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Shame on them!!! Calculus made easy: being a very-simplest introduction to those beautiful methods of reckoning which are generally called by the terrifying names of the differential calculus and the integral calculus , Macmillan.
Calculus made easy Calculus made easy: Being a very-simplest introduction to those beautiful methods of reckoning which are generally called by the terrifying names of differential calculus and the integral calculus Aug 27, , Macmillan. Calculus Made Easy: being a very simplest introduction to those beautiful methods of recknoning which are generally called by the terrifying names of the differntial calculus and the integral calculus.
Calculus made easy. Publisher unknown. Edition Notes Includes index. T45 , QA T45 External Links Contributor biographical information Publisher description. The Physical Object Pagination vi, p. Community Reviews 0 Feedback? Loading Related Books. Then, if we were told to differentiate it, that means we must find its differential coefficient with respect to t. But if we agree that we may ignore small quantities of the second order, 1 may be rejected as compared with 10, ; so we may round off the enlarged y to 10, Well, try again, making dx a still smaller bit.
Now the last figure 1 is only one-millionth part of the 10, , and is utterly negligible; so we may take 10, without the little decimal dy 20 at the end. Case 2. Let us collect the results to see if we can infer any general rule. Put them in two columns, the values of y dy in the other: thus in one and the corresponding values found for dx y dy dx x2 2x x3 3x2 x4 4x3 Just look at these results: the operation of differentiating appears to have had the effect of diminishing the power of x by 1 for example in the last case reducing x4 to x3 , and at the same time multiplying by a number the same number in fact which originally appeared as the power.
Now, when you have once seen this, you might easily conjecture how the others will run. You would expect that differentiating x5 would give 5x4 , or differentiating x6 would give 6x5. If you hesitate, try one of these, and see whether the conjecture comes right. Case of a negative power. Case of a fractional power. Agreeing with the general rule. Let us see how far we have got. Exercises I. See p. How easy it is! In our equations we have regarded x as growing, and as a result of x being made to grow y also changed its value and grew.
We usually think of x as a quantity that we can vary; and, regarding the variation of x as a sort of cause, we consider the resulting variation of y as an effect. In other words, we regard the value of y as depending on that of x. Added Constants. It added nothing to the growth of x, and does not enter into the differential coefficient. If we had put 7, or , or any other number, instead of 5, it would have disappeared.
So if we take the letter a, or b, or c to represent any constant, it will simply disappear when we differentiate. Multiplied Constants. Carefully compare the two figures, and verify by inspection that the height of the ordinate of the derived curve, Fig. To the left of the origin, where the original curve slopes negatively that is, downward from left to right the corresponding ordinates of the derived curve are negative.
Now if we look back at p. If we had taken 8x2 , the differential coefficient would have come out eight times as great as that of x2. And, what is true about multiplication is equally true about division: for if, in the example above, we had taken as the constant the same 1 7 1 7 instead of 7, we should have had come out in the result after differentiation.
Some Further Examples. As a rule an expression of this kind will need a little more knowledge than we have acquired so far; it is, however, always worth while to try whether the expression can be put in a simpler form. It means that a change of radius of 1 inch will cause a change of volume of Exercises II. Make up some other examples for yourself, and try your hand at differentiating them. Find the change of length of the rod per degree Centigrade. This formula assumes that 4t is small compared to D.
Compare the rate at which P varies for a small change of thickness and for a small change of diameter taking place separately. How are we to go to work on this new job? And this is the way. You differentiate each function separately and add the results. So if now we take the example of the preceding paragraph, and put in the values of the two functions, we shall have, using the notation shown p. Now there are two ways in which we may go to work. First way.
Do the multiplying first, and, having worked it out, then differentiate. You should note that this process amounts to the following: Treat u as constant while you differentiate v; then treat v as constant while you dy differentiate u; and the whole differential coefficient will be the sum dx of these two treatments.
Now, having found this rule, apply it to the concrete example which was considered above. Lastly, we have to differentiate quotients. So there is nothing for it but to go back to first principles, and find a rule. Multiply the divisor function by the differential coefficient of the dividend function; then multiply the dividend function by the differential coefficient of the divisor function; and subtract.
Lastly divide by the square of the divisor function. Some further examples fully worked out are given hereafter. Same remarks as for preceding example. This is not, however, always possible; see, for instance, p.
The side of the bottom is feet. Find an expression for the quantity pouring in or out when the depth of water varies by 1 foot; hence find, in gallons, the quantity withdrawn hourly when the depth is reduced from 14 to 10 feet in 24 hours. Gallons per hour corresponding to a change of depth of 4 ft. Expand the numerator by the binomial theorem see p. Exercises III. See the Answers on p. Find an expression giving the variation of the current corresponding to a variation of temperature.
Find the rate of variation of the resistance with regard to temperature as given by each of these formulae. Find an expression for the variation of the electromotive-force a with regard to the length of the arc; b with regard to the strength of the current.
Let us try the effect of repeating several times over the operation of differentiating a function see p. Begin with a concrete case. First differentiation, 5x4. There is a certain notation, with which we are already acquainted see p. This is to employ the general symbol f x for any function of x. Now let us generalize. But this is not the only way of indicating successive differentiations. See page for Answers.
Some of the most important problems of the calculus are those where time is the independent variable, and we have to think about the values of some other quantity that varies when the time varies. Some things grow larger as time goes on; some other things grow smaller. The distance that a train has got from its starting place goes on ever increasing as time goes on. Trees grow taller as the years go by.
Which is growing at the greater rate; a plant 12 inches high which in one month becomes 14 inches high, or a tree 12 feet high which in a year becomes 14 feet high? In this chapter we are going to make much use of the word rate. Nothing to do with poor-rate, or water-rate except that even here the word suggests a proportion—a ratio—so many pence in the pound.
Nothing to do even with birth-rate or death-rate, though these words suggest so many births or deaths per thousand of the population. When a motor-car whizzes by us, we say: What a terrific rate!
When a spendthrift is flinging about his money, we remark that that young man is living at a prodigious rate. What do we mean by rate? In both these cases we are making a mental comparison of something that is happening, and the length of time that it takes to happen. Now in what sense is it true that a speed of 10 yards per second is the same as yards per minute? Ten yards is not the same as yards, nor is one second the same thing as one minute.
What we mean by saying that the rate is the same, is this: that the proportion borne between distance passed over and time taken to pass over it, is the same in both cases. Take another example. A man may have only a few pounds in his possession, and yet be able to spend money at the rate of millions a year—provided he goes on spending money at that rate for a few minutes only.
Suppose you hand a shilling over the counter to pay for some goods; and suppose the operation lasts exactly one second. Now try to put some of these ideas into differential notation. Let y in this case stand for money, and let t stand for time. A more apt illustration of the idea of a rate is furnished by the speed of a moving body.
From London Euston station to Liverpool is miles. You are dividing one by the other. If y is the whole distance, and t the whole time, clearly y the average rate is. Now the speed was not actually constant all the t way: at starting, and during the slowing up at the end of the journey, the speed was less. Probably at some part, when running downhill, the speed was over 60 miles an hour. If, during any particular element of time dt, the corresponding element of distance passed over was dy, dy then at that part of the journey the speed was.
The rate at which dt one quantity in the present instance, distance is changing in relation to the other quantity in this case, time is properly expressed, then, by stating the differential coefficient of one with respect to the other. The rate at which a velocity is increasing is called the acceleration.
When a railway train has just begun to move, its velocity v is small; but it is rapidly gaining speed—it is being hurried up, or accelerated, d2 y by the effort of the engine. So its 2 is large. When it has got up its dt d2 y top speed it is no longer being accelerated, so that then 2 has fallen dt to zero.
The force necessary to accelerate a mass is proportional to the mass, and it is also proportional to the acceleration which is being imparted. If we differentiate momentum with d mv for the rate of change of momentum. That is to say, force may be expressed either as mass times acceleration, or as rate of change of momentum. Again, if a force is employed to move something against an equal and opposite counter-force , it does work ; and the amount of work done is measured by the product of the force into the distance in its own direction through which its point of application moves forward.
But as dy is only an element of length, only an element of work will be done. In this last sentence the word rate is clearly not used in its time-sense, but in its meaning as ratio or proportion. Sir Isaac Newton, who was along with Leibnitz an inventor of the methods of the calculus, regarded all quantities that were varying as flowing; and the ratio which we nowadays call the differential coefficient he regarded as the rate of flowing, or the fluxion of the quantity in question.
He did not use the notation of the dy and dx, and dt this was due to Leibnitz , but had instead a notation of his own. But this notation does not tell us what is the independent variable with dy respect to which the differentiation has been effected. When we see dt dy we know that y is to be differentiated with respect to t. If we see dx we know that y is to be differentiated with respect to x.
So, therefore, to mean dx dt dz this fluxional notation is less informing than the differential notation, and has in consequence largely dropped out of use. But its simplicity gives it an advantage if only we will agree to use it for those cases exclusively where time is the independent variable.
Find also the average velocity during the first 10 seconds of its motion. Suppose distances and motion to the right to be positive. The body started from a point The body began moving 7. To find the distance travelled during the 10 first seconds of the motion one must know how far the body was from the point O when it started.
So, in 10 seconds, the distance travelled was The acceleration is no more constant. The body is at rest, but just ready to move with a negative acceleration, that is to gain a velocity towards the point O. The body is moving towards the point O with a velocity of 3 ft. In the last two cases the mean velocity during the first 10 seconds and the velocity 5 seconds after the start will no more be the same, because the velocity is not increasing uniformly, the acceleration being no longer constant.
At what time is it at rest, and how many revolutions has it performed up to that instant? This is a retardation; the wheel is slowing down. The velocity is reversed. Draw a curve showing the relation between s and t. Is the acceleration the same for all values of t? Find the angular velocity in radians per second of that wheel when 1 21 seconds have elapsed. Find also its angular acceleration. Find the expression for the velocity and the acceleration at any time; and hence find the velocity and the acceleration after 3 seconds.
Find an expression for the velocity and the acceleration at any time. Draw curves to show the variation of height, velocity and acceleration during the first ten minutes of the ascent.
Find the velocity and acceleration after 10 seconds. Find the value of n when the velocity is doubled from the 5th to the 10th second; find it also when the velocity is numerically equal to the acceleration at the end of the 10th second. Sometimes one is stumped by finding that the expression to be differentiated is too complicated to tackle directly.
By and bye, when you have learned how to deal with sines, and cosines, and exponentials, you will find this dodge of increasing usefulness. Let us practise this dodge on a few examples. See Ex. Exercises VI. It is useful to consider what geometrical meaning can be given to the differential coefficient. Let P QR, in Fig. 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. 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 dy is the slope of the curve along QT. The straight line QT the ratio dx 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.
This tangent to the curve has evidently the same slope as QT , so dy that is the slope of the tangent to the curve at the point Q for which dx dy the value of is found. We shall hereafter make considerable use of this circumstance that dy represents the slope of the curve at any point. In other words its value of dx slope is constant.
If a curve is one that turns more upwards as it goes along to the dy right, the values of will become greater and greater with the indx creasing steepness, as in Fig. If a curve is one that gets flatter and flatter as it goes along, the dy values of will become smaller and smaller as the flatter part is dx Y Q dx dy y O x dx Fig.
If a curve first descends, and then goes up again, as in Fig. In dx 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 Fig. O X Fig. The characteristic of a minimum is that y must increase on either side of it. 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.
If a curve has the form of Fig. It is plotted out in Fig. X Fig. The line has a gradient of 1 in 1. And this slope is constant—always the same slope. The slope will be constant, slope by differentiating; which gives dx at an angle, the tangent of which is here called a. Let us assign to a some numerical value—say 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 Fig. So, draw the line in Fig.
Again the curve will start on the y-axis at a height b above the origin. Now differentiate. On the left of the origin, where x dy has negative values, will also have negative values, or will descend dx from left to right, as in the Figure. Let us illustrate this by working out a particular instance.
Tabulating results, we have: x 0 1 2 3 4 5 y 3 3 14 4 5 14 7 9 14 dy dx 0 1 2 1 1 12 2 2 12 Then plot them out in two curves, Figs. For any assigned value of x, the height of the ordinate in the second curve is proportional to the slope of the first curve.
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