Circles, Conics and Regions Given by Inequalities

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle first. Take the circle x squared plus y squared equals four, and the parabola y squared equals three x. Put three x for y squared in the circle: x squared plus three x equals four. So x equals one, or x equals minus four. Two meeting points, right? But the parabola never reaches the left of the y axis! So where did minus four come from? Hold that thought. Today: areas inside circles and parabolas, and regions given by inequalities. These come up often in JEE, so let's begin.

0:40Reading a region from its inequalities

So, how do we read a region like this? The curly brackets are set notation, x comma y is a point, and the colon is read as such that. It means all points satisfying every condition after the colon. Okay? One condition at a time. y greater than or equal to f of x, the sign with a line under it, means on the curve or above it. And y less than or equal to g of x means on the curve or below it. Right? Now a circle of radius r. Is x squared plus y squared less than or equal to r squared the inside of the circle, or the outside? Inside, yes, exactly! x squared plus y squared is the square of the distance from the origin, the centre. So that distance is at most r: inside the circle, or on it. Clear? Now the parabola, with a greater than zero. y squared less than or equal to four a x means y lies between minus two root of a x and plus two root of a x; root is the tick-shaped square root sign. That is the inside of the parabola, between its two arms. And here is the trap: students keep only the upper arm and lose the lower half. So always keep both arms. Got it?

2:00Circle and parabola: split where the top edge changes

The dark blue curve is the circle x squared plus y squared equals four, radius two. The orange one is the parabola y squared equals three x. The shaded region is inside both. Step one: sketch, and mark the meeting points, the black dots at one, plus or minus root three. The plus-minus sign means both values. Okay? The top half and the bottom half are mirror images. What in the equations makes this happen? Correct! y appears only as y squared, so changing y to minus y changes nothing. The picture is symmetric about the x axis, so the area is two times the top half. Right? Now walk along the top edge, from the origin to the right. Which curve is the top edge? First the orange parabola, up to the dot at x equals one. Then the dark blue circle, right up to x equals two. So the top edge changes at x equals one: split the integral there. With me?

3:03The tool: area under a circle

Now, one tool for areas under a circle. The circle x squared plus y squared equals a squared has radius a, and its upper half is y equals root of a squared minus x squared. Clear? The long S is the integral sign, and d x means with respect to x. The integral of root of a squared minus x squared, d x, equals x over two, times the same root, plus a squared over two, times sine inverse of x over a, plus capital C, the constant of integration. The little minus one means inverse, not one over sine. Got it? Quick check. What should the integral from zero to a be? Zero and a are the limits. Think of the picture. One quarter of the circle, yes! At x equals a, the first term is zero, and sine inverse of one is pi over two: a squared over two, times pi over two. At zero, it is all zero. So pi a squared over four, exactly one quarter. The formula works!

4:08Example 1: area between y = x squared and y = mod x

Example one. Find the area of the set of points x comma y with x squared less than or equal to y, less than or equal to mod x. The two straight bars are the modulus, read as mod x. Pause the video and try it first. Done? Okay, let's check. x squared less than or equal to y: on or above the parabola y equals x squared. y less than or equal to mod x: on or below the V shape. So the region is trapped between them. Right? Now think. The V goes on forever. Does our region also go on forever? No! Far out, the parabola climbs above the V and stays there. So find where they meet. For x greater than or equal to zero, mod x is just x, so x squared equals x. The double arrow means, so it follows: x is zero or one. Clear? Both mod x and x squared are even: unchanged when x becomes minus x. So the region is symmetric about the y axis: double the right half. Okay? Now, between zero and one, which is on top: x, or x squared? y equals x, yes, exactly! Test one half: x is one half, but x squared is only one quarter. So the area equals two times the integral from zero to one of x minus x squared, d x. That is two times, one half minus one third, which is one third square units. Perfect!

5:47Example 2: area inside circle x2+y2=4 and parabola y2=3x

Example two finishes our puzzle. Find the area inside both the circle x squared plus y squared equals four, and the parabola y squared equals three x. Pause the video and try it first. Done? Okay, let's check. Substitute y squared equals three x in the circle: x squared plus three x equals four. Factorise: x plus four, times x minus one, equals zero. Okay? So x equals one, or x equals minus four. Two meeting points, right? No, no, careful here! At x equals minus four, the parabola gives y squared equals minus twelve; the less-than sign says below zero. But a square is never negative, so there is no real point there. Reject it, and that solves our puzzle! So they meet only at x equals one, y plus or minus root three. Got it? Area equals two times the bracket: the integral from zero to one of root of three x, d x, under the parabola, plus the integral from one to two of root of four minus x squared, d x, under the circle. The two in front adds the bottom half. Right? First piece: root x integrates to two thirds from zero to one, so we get root three times two thirds. That is two over root three. With me? Second piece: the circle tool, with a equals two. At two: zero, plus two times sine inverse of one, that is pi. At one: root three over two, plus two times sine inverse of one half, that is pi over three. So pi, minus root three over two, minus pi over three: two pi over three, minus root three over two. Clear? Now two times the bracket: four over root three, plus four pi over three, minus root three. And four over root three, minus root three, is one over root three. So the area is four pi over three, plus one over root three square units. Very good!

7:59Quick revision: meeting points, split, symmetry, circle area

Quick revision. Read each inequality: above or below a curve, inside or outside a circle. Find the meeting points, and reject any that are not real points. Split where the top edge changes, and use symmetry to double. And under a circle, use the root of a squared minus x squared formula. Now go and try the practice problems on this page. All the best!