It is useful to consider what geometrical meaning can be given to the differential coefficient.
In the first place, any function of x, such, for example, as x2, or √x, or ax+b, can be plotted as a curve; and nowadays every schoolboy is familiar with the process of curve-plotting.
Let PQR, in Figure 7, be a portion of a curve plotted with respect to the axes of coordinates OX and OY. Consider any point Q on this curve, where the abscissa of the point is x and its ordinate is y. Now observe how y changes when x is varied. If x is made to increase by a small increment dx, to the right, it will be observed that y also (in this particular curve) increases by a small increment dy (because this particular curve happens to be an ascending curve). Then the ratio of dy to dx is a measure of the degree to which the curve is sloping up between the two points Q and T. As a matter of fact, it can be seen on the figure that the curve between Q and T has many different slopes, so that we cannot very well speak of the slope of the curve between Q and T. If, however, Q and T are so near each other that the small portion QT of the curve is practically straight, then it is true to say that the ratio dydx is the slope of the curve along QT. The straight line QT produced on either side touches the curve along the portion QT only, and if this portion is indefinitely small, the straight line will touch the curve at practically one point only, and be therefore a tangent to the curve.
If a curve is sloping up at 45° at a particular point, as in Figure 8, dy and dx will be equal, and the value of dydx=1.
If the curve slopes up steeper than 45° (Figure 9), dydx will be greater than 1.
If the curve slopes up very gently, as in Figure 10, dydx will be a fraction smaller than 1.
For a horizontal line, or a horizontal place in a curve, dy=0, and therefore dydx=0.
If a curve slopes downward, as in Figure 11, dy will be a step down, and must therefore be reckoned of negative value; hence dydx will have negative sign also.
If the “curve” happens to be a straight line, like that in Figure 12, the value of dydx will be the same at all points along it. In other words its slope is constant.
If a curve is one that turns more upwards as it goes along to the right, the values of dydx will become greater and greater with the increasing steepness, as in Figure 13.
If a curve is one that gets flatter and flatter as it goes along, the values of dydx will become smaller and smaller as the flatter part is reached, as in Figure 14.
If a curve first descends, and then goes up again, as in Figure 15, presenting a concavity upwards, then clearly dydx will first be negative, with diminishing values as the curve flattens, then will be zero at the point where the bottom of the trough of the curve is reached; and from this point onward dydx will have positive values that go on increasing. In such a case y is said to pass by a minimum. The minimum value of y is not necessarily the smallest value of y, it is that value of y corresponding to the bottom of the trough; for instance, in Figure 28 (the value of y corresponding to the bottom of the trough is 1, while y takes elsewhere values which are smaller than this. The characteristic of a minimum is that y must increase on either side of it.
Note–For the particular value of x that makes y a minimum, the value of dydx=0.
If a curve first ascends and then descends, the values of dydx will be positive at first; then zero, as the summit is reached; then negative, as the curve slopes downwards, as in Figure 16. In this case y is said to pass by a maximum, but the maximum value of y is not necessarily the greatest value of y. In Figure 28, the maximum of y is 213, but this is by no means the greatest value y can have at some other point of the curve.
Note–For the particular value of x that makes y a maximum, the value of dydx=0.
If a curve has the peculiar form of Figure 17, the values of dydx will always be positive; but there will be one particular place where the slope is least steep, where the value of dydx will be a minimum; that is, less than it is at any other part of the curve.
If a curve has the form of Figure 18, the value of dydx will be negative in the upper part, and positive in the lower part; while at the nose of the curve where it becomes actually perpendicular, the value of dydx will be infinitely great.
(1) As the simplest case take this: y=x+b.
It is plotted out in Figure 19, using equal scales for x and y. If we put x=0, then the corresponding ordinate will be y=b; that is to say, the “curve” crosses the y-axis at the height b. From here it ascends at 45°; for whatever values we give to x to the right, we have an equal y to ascend. The line has a gradient of 1 in 1.
Now differentiate y=x+b, by the rules we have already learned (here and here), and we get dydx=1.
The slope of the line is such that for every little step dx to the right, we go an equal little step dy upward. And this slope is constant–always the same slope.
(2) Take another case: y=ax+b. We know that this curve, like the preceding one, will start from a height b on the y-axis. But before we draw the curve, let us find its slope by differentiating; which gives dydx=a. The slope will be constant, at an angle, the tangent of which is here called a. Let us assign to a some numerical value–say 13. Then we must give it such a slope that it ascends 1 in 3; or dx will be 3 times as great as dy; as magnified in Figure 21. So, draw the line in Figure 20 at this slope.
(3) Now for a slightly harder case. Lety=ax2+b.
Again the curve will start on the y-axis at a height b above the origin.
Now differentiate. [If you have forgotten, turn back to here; or, rather, don't turn back, but think out the differentiation.] dydx=2ax.
This shows that the steepness will not be constant: it increases as x increases. At the starting point P, where x=0, the curve (Figure 22) has no steepness–that is, it is level. On the left of the origin, where x has negative values, dydx will also have negative values, or will descend from left to right, as in the Figure.
Let us illustrate this by working out a particular instance. Taking the equation y=14x2+3, and differentiating it, we get dydx=12x. Now assign a few successive values, say from 0 to 5, to x; and calculate the corresponding values of y by the first equation; and of dydx from the second equation. Tabulating results, we have:
x | 0 | 1 | 2 | 3 | 4 | 5 |
y | 3 | 314 | 4 | 514 | 7 | 914 |
d | 0 | 12 | 1 | 112 | 2 | 212 |
If a curve comes to a sudden cusp, as in Figure 25, the slope at that point suddenly changes from a slope upward to a slope downward. In that case dydx will clearly undergo an abrupt change from a positive to a negative value.
The following examples show further applications of the principles just explained.
The slope of the tangent is the slope of the curve at the point where they touch one another (see here); that is, it is the dydx of the curve for that point. Here dydx=−12x2 and for x=−1, dydx=−12, which is the slope of the tangent and of the curve at that point. The tangent, being a straight line, has for equation y=ax+b, and its slope is dydx=a, hence a=−12. Also if x=−1, y=12(−1)+3=212; and as the tangent passes by this point, the coordinates of the point must satisfy the equation of the tangent, namely y=−12x+b, so that 212=−12×(−1)+b and b=2; the equation of the tangent is therefore y=−12x+2.
Now, when two curves meet, the intersection being a point common to both curves, its coordinates must satisfy the equation of each one of the two curves; that is, it must be a solution of the system of simultaneous equations formed by coupling together the equations of the curves. Here the curves meet one another at points given by the solution of y=2x2+2,y=−12x+2or2x2+2=−12x+2; that is, x(2x+12)=0.
This equation has for its solutions x=0 and x=−14. The slope of the curve y=2x2+2 at any point is dydx=4x.
For the point where x=0, this slope is zero; the curve is horizontal. For the point where x=−14,dydx=−1; hence the curve at that point slopes downwards to the right at such an angle θ with the horizontal that tanθ=1; that is, at 45° to the horizontal.
The slope of the straight line is −12; that is, it slopes downwards to the right and makes with the horizontal an angle ϕ such that tanϕ=12; that is, an angle of 26°34′. It follows that at the first point the curve cuts the straight line at an angle of 26°34′, while at the second it cuts it at an angle of 45°−26°34′=18°26′.
(5) A straight line is to be drawn, through a point whose coordinates are x=2, y=−1, as tangent to the curve y=x2−5x+6. Find the coordinates of the point of contact.
The slope of the tangent must be the same as the dydx of the curve; that is, 2x−5.
The equation of the straight line is y=ax+b, and as it is satisfied for the values x=2, y=−1, then −1=a×2+b; also, its dydx=a=2x−5.
The x and the y of the point of contact must also satisfy both the equation of the tangent and the equation of the curve.
We have then y=x2−5x+6,(i) y=ax+b,(ii) −1=2a+b,(iii) a=2x−5,(iv) four equations in a, b, x, y.
Equations (i) and (ii) give x2−5x+6=ax+b.
Replacing a and b by their value in this, we get x2−5x+6=(2x−5)x−1−2(2x−5), which simplifies to x2−4x+3=0, the solutions of which are: x=3 and x=1. Replacing in (i), we get y=0 and y=2 respectively; the two points of contact are then x=1, y=2, and x=3, y=0.
Note.–In all exercises dealing with curves, students will find it extremely instructive to verify the deductions obtained by actually plotting the curves.
(1) Plot the curve y=34x2−5, using a scale of millimetres. Measure at points corresponding to different values of x, the angle of its slope.
Find, by differentiating the equation, the expression for slope; and see, from a Table of Natural Tangents, whether this agrees with the measured angle.
(2) Find what will be the slope of the curve y=0.12x3−2, at the particular point that has as abscissa x=2.
(3) If y=(x−a)(x−b), show that at the particular point of the curve where dydx=0, x will have the value 12(a+b).
(4) Find the dydx of the equation y=x3+3x; and calculate the numerical values of dydx for the points corresponding to x=0, x=12, x=1, x=2.
(5) In the curve to which the equation is x2+y2=4, find the values of x at those points where the slope =1.
(6) Find the slope, at any point, of the curve whose equation is x232+y222=1; and give the numerical value of the slope at the place where x=0, and at that where x=1.
(7) The equation of a tangent to the curve y=5−2x+0.5x3, being of the form y=mx+n, where m and n are constants, find the value of m and n if the point where the tangent touches the curve has x=2 for abscissa.
(8) At what angle do the two curves y=3.5x2+2andy=x2−5x+9.5 cut one another?
(9) Tangents to the curve y=±√25−x2 are drawn at points for which x=3 and x=4. Find the coordinates of the point of intersection of the tangents and their mutual inclination.
(10) A straight line y=2x−b touches a curve y=3x2+2 at one point. What are the coordinates of the point of contact, and what is the value of b?
(2) 1.44.
(4) dydx=3x2+3; and the numerical values are: 3, 334, 6, and 15.
(5) ±√2.
(6) dydx=−49xy. Slope is zero where x=0; and is ∓13√2 where x=1.
(7) m=4, n=−3.
(8) Intersections at x=1, x=−3. Angles 153°26′, 2°28′.
(9) Intersection at x=3.57, y=3.50. Angle 16°16′.
(10) x=13, y=213, b=−53.