Errors in General Formulae

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a squared minus b squared. From algebra, it is the same as a plus b, times a minus b, right? Put in the same measured values, and find its error both ways. One way gives ten percent. The other way gives twelve and a half percent! Same quantity, two different errors? Hmm, both cannot be right, can they? Today you will see one method that works for any formula, and find exactly where the extra error sneaks in. And look, in JEE, parallel resistors and the mirror formula hide exactly this trap. So, let's begin.

0:42One method for any formula

So far, you have separate rules for sums, and for products and powers, okay? But what if Z is some general function, f of A, B and so on, like you see on screen? We need one method for everything. Now think. An error is just a small change in a measured quantity, isn't it? So which tool from maths tells you how a small change in A changes Z? Yes, differentiation! Exactly. So, step one: differentiate to get d Z, a small change in Z, okay? For products and powers, take the log first; the l n on screen is the natural log. That is exactly where the power rule came from, remember? Step two: replace d A and d B by the errors, delta A and delta B, okay? This little triangle, the capital delta, here means the error in that quantity. Alright, now suppose one term comes with a minus sign. Do we subtract it? No! Each error can push the answer either way, and we want the maximum possible error. So, step three: add the absolute values of all the terms. Every term becomes positive. Three steps, and any formula is done. Really simple.

2:01The trap: one quantity in two places

Now, the trap. Two resistors, R one and R two, in parallel. The equivalent resistance is R one times R two, divided by R one plus R two. Simple quotient, isn't it? So just add the percentage errors of the top and the bottom, like the product rule says. Right? No, no, careful! This is the classic JEE trap. Look closely. R one sits in the numerator, and R one sits in the denominator too. Same measurement, in two places, okay? Now, what if R one comes out a little too big? The top gets bigger, so R goes up. But the bottom also gets bigger, so R comes down. The two effects partly cancel, right? But the product rule treats top and bottom as separate measurements, and simply adds both errors. So it counts the same error twice, and it overestimates. Remember: when one quantity appears in several places, do not split the formula into separate factors. Got it?

3:03The fix: differentiate the whole formula

So what do we do? Can you write the same parallel formula so that each resistor appears only once? Yes, the reciprocal form! You already know it. One over R equals one over R one, plus one over R two. Same formula, but now each resistor appears in just one place. Clear? Now differentiate the whole thing. The derivative of one over R is minus d R over R squared. So, minus d R over R squared equals minus d R one over R one squared, minus d R two over R two squared, okay? Now steps two and three. Replace each d by delta, and take absolute values, so the minus signs go. Delta R over R squared equals delta R one over R one squared, plus delta R two over R two squared. This is the error formula for parallel resistors. And here is a topper's trick. The mirror formula, one over f equals one over u plus one over v, has exactly the same shape. The double arrow means so: delta f over f squared equals delta u over u squared, plus delta v over v squared. Same shape, same error formula. That's it!

4:18Example 1: error in two resistors in parallel

Now, example one, a proper JEE pattern question. Two resistors, one hundred ohm plus minus three, and two hundred ohm plus minus four, are joined in parallel. The omega on screen is the symbol for ohm, okay? Find the equivalent resistance with its error. Pause the video, and try it first. Done? Okay, let's check. First, the value: one hundred times two hundred, divided by three hundred. Sixty six point seven ohm. Now our error formula. R one squared is ten to the four, and R two squared is four times ten to the four. So, three over ten to the four, plus four over four times ten to the four. That second term is just one over ten to the four, right? Total: four over ten to the four. Now, careful. This is delta R over R squared, okay? So we multiply by R squared, not by R alone. Sixty six point seven squared is about four thousand four hundred and forty nine. Times four into ten to the minus four. That gives about one point eight ohm. The wavy equals sign means approximately. So, R equals sixty six point seven, plus minus one point eight ohm. Notice, we never split the formula into top and bottom. The whole formula handled the cancelling for us. Clear?

5:48Example 2: error in a squared minus b squared over c

Now a harder one, and this one solves our puzzle. The quantity x equals a squared minus b squared, over c. Here, a is five point zero, plus minus zero point one; b is three point zero, plus minus zero point one; and c is two point zero, plus minus zero point zero two. Find x with its maximum error. Pause and try. Hint: keep the numerator whole. Done? Okay, let's check. Twenty five minus nine is sixteen, divided by two point zero: eight point zero. Easy. Now the error in a squared minus b squared. Many students factorise it into a plus b, times a minus b, and use the product rule. Sounds smart, doesn't it? Hmm, wait. Where do a and b appear now? In both brackets! Our trap again. So, we differentiate the numerator as a whole. Its error is two a delta a, plus two b delta b; the minus sign becomes plus, because we add absolute values. Two times five times zero point one is one point zero, and two times three times zero point one is zero point six. Total: one point six. Now, one point six out of sixteen is ten percent. And c? Zero point zero two out of two point zero is one percent, and since c just divides, it simply adds. Ten plus one: eleven percent. Eleven percent of eight is about zero point nine. So, x equals eight point zero, plus minus zero point nine. But is ten percent really right for the numerator? Let's check with the extremes. Take a as five point one and b as two point nine: seventeen point six. Take a as four point nine and b as three point one: fourteen point four. Exactly sixteen, plus minus one point six. Ten percent, absolutely correct! And the factorised way? Here a plus b is eight, with error zero point two: two point five percent. And a minus b is two, with the same error zero point two: ten percent. Add them: twelve and a half percent. Too large! Those two errors are not independent: the same a and b sit in both brackets. So, puzzle solved. Ten percent is right; twelve and a half percent counts the same error twice. Got it?

8:20Quick revision: differentiate, then add absolute errors

So, quick revision. For any formula: differentiate, replace d by delta, and add the absolute values. When the same quantity appears in several places, differentiate the whole formula; do not split it into separate factors. And for parallel resistors, use the one over R form, not the product rule. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!