Rate of Change

Transcript

0:00Introduction

Hello everyone! Let us start with a balloon. You pump air into it at a steady rate: nine hundred cubic centimetres, every second. So the radius also grows at a steady rate, right? Hmm, not so fast. In fact, the radius grows more and more slowly. Why? Today we find out, with one small idea: rate of change. Questions on rates come up often in JEE, so let's begin.

0:29Rate of change and the chain rule

So, first the symbol on screen: d y by d x. It is the derivative of y with respect to x, and it is not a division, okay? It tells us the rate of change of y with respect to x: how fast y changes as x changes. Clear? Now, in real problems, both x and y change with time, t. If we know how fast x is changing, how do we get how fast y is changing? The chain rule! Yes, exactly. d y by d t equals d y by d x, times d x by d t. The small dot in the middle means times. We call this related rates: two rates, linked by one equation. With me? Let us try it on a circle. Its area, capital A, equals pi r squared, where r is the radius, and both change with time. So, what is d A by d t? Careful now. Two pi r, times d r by d t. Correct! If you wrote just two pi r, oh no, careful! That is only d A by d r. For the rate with respect to time, d r by d t must be there. And the double arrow means this gives that. Got it? Now some numbers. The radius grows at three centimetres per second; the minus one in the power on s means per second. When the radius is ten centimetres, d A by d t is two pi, times ten, times three: sixty pi square centimetres per second. The small two makes them square centimetres. Right?

2:02The method for any rates question

Alright, now the method for every rates question. Just four steps. Step one: draw the figure, and label the quantities that change. Okay? Step two: write a relation between them that is true at all times, not just at one moment. Right? Step three: differentiate both sides with respect to time, t. Each changing quantity gives its rate. Clear? And step four? First, a question for you. Why not put the numbers in first, and differentiate after? It looks faster, doesn't it? Because once you put r equals ten, r becomes a constant, and the derivative of a constant is zero! The rates vanish, and the answer goes wrong. So, step four: substitute the values at that instant, only at the end. Don't fall for this one! Got it?

2:54Differentials: small changes and errors

Now, the same derivative also measures small changes. The capital Greek letter delta, the little triangle, means a small change; so delta x is a small change in x. The wavy equals sign means approximately equal. So, delta y is approximately f prime of x, times delta x. The small dash, read as prime, marks the derivative. Okay? Next, the relative error: delta y over y, the change compared with the whole quantity. Multiply it by a hundred, and you get the percentage error. Clear? And now a lovely shortcut. If y equals x to the power n, then delta y over y is approximately n times delta x over x. So the relative error simply gets multiplied by the power, n. With me? Quick check. The radius of a sphere is measured with a one percent error. What is the percentage error in its volume? Three percent! Very good. The volume of a sphere is proportional to r cubed; the sign like a sideways alpha means proportional to. So the power is three, and three times one percent is three percent. Right?

4:08Example 1: rate of growth of balloon radius at r = 15 cm

Now, remember our balloon? Let's solve it. Air goes into a spherical balloon at nine hundred cubic centimetres per second; the small three means cubic. How fast is the radius increasing when the radius is fifteen centimetres? Pause the video, and try it with the four steps. Done? Okay, let's check. Step one, the figure: a sphere of radius r and volume capital V, both growing. Step two, the relation true at all times: capital V equals four over three, pi r cubed. Okay? Step three: differentiate both sides with respect to t. What do you get? d V by d t equals four pi r squared, times d r by d t. Correct! Three times four over three is four, and the chain rule brings in d r by d t. Right? Step four, and only now the numbers. d V by d t is nine hundred, and r is fifteen, so r squared is two hundred and twenty five. So, nine hundred equals four pi times two hundred and twenty five, times d r by d t. Clear? So, d r by d t equals nine hundred, over four pi times two hundred and twenty five. And four times two hundred and twenty five is nine hundred! The nine hundreds cancel. One over pi centimetres per second. Lovely! Now, back to our puzzle. In general, d r by d t equals nine hundred over four pi r squared. See, r squared sits in the denominator. So the bigger the radius, the slower it grows. Mystery solved! With me?

5:45Example 2: how fast the top of a sliding ladder falls

Now a harder one, and a classic. Here, a five metre ladder leans against a wall. Its foot slides away at two metres per second. How fast does the top slide down, when the foot is four metres from the wall? Pause the video, and try it. Done? Okay, first a quick guess. The foot moves at two metres per second. So the top also slides down at two metres per second, right? Same ladder! No, no, wait! Let us not guess; let us follow the four steps. Let x be the distance of the foot from the wall, and y the height of the top. The wall, the ground and the ladder make a right triangle, so by Pythagoras, x squared plus y squared equals twenty five, at every moment. Okay? At the instant asked, x is four. So y is the square root of twenty five minus sixteen, which is three. But careful: we keep these numbers aside for step four. Got it? Step three: differentiate with respect to t. New notation here: x with a dot on top, called x dot, means d x by d t, and y dot means d y by d t. We get two x times x dot, plus two y times y dot, equals zero. Divide by two: x times x dot, plus y times y dot, equals zero. Clear? Now step four. x is four, y is three, and x dot is two. Find y dot. y dot equals minus x times x dot, over y: minus four times two, over three. Minus eight over three metres per second. Correct! Now, why the minus sign? Because y is decreasing: the top is moving down. So, the top slides down at eight over three metres per second. That is more than two! Our quick guess was wrong. Always trust the relation, not the guess.

7:47Quick revision: related rates and percentage error

Okay? So, quick revision. Related rates: d y by d t equals d y by d x, times d x by d t. Write the relation first, differentiate, and substitute the numbers last. And for errors: if y equals x to the power n, the relative error gets multiplied by n. Now go and try the practice problems on this page. All the best!