C. Equation (1) in figure 1.1 is the bridge between differential calculus and integral calculus.
Applications Of The AntiDerivative
Problem I: The velocity of a rocket fired vertically into the air is given by v(t) = (6t + 0.5) m/sec, t seconds after lift-off. Before lift-off (rocket still at launch pad), the top of the rocket is 8 meters above the launch pad. What is the height of the rocket (measured from the top of the rocket to the launch pad) at time t?
Solution to Problem I: let s(t) be the height of the rocket at time t.
Then s'(t) is the rate at which the height is changing with respect to t.
So, s'(t) = v(t). Hence s(t) is an antiderivative of v(t).
Therefore, we can apply equation (1) as follows:
s(t) = ∫ v(t)dt = ∫ (6t + 0.5)dt.
=3t2 + 0.5t + C (where C is a constant).
C = 8 since s(0) = 8.
Therefore, s(t) = t2 + 0.5t + 8.
Problem II: A company's marginal cost function is 0.015x2 - 2x + 80 dollars (x is the number of units produced per day).
Company's fixed cost = $1000 per day.
(a) What is the cost of producing x units per day?
(b) Determine the increase in cost if the current daily production of 30 units rises to 60 units per day.
Solution to Problem II: let C(x) be the cost of producing x units per day. The derivative C'(x) is the marginal cost.
So, C(x) is an antiderivative of the marginal cost function.
Therefore, we can apply equation (1) as follows:
(a) C(x) = ∫ (0.015x2 - 2x + 80)dx.
= 0.005x3 - x2 + 80x + C (where C is a constant). C = 1000 since C(0) = 1000.
Therefore, C(x) = 0.005x3 - x2 + 80x + 1000.
(b) The cost when x = 30 is C(30), and the cost when x = 60 is C(60).
The increase in cost when production is raised from x = 30 to x = 60 is C(60) - C(30).
C(60) = 0.005(60)3 - (60)2 + 80(60) + 1000 = 3280
C(30) = 0.005(30)3 - (30)2 + 80(30) + 1000 = 2635
Therefore, increase in cost = 3280 - 2635 = 645.
Problems I and II are examples of why antiderivatives are important. Often, the functions given or known are derivatives whereas the functions desired are their antiderivatives. This is the case in the branch of mathematics called partial differential equations(PDE) where many important natural phenomena can only be expressed as derivatives. The problem then becomes that of determining the antiderivatives that satisfy the partial differential equations (partial derivatives occur when there are more than one independent variables in a function).
Antiderivatives, like derivatives, are expressions of Pj problems. The types of Pj problems that the antiderivatives express, depend on the types of Pj problems expressed by their derivatives. For example, problem I is of the type motion problems and change problems. Problem II is of the type change problems.