Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. There is a tank of one hundred litres of salty water, with ten kilograms of salt dissolved in it. Pure water flows in at five litres per minute, and the mixture flows out at the same rate. So in twenty minutes, one hundred litres flow out, the whole tank! Is all the salt gone after twenty minutes? By the end of this video, you will see why the answer is no, and find exactly how much salt is left. JEE Main often asks questions in this style: solve a differential equation, then use its solution. So, let's begin!

0:39Set 3: first, spot the tool

So, this is Set three, the last set of mixed problems. The pattern stays the same: solve the initial-value problem, then use y. But first, you must spot the right tool. Okay? Let me ask you this. One problem in this set says that the area under a curve, from zero to x, is always one third of x times y. What is your first move? Differentiate both sides, yes, exactly! The tall S is the integral sign, from zero to x, and t is just the variable inside it. Differentiating the integral gives back y, so the area condition turns into a differential equation. Clear? Next, if you see x plus y, times d x minus d y, equals d x plus d y. d x minus d y is the change in x minus y, and d x plus d y is the change in x plus y. So put t equals x plus y, and u equals x minus y. With me? And orthogonal trajectories are curves that cut every member of a family at right angles. First write the differential equation of the family, y prime equals f of x, y; the dash, called prime, means the derivative. Then solve y prime equals minus one over f of x, y, because perpendicular slopes multiply to minus one. Got it?

2:02Rate problems: growth and decay

Now, the tank in our puzzle is a rate problem. The standard model is d N by d t equals k times N, okay? It says the rate of change is proportional to the amount present. So tell me, what is the quickest way to solve this equation? Separate the variables, yes, exactly! It is also linear, but separating is the quickest. Solving it gives N equals N naught times e to the power k t. N naught, with the small zero below, is the amount at t equals zero. Right? If k is negative, the amount decays. The half-life, the time to fall to half, is the natural log of two, over mod k. Here l n means natural log, and the two vertical bars, mod, give the size of k without its sign. Clear?

2:52Separable: watch what you divide by

One more tool before the examples. Separable means d y by d x equals g of x times h of y. The double arrow means implies: the integral of d y over h of y equals the integral of g of x, d x, plus capital C, with one constant on one side only. Okay? But to get there, we divided by h of y. What could go wrong? Yes, exactly, h of y could be zero! So dividing assumes that h of y is not equal to zero; that is the slashed equals sign. Right? So if h of y naught is zero, the constant y equals y naught is a solution that the division can lose. Always check such solutions separately. Got it?

3:38Example 1: salt tank, 10 kg in 100 L, salt left after 20 min

Remember our puzzle? Let's solve it. Example one: a tank of one hundred litres holds brine, that is, salty water, with ten kilograms of salt. On screen, capital L means litres, k g means kilograms, and min means minutes. Pure water flows in at five litres per minute, and the well-mixed solution flows out at the same rate. Find the salt left after twenty minutes. Pause the video, and try it first. Done? Okay, let's check. Let capital S be the salt in kilograms at time t, in minutes. In and out are equal, so the volume stays one hundred litres, and each litre holds S over one hundred kilograms of salt. Five litres leave every minute, so five times S over one hundred kilograms of salt leave per minute; the cross means times. With me? Pure water brings in no salt. So d S by d t equals minus five S over one hundred, which is minus S over twenty, and S of zero is ten. Why the minus sign? Because the salt is leaving. Okay? This is our decay model, with k equal to minus one over twenty. So capital S equals ten times e to the power minus t over twenty; the slash in the power means over. Right? In twenty minutes, one hundred litres have flowed out. So is capital S zero now? That is the trap, and the answer is no. The water leaving gets weaker and weaker in salt, so less and less salt leaves each minute. At t equals twenty, capital S is ten times e to the power minus one. The wavy equals sign means approximately: about three point six eight kilograms. And e to the power minus t over twenty is never zero, so the salt never fully runs out. And that solves our puzzle!

5:32Example 2: xy' = y2 - y, y(1) = 2, find y(3/2) and blow-up

Now, example two, a harder one. x times y prime equals y squared minus y, and y of one equals two. Find y of three halves, and the point where the solution blows up, that is, where y shoots off to infinity. Pause the video, and try it first. Done? Okay, let's check. We see y prime and y here, so is it a linear equation? That is the trap: y squared is there, so it is not linear. But x and y can be pulled apart, so it is separable: d y over y times y minus one, equals d x over x. And this step assumes y is not zero and not one. Okay? Now, we divided by y times y minus one, which is zero at y equals zero and at y equals one. So did this division lose our answer? No, it did not. The constants y equals zero and y equals one are solutions, but neither passes through the point where x is one and y is two. So nothing is lost here. Clear? Now split into partial fractions: one over y times y minus one, equals one over y minus one, minus one over y. Integrate both sides: the natural log of mod of y minus one over y, equals the natural log of mod x, plus small c, a constant. With me? Remove the logs: y minus one over y equals capital C times x, with capital C a new constant. Put x equals one and y equals two, and capital C comes out as one half. Solve for y, and you get y equals two over two minus x. Right? So, y of three halves. Two minus three halves is one half, and two over one half is four. Okay? Now, where does this solution blow up? Where the denominator is zero, exactly: two minus x equals zero, so at x equals two. As x comes close to two from the left, y shoots up without limit. So the answers are four, and x equals two. Got it?

7:46Quick revision: spot the tool, decay model, lost solutions

So, quick revision. First, spot the tool: differentiate an area condition, substitute, or for orthogonal trajectories, use minus one over f of x, y. Rates: d N by d t equals k N gives N equals N naught times e to the power k t, and a decaying amount never reaches zero. Separable: check the zeros of what you divide by, and where y blows up. Now try the practice problems on this page. One of them asks when our tank is down to five kilograms, so remember the half-life. All the best!