Growth, Decay and Other Rate Problems

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle: a body at eighty degrees Celsius is kept in a room at twenty degrees Celsius, and in ten minutes it cools to sixty degrees. When will it reach forty degrees? Most students say twenty minutes in all, twenty degrees every ten minutes. It sounds perfect, but it is wrong, and by the end of this video you will see exactly why. Rate problems like this do come up in JEE, so let's begin.

0:30Rate proportional to the amount

So, the first model. N is an amount, like a population or a radioactive mass, and t is time. d N by d t, the rate at which N changes, equals k times N, with k a constant. So the bigger the amount, the faster it changes. Okay? This is a separable equation, just like in the last lesson. So how do we separate the variables here? Yes, exactly, divide by N and multiply by d t, to get d N over N equals k, d t. Dividing by N assumes N is not zero; N is a positive amount, so that is safe. The double arrow means implies. Integrating, l n N, the natural log of N, equals k t, plus small c, with no modulus, since N is positive. Right? Put t equals zero: the starting amount is N naught, with a small zero below the N, so small c is l n of N naught. Taking e to the power of both sides, N equals N naught times e to the power k t. And k greater than zero means growth, while k less than zero means decay. Clear?

1:37Doubling time and half-life

Next, the doubling time. This is growth, so k is positive. We want N to become two times N naught. So e to the power k t equals two, k t equals l n two, and t equals l n two over k. Okay? Now decay, so k is negative. The half-life is the time in which N falls to half of N naught. Can you find it, the same way? Let's see, e to the power k t equals one half, so k t equals minus l n two. Since k is negative, t equals l n two over mod k; the two vertical bars mean the modulus. So the time comes out positive. Got it? And notice, neither formula has N naught in it. So a big sample and a small sample of the same substance have the same half-life. With me?

2:25Newton's law of cooling

Now, Newton's law of cooling. T is the temperature of a body, and T s, with a small s below it, is the temperature of the surroundings. d T by d t equals minus k times, T minus T s, with k positive. So a bigger gap means faster cooling. The figure on the right shows two cooling bodies; we will come back to it in a moment. Okay? Since T s is a constant, the gap, T minus T s, changes exactly as fast as T does. So which quantity follows the decay law here? Yes, exactly, the gap itself! Its rate is minus k times the gap. So T minus T s equals T naught minus T s, times e to the power minus k t, where T naught is the starting temperature. Right? Move T s across, and T equals T s, plus T naught minus T s, times e to the power minus k t. Clear? Here is a quick check: as time goes on, does the temperature itself decay to zero? No, and the figure shows why. Bodies starting at eighty and at fifty, in a room at twenty, both flatten towards the dashed line at twenty. The arrow means tends to, and the sideways eight is infinity. So as t tends to infinity, T tends to T s, not to zero. Got it?

3:43Two more patterns: drift and geometry

Next, linear drift. d P by d t equals a times P, minus b, where a and b are constants. Okay? Tell me, for which value of P does P stop changing? Yes, when the rate is zero, so a times P equals b, and P equals b over a. This is P star, the equilibrium; the little star is just part of its name. Right? And the gap, P minus P star, follows the exponential law: P minus P star equals P naught minus P star, times e to the power a times t. Same trick as cooling: measure from the equilibrium. Clear? Last, geometry. The small dash after y is a prime, and y prime, d y by d x, is the slope of the curve. If a question gives the slope at every point of a curve, that rule is a differential equation for the curve, and in the problems of this lesson it is a separable one. With me?

4:41Example 1: population doubles in 50 years, when it triples

Now, Example one. Here is the question: a population grows at a rate proportional to its size, and doubles in fifty years. When does it triple? Pause the video and try it first. Done? Let us first make a quick guess. It gained N naught in fifty years, so it gains two times N naught in one hundred years, and triples then, right? No, no, that is the trap, because it is linear thinking. In one hundred years it doubles twice, so it becomes four times N naught, not three times. So the answer must be less than one hundred years. With me? Now let us solve it properly. N equals N naught times e to the power k t, and doubling in fifty years gives e to the power fifty k equals two. Okay? Tripling at time t gives e to the power k t equals three. Taking l n of both, fifty k equals l n two, and k t equals l n three. Right? Divide, and k cancels: t equals fifty times l n three over l n two. The wavy equals sign means approximately equal, and t is about seventy nine point two years. Less than one hundred years, just as we said, perfect.

5:58Example 2: body cools 80 to 60 C in 10 min, when is it 40 C

Now, Example two, and this is our puzzle from the start. Here it is: a body at eighty degrees Celsius, in a room at twenty degrees Celsius, cools to sixty degrees Celsius in ten minutes. When does it reach forty degrees Celsius? The small circle means degrees, and m i n means minutes. Pause the video and try it first. Done? Okay, let's check. T s is twenty and T naught is eighty, so T minus twenty equals eighty minus twenty, times e to the power minus k t, which is sixty times e to the power minus k t. Okay? At t equals ten, T is sixty, so the gap is forty. Forty equals sixty times e to the power minus ten k, so e to the power minus ten k equals two thirds. With me? Now we want T equal to forty. So tell me, what is the gap then? Yes, twenty degrees. So twenty equals sixty times e to the power minus k t, and e to the power minus k t equals one third. Right? Taking l n, ten k equals l n of three over two, and k t equals l n three. Divide, and t equals ten times l n three, over l n of three over two, which is about twenty seven point one minutes. So it is not twenty minutes, as most students guess. Every ten minutes the gap is multiplied by two thirds, so the cooling slows down. Clear?

7:26Quick revision: doubling time, half-life and the cooling gap

So, quick revision. Growth and decay: N equals N naught times e to the power k t, with doubling time l n two over k, and half-life l n two over mod k. Cooling: the gap, T minus T s, decays, and T tends to T s. And in equal times, N, or the cooling gap, is multiplied by equal factors, not changed by equal amounts. Now go and try the practice problems on this page. All the best!