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A practical tutorial on precalculus for high school students who have completed Algebra 1.;useful for self-learning and also as review for starting calculus course...many advanc d concepts are introduced.in a simple manner; The tutorial has several simple examples and exercise problems to practise...Several applications are given in each topic, including exercise using graphing uitlity for polar equations.

Adventures in Precalculus - 1Dr N K Srinivasan

What is Precalculus?If you scan through a 'Precalculus' book, you will note that it is a course with a jumble of topics--more like a bridge course from Algebra 2 to regular Calculus. But a careful study will show you something: It teaches you several mathematical functions and how to manipulate them and use them for 'modeling' real-world problems.

Well, you have to learn many mathematical functions because these functions are the 'ingredients' for using Calculus.While learning these functions, you will use a bag of clever tricks,call them 'tools of the trade" if you like. This article teaches you some of the clever tricks you will use--so that 'precalculus' becomes easy and interesting to study and develops your 'calculus muscles' to handle tougher math problems later.! Note: Get a graphing utility and use as required.

1 Completing the square:

Take a simple example: Solve

x +6x -16 =02

You can, of course, use the quadratic formula ,if you remember it ! Is there a simple way to solve? Recall : 2ax + a2

(x+a)

2

= (x+a)(x+a) = x

2

+

Keep this equation in mind. We will proceed from right to left while "completing the square". Take the example and rewrite this equation as follows:

x2

+ 6x

= 16

Take the coefficient of x, which is 6 in this example; divide this by 2 and

square the number: that is 6/2=3 and 3 x3 =9 Add this number both sides:+9 You can factor the left side using the relation given earlier:

x

2

+ 6x +9 = 16

(x + 3)2

= 25

Take the square root on both sides: x+3 = +/- 5 So the roots are: x+3 =5 or x=2 or

and x+3 = -5 x=-8

This method is far easier than using quadratic formula.

Try the following problems before you proceed: 1 solve x2

+ 4x -32 =0

[Ans ; x=

4, x=-8 2 Solve x - 8x -9=02

[Ans:

x=9;x=-1] 3 Solve 3x +6x -24 =02

[Ans

x=2

;x=-4]

Example 2 This method--"completing the square has other uses too. We will see one application: Solve x2

+ y +2x +6y -6 =02

This problem appears 'tricky'...well -this is a 'standard' equation for a circle. Recall that the equation for a circle with centre C(0,0) at the origin is : x + y = r2 2 2

where r is the radius.

The equation changes when the center is shifted to C'(h,k) position.

The equation becomes:

(x-h)

2 + (y-k)2 = r2

Now the given equation can be factored by "completing

the square" for both x and y . Rewrite the expression:

(x2

+ 2x + ) + ( y2+6y + ) = -6

Add 1 to the first bracket and 9 to the second bracket and the same numbers on the right side

too.

(x

2 + 2x +1) + (y2 + 6y + 9) = -6 +1+9

Factoring for x and y we get:

(x+1)2 + (y+3)2 = 4

The problem is solved : this is a circle with centre at C(-

1,-3) and radius 2.

Problem to solve:

1 find the center and radius of the circle: x2 + y2 -4x

+10y-7=0

[Ans center C(2.-5) radius 6]

2 Find the center and radius of this circle: x2

+x + y2 -6y +1/4 =0

[Ans center (-1/2,3), radius 3]

Example 3 Find the vertex of the parabola whose equation is:

x2

-6x-y +10 =0

Let us rewrite this first: )+10 = y

(x -6x2

The coefficient of x is 6; so let us put 9 inside the bracket and subtract 9 outside: ( x +10 =y (x -3) 1 This is a parabola, with vertex at [3,1] Try this:2 2

-6x +9) -9

= y -

1 Find the vertex of the parabola : 2x2

-8x -y + 7 =0

We can extend this method to find the equation of an ellipse from

the equation in

standard form:Read this from your text book .

2 Function of a functionSuppose you have a function: y= f(x) = 2x +3

You have another function : z = g(x) = x2

+1

you can generate another function f [ g(x)] Write this as: f[ g(x)] = f [x +1]2

Then write the function f(x) replacing x by x2

+ 1.2

We get f[g(x)] = 2(x +1) + 3

You also get f[f(x)] by writing ;

f [ 2x+3] = 2 (2x+3) + 3 = 4x + 9Try writing g[g (x)] Example: g[f(x)] Are they same? no. Given f(x) = 1/x + 3. g(x)= x -1. write f[g(x)] and

Complex formulae can be created by using this method:Example 1: The electrical resistance of a wire increases with temperature as follows: R = a + bT + c T2Now the temperature of an electrical heater varies with time t as follows:

T = d + 4 t Find the rate of heating by using the relation : Heat produced H = I2

R as a function of time:

R =

f(T) = a + bT + c T.T

T (t) = d + 4t So f [ T (t)] = f [ d+4t] = a + b

[d+4t] + c [ d+4t][d+4t] Now you see the usefulness of function of function. Here is another example: example 2: The atmospheric air pressure decreases almost linearly [like a straight line] as we ascend to about 10000 feet.The pressure P = P0 2 (h) where h is height in feet.

A baloon ascends at the rate of 50 feet per minute. where t is time in minutes from the ground level.

h = 50 t

Write the equation for change in pressure with respect to time for a traveller going up in the balloon.

3 Inverse functionsTake the function y = f(x) = 2x + 3 You can find the inverse function by switching x and y: Replace x by y abd y by x: Now solve for Y : x= 2y + 3 2y = x - 3 y= (x-3)/2 Note that this function is an inverse function of f(x) . Let us write this as f-1 (x)

Note that the inverse function is not the reciprocal of f(x); it is not 1/f(x). The inverse function is the reflection to the original function across the line y=x. Y=x is the diagonal line that passes through the origin.

Draw the lines y= 2x +3 and the line y=(x-3)/2 in a graph paper. You will see the two lines as mirror images of each other. Example 2 Find the inverse function for the parabola: y = 4 x.x Switching x and y ; x= 4 y.y

or y= sqrt(x)/2 Plot again ,using your graphing utility, the two curves: y= 4xx and y= sqrt(x)/2 .Do they form reflection across the diagonal line y=x? Check this.

What if we find f[ f-1

(x)] ?-1

Take the example of f(x)= 2x+3 , and f (x) = (x-3)/2 f[ (x-3)/2] = 2[(x-3)/2] + 3 = x-3 +3 =x

Are you surprised with the result? We took x, transformed it into f(x) ,then took its inverse and again revert it back by f(x) ,getting x again. So, you can always check your result of inverse function by forming f [ f-1

(x)]----if you get x, you are right.Exponential function and logarithmic function are inverses of each other. y =f(x)= e Switch y and x:x

x = ey

Solve for y:

ln x = y= f-1

(x)

So the two functions are inverses. Plot the two functions using your graphing utility; you will find that they are curves reflected over y=x. Try these problems: 1 If f(x) = sqrt(2x+3), find its inverse function. 2 If f(x) = 2ex find its inverse function.

ApplicationWe can use inverse functions to find the conversion formula for temperature in Centigrade scale to Fahrenheit and the reverse one: If x is the temp in centigrade and y the temp in Fahrenheit, we have the relation as follows: y = 32 + (9/5) x Let us find the inverse function; switch x and y in the above equation: x = 32 + (9/5) y Solve for y: y = (x-32) (5/9)

This formula gives the conversion fromula when x is in Fraherheit and y the centigrade.! Example: If 212 F is the boiling point of water, find the temperature in centigrade scale: deg C = (212-32) (5/9) = (180 x 5 )/9 = 100 deg C So, the boiling point of water is 212 deg F or 100 deg C.

3 Shifting and scalingMany functions can be shifted around and scaled up or down [multiplied or divided] by suitable manipulation. Consider a parabola with vertex at the origin V(0,0): y = x2Suppose you want to shift the vertex to V(2,0) , change x to x-2. y = (x-2)2 You want to shift the vertex to V(2,1), change y to y-1. y-1 = (x-2)2Consider a circle with center at the origin C(0,0) and radius 2 units>

The equation for the cirlce is :

x + y = 42 2

Suppose you want to shift the center of the circle to P(-2,1) with the same radius,

(x+2)2

+ (y-1)

2

=

4x

Another transformation you do is to enlarge or shrink by multiplying or dividng or y.

y = k x.x is a parabola that rises very fast along the y axis--if k is greater than 1.

Take y = 2 x2Compare this parabola with y = x2 We call this process "scaling" Again consider the graph of y = You can choose k=1, k=2 k=1/2 shapes you get. 1 Write the equation for a ellipse with centre at (-2,2) k . Draw the graphs and study the

2 Write the equation for a cirlce with center at (2,-2) and radius 3 units.

4 parametric equationsWhen we relate x to y through the function , y = f(x) , we have a single graph of y versus x. Often it is convenient to use another variable which is intermediate and common to both: Let us call this intermediate variable t; then x = f(t) and y= g(t) We use two functions now, both functions of t. Consider a sphere whose surface area is related to radius: A = f (r) = 4 .pi.r.r the volume is related to radius: V = g(r)= 4.pi.r.r.r/3

Here the radius is the common parameter. We can take the ratio of surface area to volume: A/V = f(r)/g(r)= 3/r So writing parametric equations help us to see new relations.Thus if you want to study cooling of a sphere [like our Earth] we want to study the ratio of surface area to

volume.As r increases, the ratio of surface area to volume decreases, so the cooling rate could get lower. If your housetop is like a hemispherical dome, it would cool slo