Calculus Made Easy · Silvanus P. Thompson
RETURNING to the process of successive differentiation, it may be asked: Why does anybody want to differentiate twice over? We know that when the variable quantities are space and time, by differentiating twice over we get the acceleration of a moving body, and that in the geometrical interpretation, as applied to curves, (d y)/(d x) means the slope of the curve. But what can (d^2 y)/(d x^2) mean in this case? Clearly it means the rate (per unit of length x ) at which the slope is changing-in brief, it is a measure of the curvature of the slope .
Suppose a slope constant, as in Fig. 31 .
Here, (d y)/(d x) is of constant value.
Suppose, however, a case in which, like Fig. 32 , the slope itself is getting greater upwards, then (d((d y)/dx))/(d x) , that is, (d^2 y)/(d x^2) , will be positive.
If the slope is becoming less as you go to the right (as in Fig. 14 , p. 80), or as in Fig. 33 , then, even though the curve may be going upward, since the change is such as to diminish its slope, its (d^2y)/(d x^2) will be negative .
It is now time to initiate you into another secret-how to tell whether the result that you get by "equating to zero" is a maximum or a minimum. The trick is this: After you have differentiated (so as to get the expression which you equate to zero), you then differentiate a second time, and look whether the result of the second differentiation is positive or negative . If (d^2 y)/(d x^2) comes out positive , then you know that the value of y which you got was a minimum ; but if (d^2 y)/(d x^2) comes out negative , then the value of y which you got must be a maximum . That's the rule.
The reason of it ought to be quite evident. Think of any curve that has a minimum point in it (like Fig. 15 , p. 80), or like Fig. 34 , where the point of minimum y is marked M , and the curve is concave upwards. To the left of M the slope is downward, that is, negative, and is getting less negative. To the right of M the slope has become upward, and is getting more and more upward. Clearly the change of slope as the curve passes through M is such that (d^2 y)/(d x^2) is positive , for its operation, as x increases toward the right, is to convert a downward slope into an upward one.
Similarly, consider any curve that has a maximum point in it (like Fig. 16, p. 81), or like Fig. 35, where the curve is convex, and the maximum point is marked M . In this case, as the curve passes through M from left to right, its upward slope is converted into a downward or negative slope, so that in this case the "slope of the slope" (d^2 y)/(d x^2) is negative .
Go back now to the examples of the last chapter and verify in this way the conclusions arrived at as to whether in any particular case there is a maximum or a minimum. You will find below a few worked out examples.
(1) Find the maximum or minimum of
(a) y=4 x^2-9 x-6 ;
(b) y=6+9 x-4 x^2 ;
and ascertain if it be a maximum or a minimum in each case.
(a) (d y)/(d x) =8 x-9=0; x=1 1/8 , and y=-11.065.
(d^2 y)/(d x^2) =8 ; it is +; hence it is a minimum.
(b) (d y)/(d x)=9-8 x=0; x=1 1/8 and y=+11.065 . (d^2 y)/(d x^2)=-8 it is - hence it is a maximum.
(2) Find the maxima and minima of the function y=x^3-3 x+16 . (d y)/(d x)=3 x^2-3=0; x^2=1; and x= ± 1. ; (d^2 y)/(d x^2)=6 x; for x=1; it is + hence x=1 corresponds to a minimum y=14 . For x=-1 it is -; hence x=-1 corresponds to a maximum y=+18 .
(3) Find the maxima and minima of y=(x-1)/(x^2+2) . (d y)/(d x)=((x^2+2) × 1-(x-1) × 2 x)/((x^2+2)^2)=(2 x-x^2+2)/((x^2+2)^2)=0 or x^2-2 x-2=0 , whose solutions are x=+2.73 and x=-0.73 . (d^2 y)/(d x^2) =-((x^2+2)^2 ×(2 x-2)-(x^2-2 x-2)(4 x^3+8 x))/((x^2+2)^4) ; =-(2 x^5-6 x^4-8 x^3-8 x^2-24 x+8)/((x^2+2)^4).
The denominator is always positive, so it is sufficient to ascertain the sign of the numerator.
If we put x=2.73 , the numerator is negative; the maximum, y=0.183 .
If we put x=-0.73 , the numerator is positive; the minimum, y=-0.683 .
(4) The expense C of handling the products of a certain factory varies with the weekly output P according to the relation C=a P+ b/(c+P)+d , where a, b, c, d are positive constants. For what output will the expense be least? (d C)/(d P)=a-b/((c+P)^2)=0 for maximum or minimum hence a=b/((c+P)^2) and P= ± √(b/a)-c .
As the output cannot be negative, P=+√(b/a)-c .
Now (d^2 C)/(d P^2)=+(b(2 c+2 P))/((c+P)^4), which is positive for all the values of P ; hence P=+√(b/a)-c corresponds to a minimum.
(5) The total cost per hour C of lighting a building with N lamps of a certain kind is C=N(C_l/t+(E P C_e)/1000) where E is the commercial efficiency (watts per candle), P is the candle power of each lamp, ; t is the average life of each lamp in hours, ; C_l= cost of renewal in pence per hour of use, ; C_e= cost of energy per 1000 watts per hour.
Moreover, the relation connecting the average life of a lamp with the commercial efficiency at which it is run is approximately t=mE^n , where m and n are constants depending on the kind of lamp.
Find the commercial efficiency for which the total cost of lighting will be least.
We have C =N(C_l/m E^(-n)+(P C_e)/1000 E), ; (d C)/(d E) =(P C_e)/1000-(n C_l)/m E^(-(n+1))=0 for maximum or minimum. E^(n+1)=(1000 × n C_l)/(m P C_e) and E=sqrt[n+1](1000 × n C_l)/(m P C_e).
This is clearly for minimum, since (d^2 C)/(d E^2)=(n+1) (n C_l)/m E^(-(n+2)), which is positive for a positive value of E .
For a particular type of 16 candle-power lamps, C_l=17 pence, C_e=5 pence; and it was found that m=10 and n=3.6 . E=sqrt[4.6](1000 × 3.6 × 17)/(10 × 16 × 5)=2.6 watts per candle-power.
Exercises X . (You are advised to plot the graph of any numerical example.) (See p. 258 for the Answers.)
(1) Find the maxima and minima of y=x^3+x^2-10 x+8.
(2) Given y=b/a x-c x^2 , find expressions for dy/(d x) , and for (d^2 y)/(d x^2) , also find the value of x which makes y a maximum or a minimum, and show whether it is maximum or minimum.
(3) Find how many maxima and how many minima there are in the curve, the equation to which is y=1-(x^2)/2+(x^4)/24 and how many in that of which the equation is y=1-(x^2)/2+(x^4)/24-(x^6)/720.
(4) Find the maxima and minima of y=2 x+1+5/(x^2).
(5) Find the maxima and minima of y=3/(x^2+x+1).
(6) Find the maxima and minima of y=(5 x)/(2+x^2).
(7) Find the maxima and minima of y=(3 x)/(x^2-3)+x/2+5.
(8) Divide a number N into two parts in such a way that three times the square of one part plus twice the square of the other part shall be a minimum.
(9) The efficiency u of an electric generator at different values of output x is expressed by the general equation: u=x/(a+b x+c x^2) where a is a constant depending chiefly on the energy losses in the iron and c a constant depending chiefly on the resistance of the copper parts. Find an expression for that value of the output at which the efficiency will be a maximum.
(10) Suppose it to be known that consumption of coal by a certain steamer may be represented by the formula y=0.3+0.001 v^3 ; where y is the number of tons of coal burned per hour and v is the speed expressed in nautical miles per hour. The cost of wages, interest on capital, and depreciation of that ship are together equal, per hour, to the cost of 1 ton of coal. What speed will make the total cost of a voyage of 1000 nautical miles a minimum? And, if coal costs 10 shillings per ton, what will that minimum cost of the voyage amount to?
(11) Find the maxima and minima of y= ± x/6 √(x(10-x)).
(12) Find the maxima and minima of y=4 x^3-x^2-2 x+1.
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End of part 15 of 27
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