Sample Module

Calculus I

Introduction

The derivative plays a very important role in almost all heavily quantitative disciplines. Engineering, physics, economics, and of course mathematics, all have principles and concepts to which differentiation and derivatives are essential. The derivative yields the slope of a curve at a particular point. Or more precisely, the slope of the straight line that is tangent to the curve at that particular point. A slope represents the direction and number of units by which a dependent variable, say Y, changes when an independent variable, say X, increases by 1 unit.

Rate of Change

One interpretation of the derivative is as the rate of change of the dependent variable Y with respect to the independent variable X. A plain example is the area of a rectangle. This area is the product of the rectangle's length and width. As the length or width of the rectangle changes, its area changes. The rate of change of the rectangle's area as its length changes is given by the derivative of the area with respect to the length of the rectangle.

Consider a square with side length S. What is the rate of change of the perimeter of the square as the length of its side changes? The perimeter (P) of a square is 4 tmes the length of its side. So P = 4S. The derivative of the perimeter with respect to side length is dP/dS = d/dS(4S) = 4. The solution is the constant 4. The perimeter of a square changes at 4 times the change of its side length for all values of the side length.

Consider a square with side length S. What is the rate of change of the area of the square as the length of its side changes? The area (A) of a square is the square of its side. So A = S2. The derivative of the area with respect to side length is dA/dS = d/dS(S2) = 2S. The solution is the non-constant 2S. The area of a square changes at a rate that is twice its side length. This rate must be evaluated at each side length value. When S = 0.5, dA/dS = 1.0. The area changes at the same rate as its side length. When S = 1.0, dA/dS = 2.0. The area changes at the twice rate of its side length. When S = 2.0, dA/dS = 4.0. The area changes at the 4 times the rate of its side length.

Example 1.

The volume of a cylinder (V) is the area of its cross section times it height. If the cylinder has a square cross section, the area of the cross section is the square of its width. If the width is given by S and the height given by H, then the volume of the cylinder is V = S2H. If the width of the cylinder remains constant, what is the rate of change of the cylinder's volume with respect to its height?

The rate of change of the volume with respect to height is dV/dH = d/dH(S2H) = S2. The volume of the cylinder changes at the rate of its cross sectional area times the rate of change of its height.

If H = 100 cm and S = 5 cm, what is the rate of change of the cylinder's volume with respect to its height? dV/dH = 52 = 25 cm2. At the dimensions of 100 cm of height and 5 cm of width, the volume of the cylinder changes at 25 times the rate of change of its height.

Example 2.

The volume of a cylinder (V) is the area of its cross section times it height. If the cylinder has a square cross section, the area of the cross section is the square of its width. If the width is given by S and the height given by H, then the volume of the cylinder is V = S2H. If the height of the cylinder remains constant, what is the rate of change of the cylinder's volume with respect to its width?

The rate of change of the volume with respect to width is dV/dS = d/dS(S2H) = 2SH. The volume of the cylinder changes at the rate of twice the product of its width and height.

If H = 100 cm and S = 5 cm, what is the rate of change of the cylinder's volume with respect to its width? dV/dS = 2 * 5 * 100 = 1,000 cm2. At the dimensions of 100 cm of height and 5 cm of width, the volume of the cylinder changes at 1,000 times the rate of change of its width.

Example 3.

The volume of a cylinder (V) is the area of its cross section times it height. If the cylinder has square cross section, the area of the cross section is the square of its width. If the width is given by S and the height given by H, then the volume of the cylinder is V = S2H. If both the cylinder's height and width are variable, at what rate must they change in relation to each other for the cylinder's volume to remain constant?

Since the volume is to remain constant, dV/dS (or dV/dH) = 0. The objective is to find dH/dS (or dS/dH). Since the cylinder's height and width are to change in relation to each other, the term must be different from zero. Since both the height and width are changing, the total derivative is required.

First finding the rate of change with respect to the width yields dV/dS = S2 * dH/dS + 2SH. The volume is remaining constant so the term is set to zero. 0 = S2 * dH/dS + 2SH. Solving for dH/dS yields -2H/S. For the cylinder volume to remain constant, the height of the cylinder must decrease (increase) at 2H/S times the rate that the width of the cylinder increases (decreases).

If V = 2,500 cm3, H = 100 cm, and S = 5 cm, what is the rate of change of the cylinder's height with respect to its width? dH/dS = -2 * 100 / 5 = -40. At the dimensions of 2,500 cm3 of volume, 100 cm of height, and 5 cm of width, the height of the cylinder must change at -40 times the rate of change of its width for the cylinder's volume to remain constant.

Now finding the rate of change with respect to the height yields dV/dH = S2 + 2SH * dS/dH. The volume is remaining constant so the term is set to zero. 0 = S2 + 2SH * dS/dH. Solving for dS/dH yields -S/2H. For the cylinder volume to remain constant, the width of the cylinder must decrease (increase) at S/2H times the rate that the height of the cylinder increases (decreases).

If V = 2,500 cm3, H = 100 cm, and S = 5 cm, what is the rate of change of the cylinder's width with respect to its height? dS/dH = -5 / 2 * 100 = -0.025. At the dimensions of 2,500 cm3 of volume, 100 cm of height, and 5 cm of width, the width of the cylinder must change at -0.025 times the rate of change of its height for the cylinder's volume to remain constant.

When using the total derivative to find a rate of change, it doesn't matter which variable is used as the independent variable. The end result will be the same.

Sample Questions about the Derivative as a Rate of Change.

Minima and Maxima

A very important application of derivatives is the determination of the value of the independent variable (X) at which a function, f(X), has its highest or lowest values. You have already learned that the sign of the first derivative of a function, f'(X), indicates whether the function, f(X), is positively sloped or increasing, is negatively sloped or decreasing, or has no slope or is flat. Based on that, a minimum can be said to be that critical point X, at which the function, evaluated before X say X-1, has a negative slope and at which the function, evaluated after X say X+1, has a positive slope. A maximum is the point (X, f(X)), at which the function, evaluated at X-1, has a positive slope and evaluated at X+1, has a positive slope. In terms of the derivative, a function f(X) has a minimum at X when f'(X-1) < 0 and f'(X+1) > 0. A function f(X) has a maximum at X when f'(X-1) > 0 and f'(X+1) < 0. This is a general rule for all functions whether or not the function is continuous and differentiable at X.

In the special case of a function that is continuous and differentiable at X, a minimum or maximum (extremum) is found where f'(X) = 0. When f'(X) = 0, the distinction between a minimum and a maximum is determined by the sign of the function's second derivative, f''(X), evaluated at X. If f''(X) > 0, the function is upwardly concave at X and therefore the extremum is a minimum. If f''(X) < 0, the function is downwardly concave at X and therefore the extremum is a maximum.

Example 1.

Let f(X) = 1/3 * X3 - 3 * X2 + 5 * X - 7. Find the critical points for X and the extrema for f(X).

The critical points for X are those values of X for which f'(X) = 0. f'(X) = X2 - 6 * X + 5. Setting f'(X) equal to zero yields X2 - 6 * X + 5 = 0, a quadratic equation. Solving for the roots of the equation, (X-5) * (X-1) = 0, yields the critical points for X; X = 5, and X = 1. The values of the function at the critical points are f(5) = -15.33 and f(1) = -4.67.

The critial points are minima or maxima depending on the values of the second derivative. f''(X) = 2 * X - 6. f''(5) = 4 > 0. Therefore, the critical point of X = 5 is a minimum. f''(1) = -4 < 0, therefore the critical point at X = 1 is a maximum.

Example 2.

Let f(X) = X2 / (1 + 2 * X). Find the critical points for X and the extrema for f(X).

The critical points for X are those values of X for which f'(X) = 0. f'(X) = [(1 + 2 * X) * 2 * X - 2 * X2] / (1 + 2 * X)2 = 2 * X * (1 + X) / (1 + 2 * X)2. Setting f'(X) equal to zero yields 2 * X * (1 + X) = 0. Solving for the roots of this equation yields the critical points for X; X = 0, and X = -1. The values of the function at the critical points are f(0) = 0 and f(-1) = -1. Note that both the function and the first derivative are undefined at X = -0.5.

The critial points are minima or maxima depending on the values of the first derivative around the critial points. f'(-2) = 4/9 and f'(-0.6) = -12. Therefore, the critical point of X = -1 is a maximum. f'(-0.4) = -12 and f'(1) = 4/9, therefore the critical point at X = 0 is a minimum.

Example 3.

A rectangle has a perimeter of 50 cm. What is the maximum possible area the rectangle can have?

This is an example of constrained maximization/minimization. The area of a rectangle is the product of its length and with; A = L * W. The constraint is that the perimeter must equal a fixed value. The trick to solving this type of problem is to write the variable to be maximized or minimized in terms of just one independent variable. The constraint is used to accomplish that.

The perimeter of a rectangle is twice the sum of its length and width; P = 2 * (L + W). In this case, 2 * (L + W) = 50. Expressing the width in terms of the length yields W = (50 - 2 * L) / 2 = 25 - L. Now substituting the constraint expression into the area equation yields A = L * (25 - L). The area to be maximized is now expressed in terms of 1 variable, the length of the rectangle. We can now differentiate the area equation with respect to length and find the length of the rectangle that maximizes its area. The perimeter constraint is then used to find the width of the rectangle.

dA/dL = d/dL(25 * L - L2) = 25 - 2 * L. Setting A'(L) to zero yields 25 - 2 * L = 0 resulting in an optimal length of 12.5 cm. Substituting that value into the perimeter constraint yields W = 25 - 12.5 = 12.5 cm. So the length and width of the rectangle that maximizes its area are L = 12.5 cm and W = 12.5 cm. The length and width are equal making this rectangle a square. The area, A = 12.5 * 12.5 = 156.25 cm2.

The critical value of the length is a maximum for the area of the rectangle if the second derivative is negative. f''(L) = -2, a constant less than zero, therefore the area is a maximum.

Example 4.

A business finds that its total cost function is described by the equation c(q) = 2 * q2 - 2 * q + 200, where q is the quantity of output produced. What is the level of production that minimizes the firm's average cost?

Average cost is total cost divided by quantity, ac(q) = c(q) / q. This yields ac(q) = 2 * q - 2 + 200 / q. ac'(q) = 2 - 200 / q2. Setting ac'(q) equal to zero yields 2 * q2 - 200 = 0, resulting in the solution of q = 10. Average cost at q = 10 is ac(10) = 2 * 10 - 2 + 200 / 10 = 38.0.

The critial point is a minimum or a maximum depending on the values of the first derivative around the critial point. ac'(9) = -0.47 and ac'(11) = 0.35. Therefore, the critical point of q = 10.0 is a minimum.

Example 5.

A straight line has the equation f(X) = 2 * X - 4. What point on this line is closest to the point (2, 10)?

The distance between points, (X1, Y1) and (X2, Y2), is given by the formula d = √[(X2 - X1)2 + (Y2 - Y1)2]. For this scenario, the two points are (X, f(X)) and (2, 10). Expressing distance in terms of these two points yields d(X) = √[(2 - X)2 + (10 - (2 * X - 4))2] = √[5 * X2 - 60 * X + 200]. The derivative, d'(X), is 0.5 * [.]-0.5 * (10 * x - 60), where [.] = [5 * X2 - 60 * X + 200]. d'(X) = 0 when 10 * X = 60. The critical value for X is 6.0. The value of f(X) is f(6) = 2 * 6 - 4 = 8.0. The point solution is therefore (6.0, 8.0). The critial points are minima or maxima depending on the values of the first derivative around the critial points. d'(5) = -1.0 and d'(7) = 1.0. Therefore, the critical point of X = 6.0 is a minimum.