8. Find the coördinates of the point at which the tangent to the parabola y ax2 at Pi cuts the x-axis. If the tangent cuts the x-axis at R, and the y-axis at T, prove that R is the middle point of PT. Find the equation of the line through R perpendicular to the tangent. Show that this line always cuts the y-axis at the same point F, no matter what point on the curve P1 is.

DEFINITION. The point F(0, 1/4a) is called the focus of the parabola ax2. The focus is the point at which the line in Exercise 8 cuts the y-axis.

9. In the figure, P1A is parallel to the y-axis. By means of Exercise 8, and methods of plane geometry, show that angle FPIN = angle NP1A, the notation being the same as in Exercises 7 and 8.

Y

A

N

F

R

T

FIG. 164.

2

NOTE: A parabolic reflector, such as is used in a headlight of an automobile, is formed by revolving a parabola about its axis of symmetry. If a source of light is placed at F, Exercise 9 shows that the rays will be reflected in parallel lines.

ax2

10. Find the equations of the lines tangent to the parabola y at the two points for which y

=

1/4a. Where do they intersect? At what angle?

11. Find the equation of the line through the focus F(0, 1/4a) perpendicular to the tangent to the

parabola at P1(x1, y1). Show that it intersects the line x = y 1/4a.

DEFINITION. The line y y = ax2.

x1 on the line

- 1/4a is called the directrix of the parabola

12. Given the parabola y ar2 and a point P1(xı, yı) on it, show that the distance from the focus F(0, 1/4a) to P1is FP1 = y1 + 1/4a. Show from this that any point on a parabola is equidistant from the focus and the directrix.

13. Find the equation of the tangent to the graph of y ax3 at any point P1. If it cuts the y-axis at T, and if M is the projection of P1 on the y-axis, show that the origin trisects TM.

14. Find the equation of the line tangent to the parabola y2 x at any point P1 on the curve. Show that its intercept on the y-axis is half the ordinate of the point of contact and that it is perpendicular to the line joining the point of intersection with the y-axis to the focus.

15. Let F be a fixed point on the x-axis and T any point on the y-axis. Through T draw the line perpendicular to TF. Choosing different positions for T, draw a number of such lines, enough so that the form of the parabola to which they are tangent becomes apparent.

=

16. If Pi is a point on the equilateral hyperbola xy a, M its projection on the y-axis, and if the tangent line cuts the y-axis at T, then M is the middle point of OT. How can this be used to construct the line tangent at a given point on the curve?

17. Show that the point of contact of a line tangent to the equilateral hyperbola xy a is the middle point of the segment included between the axes. 18. Find the area included between the axes and the line tangent to the equilateral hyperbola xy = a at any point P1. State the result as a theorem.

19. If a normal is drawn to the equilateral hyperbola xy = a at a point P1 on it, then P1 is the middle point of the segment included between the lines bisecting the coördinate axes.

101. Problems in Maxima and Minima. To solve a problem involving maximum and minimum values, it is necessary first to express the variable v which is to be a maximum or minimum in terms of a single independent variable x. The quantity which is to be made a maximum or minimum is usually apparent from the statement of the problem, but there is frequently some choice in the selection of the independent variable. In many problems, as in Example 2 below, the variable v may be expressed at once in terms of two variables, one of which may be eliminated by means of a relation between them. Having found the function v = f(x), one then differentiates v with respect to x, sets the derivative equal to zero, and solves for x. It must then be shown that at least one of the values of x so found makes v a maximum or minimum, and this maximum or minimum value can be determined by substitution in v = f(x).

EXAMPLE 1. A box is to be made out of a square piece of cardboard, four inches on a side, by cutting out equal squares from the corners and

<-4-2x

FIG. 165.

then turning up the sides. Find the dimensions of the largest box that can be made in this way.

If the squares cut out are small, the box will have a large base and a shallow depth. If the squares are large, the base will be small and the

box deep. In either case the volume will be small. Somewhere between these two extremes will be a box whose volume is greater than that of any other, that is, a box of maximum volume.

Let x be the side of the square cut out. Then the depth of the box is x, and the side of the base is 4 - 2x. Hence the volume, expressed as a function of x is

V = x(4 – 2x)2 = 16x · 16x2 + 4x3.

The graph of V is readily plotted. From it, the value of x which makes √ a maximum appears to be somewhat less than unity. By computing V for a large number of values of x, we could approximate the best value of x. By means of the derivative we can avoid this labor, and obtain the

-12

exact value.

At the maximum point the tangent line is horizontal, and hence its slope is zero, so that DxV = 0.

We therefore compute the derivative, set it equal to zero, and find the value of x which produces this result. We then have

From the figure, we see that x = makes V a maximum, and x = 2 gives a minimum. That x = 2 gives a minimum follows also from the fact that if x = 2, then V = 0, that is, the box will not hold anything. The preliminary discussion showed the existence of a maximum, and as x = is the only other possibility it must be the value of x for which V is a maximum. Either of the criteria in Section 96 may also be applied to show that V is a maximum if x = eo

23

In order then, to have a box of maximum capacity, we must cut out squares from the corners of an inch on a side. The depth of the box will be of an inch, the side of the base will be 4 - 2 × 3 = 23 inches, and the capacity will be V = (23)2 × 3 = 427 cu. in.

EXAMPLE 2. As large a rectangular stick of timber as possible is to be sawed from a log 10 inches in diameter at the smaller end, the length of the stick to be the same as that of the log. Find its other dimensions.

Let V denote the volume of the stick of timber, l its length, and A the area of an end. Then V = lA, and since I is constant, the volume V will be a maximum if and only if A is a maximum. The end of the stick is a rectangle inscribed in a circle whose diameter is 10, and the problem reduces to a determination of the dimensions of the maximum rectangle which can be inscribed in this circle.

The area of the rectangle is

A = xy,

(1)

which involves the two variable dimensions x and y. These are connected by the relation x2 + y2 = 100.

In order to find the derivative of A we write it in the form

(2)

(3)

lie between these extremes. Hence as x increases from 0 to 10, A first increases and then decreases, and as A does not become infinite it therefore has a maximum in this interval. But x 5 √2 is the only point in this interval at which the tangent to the graph of A is horizontal, and it must therefore give the maximum value.

The corresponding value of y, from (2), is y = 5 √2. Hence the end of the largest stick of timber will be a square whose side is 5 √2 inches, or very nearly 7 inches.

EXERCISES

1. A box is to be made by cutting squares from the corners of a piece of cardboard 6 by 8 inches, and folding up the sides. Find the dimensions if the capacity is to be a maximum.

2. A chicken yard is to be made from 36 feet of poultry fencing, the side of a barn being used for one side of the yard. Find the dimensions in order that the yard may be as large as possible.

3. Find the dimensions and capacity of the largest box which can be made with a square base and no top if the total amount of cardboard in the box is 48 square inches.

4. What should be the dimensions of a rectangular garden plot with a perimeter of 12 rods, in order to have the greatest area possible?

5. A two acre pasture in the form of a rectangle is to be fenced off along the bank of a straight river, no fence being needed along the river. Find the dimensions, in rods, in order that the fence may cost as little as possible. 6. The legs of an isosceles triangle are 6 inches long. How long must the base be in order that the area may be a maximum?

7. The height of a rifle ball fired vertically upward with an initial velocity of 1200 feet per second is s = 1200t - 16ť2. How high will it rise?

8. By the Parcel Post regulations, the combined length and girth of a package must not exceed 6 feet. Find the dimensions and volume of the largest parcel which can be sent in the shape of a box with square ends.

9. A farmer has 150 rods of fencing. Find the dimensions and area of the largest rectangular field he can enclose and divide into two equal parts by a fence parallel to two of the sides.

10. If the total area of the field in the preceding Exercise is to be 150 square rods, find the dimensions and the amount of fencing needed if the latter is to be a minimum.

11. A rectangular cistern is to be built with a square base and open top. Find the proportions if the amount of material used is to be a minimum.

12. A rectangular piece of ground is to be fenced off and divided into three equal parts by fences parallel to one of the sides. What should the dimensions be in order that as much ground as possible may be enclosed with 16 rods of fence?

13. If the total area enclosed in Exercise 12 is an acre, find the dimensions in order that the total length of the fence should be a minimum.

14. The number of tons of coal consumed per hour by a certain ship is 0.3 +0.001 V3, where V is the speed in knots. For a voyage of 1000 knots at V knots per hour, find the total consumption of coal. For what speed is the consumption of coal least?

15. Divide a string 16 inches long into two parts, so that the combined area of the square and circle with perimeters equal to the parts shall be a minimum.