Adventures in Precalculus

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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 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.


<p>Adventures in Precalculus - 1Dr N K Srinivasan</p> <p>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.</p> <p>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.</p> <p>1 Completing the square:</p> <p>Take a simple example: Solve</p> <p>x +6x -16 =02</p> <p>You can, of course, use the quadratic formula ,if you remember it ! Is there a simple way to solve? Recall : 2ax + a2</p> <p>(x+a)</p> <p>2</p> <p>= (x+a)(x+a) = x</p> <p>2</p> <p>+</p> <p>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:</p> <p>x2</p> <p>+ 6x</p> <p>= 16</p> <p>Take the coefficient of x, which is 6 in this example; divide this by 2 and</p> <p>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:</p> <p>x</p> <p>2</p> <p>+ 6x +9 = 16</p> <p>(x + 3)2</p> <p>= 25</p> <p>Take the square root on both sides: x+3 = +/- 5 So the roots are: x+3 =5 or x=2 or</p> <p>and x+3 = -5 x=-8</p> <p>This method is far easier than using quadratic formula.</p> <p>Try the following problems before you proceed: 1 solve x2</p> <p>+ 4x -32 =0</p> <p>[Ans ; x=</p> <p>4, x=-8 2 Solve x - 8x -9=02</p> <p>[Ans:</p> <p>x=9;x=-1] 3 Solve 3x +6x -24 =02</p> <p>[Ans</p> <p>x=2</p> <p>;x=-4]</p> <p>Example 2 This method--"completing the square has other uses too. We will see one application: Solve x2</p> <p>+ y +2x +6y -6 =02</p> <p>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</p> <p>where r is the radius.</p> <p>The equation changes when the center is shifted to C'(h,k) position.</p> <p>The equation becomes:</p> <p>(x-h)</p> <p>2 + (y-k)2 = r2</p> <p>Now the given equation can be factored by "completing</p> <p>the square" for both x and y . Rewrite the expression:</p> <p>(x2</p> <p>+ 2x + ) + ( y2+6y + ) = -6</p> <p>Add 1 to the first bracket and 9 to the second bracket and the same numbers on the right side</p> <p>too.</p> <p>(x</p> <p>2 + 2x +1) + (y2 + 6y + 9) = -6 +1+9</p> <p>Factoring for x and y we get:</p> <p>(x+1)2 + (y+3)2 = 4</p> <p>The problem is solved : this is a circle with centre at C(-</p> <p>1,-3) and radius 2.</p> <p>Problem to solve:</p> <p>1 find the center and radius of the circle: x2 + y2 -4x</p> <p>+10y-7=0</p> <p>[Ans center C(2.-5) radius 6]</p> <p>2 Find the center and radius of this circle: x2</p> <p>+x + y2 -6y +1/4 =0</p> <p>[Ans center (-1/2,3), radius 3]</p> <p>Example 3 Find the vertex of the parabola whose equation is:</p> <p>x2</p> <p>-6x-y +10 =0</p> <p>Let us rewrite this first: )+10 = y</p> <p>(x -6x2</p> <p>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</p> <p>-6x +9) -9</p> <p>= y -</p> <p>1 Find the vertex of the parabola : 2x2</p> <p>-8x -y + 7 =0</p> <p>We can extend this method to find the equation of an ellipse from</p> <p>the equation in</p> <p>standard form:Read this from your text book .</p> <p>2 Function of a functionSuppose you have a function: y= f(x) = 2x +3</p> <p>You have another function : z = g(x) = x2</p> <p>+1</p> <p>you can generate another function f [ g(x)] Write this as: f[ g(x)] = f [x +1]2</p> <p>Then write the function f(x) replacing x by x2</p> <p>+ 1.2</p> <p>We get f[g(x)] = 2(x +1) + 3</p> <p>You also get f[f(x)] by writing ;</p> <p>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</p> <p>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:</p> <p>T = d + 4 t Find the rate of heating by using the relation : Heat produced H = I2</p> <p>R as a function of time:</p> <p>R =</p> <p>f(T) = a + bT + c T.T</p> <p>T (t) = d + 4t So f [ T (t)] = f [ d+4t] = a + b</p> <p>[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.</p> <p>A baloon ascends at the rate of 50 feet per minute. where t is time in minutes from the ground level.</p> <p>h = 50 t</p> <p>Write the equation for change in pressure with respect to time for a traveller going up in the balloon.</p> <p>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)</p> <p>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.</p> <p>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</p> <p>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.</p> <p>What if we find f[ f-1</p> <p>(x)] ?-1</p> <p>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</p> <p>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</p> <p>(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</p> <p>x = ey</p> <p>Solve for y:</p> <p>ln x = y= f-1</p> <p>(x)</p> <p>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.</p> <p>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)</p> <p>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.</p> <p>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&gt;</p> <p>The equation for the cirlce is :</p> <p>x + y = 42 2</p> <p>Suppose you want to shift the center of the circle to P(-2,1) with the same radius,</p> <p>(x+2)2</p> <p>+ (y-1)</p> <p>2</p> <p>=</p> <p>4x</p> <p>Another transformation you do is to enlarge or shrink by multiplying or dividng or y.</p> <p>y = k x.x is a parabola that rises very fast along the y axis--if k is greater than 1.</p> <p>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</p> <p>2 Write the equation for a cirlce with center at (2,-2) and radius 3 units.</p> <p>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</p> <p>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</p> <p>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 slower if the radius of the dome increases. Did the cathedral builders think of this factor...may be!</p> <p>5 Solving exponential equationsLet us start with a simple equation to solve for x using exponents:</p> <p>Solve</p> <p>:</p> <p>2x</p> <p>= 8</p> <p>Let us write the right side number '8' in terms of exponent of base 2. 8=2 .2.2 = 23Equate the powers or indices both sides, since they have the same base: x =3 Try the following: is the answer.</p> <p>1 Solve</p> <p>3x</p> <p>= 81</p> <p>[Ans:</p> <p>x=4]</p> <p>2 Solve 4x = 256 3 Solve 5x = 125 3] 4 solve 3(2x+1) =</p> <p>[Ans : x=4] [Ans : x=</p> <p>243</p> <p>[Ans:</p> <p>x=2]</p> <p>---------------------------------------------Method of substitution Solve : e2x</p> <p>+ 5 e</p> <p>x</p> <p>+ 6 = 0</p> <p>This equation suggests that it can be converted into a simple algebraic equation without exponential function</p> <p>by 'substituting" exponential function by a variable u. Let u = ex</p> <p>Then the equation becomes: u2</p> <p>- 5u + 6 = 0 (u -3)(u-2) =0</p> <p>Solving we get So, u = 3 Therefore:</p> <p>and u = 2 u = ex</p> <p>= 3</p> <p>e</p> <p>x</p> <p>= 2</p> <p>The answer is Try the following: Solve:</p> <p>x = ln3 or ln2</p> <p>e -2 e -15 =02x x</p> <p>[ans: x= ln 5 or ln -3</p> <p>Discard the second one!There is no log function for a negative number] Taking log on both sides: Many equations can be solved by converting to log function on BOTH SIDES: 1 Solve e = 92x</p> <p>Take natural log on both sides: ) = ln 9</p> <p>ln (e</p> <p>2x</p> <p>2x =ln 9</p> <p>x = (1/2) ln9 2 Solve e3k</p> <p>=2</p> <p>[Ans</p> <p>k = ln2/3]Exponential GrowthWE use exponential function to "model" growth of populations. "Population" is a general word; it may mean the human population or population of rats, population of birds, population of houses or population of cell phones. To write an equation for growth of population, we start with an initial time t=0 when the population is "n".Then the population at a later time t is given by N(t):</p> <p>N(t) = n e ktwhere k is a constant, called 'growth constant" and t is the time measured from the start time. Let us take an example. Joe is a realtor who wants to model the growth of houses in the new neighborhood called Chico town.</p> <p>Chico had a population of 1200 homes in the year 2000. So n=1200. Joe finds that by the year 2005, that is ,five years later, the number of houses has grown to 3000. So N(t) is 3000 where t is five years. Now we can write the growth equation: 3000 = 1200 ek.5</p> <p>Joe has to calculate the growth constant first. What is the value of 'k'? The trick is to take log on both sides which you have learned in the previous section....Taking the natural logarithm 'ln': ln 3000 = ln 1200 + 5k 5k = ln 3000 - ln 1200 = ln (3000/1200) = ln 2.5 Using the calculator: ln 2.5 =0.916 k = 0.184 So the growth equation becomes: N(t) = 1200 e0.18t</p> <p>Now Joe can "predict" the future population by plugging t , provided the growth follows the same trend;that is there is no major catastrophe [like earth quake] or economic depression which will change the trend.Let us predict the population for the year 2015. t=15 N(t) = 1200 e0.18 .15</p> <p>= 1200 e</p> <p>2.7</p> <p>= 1200 x 14.9 = 17855 or roughly 18000 So, Joe can expect about 18000 homes in Chico town by the year 2015. You see that exponential growth models can be used to predict future growth, once you find the growth constant.[ All models are constructed to help you to predict the future.]</p> <p>Doubling Time It is easy to understand the rate of growth if we calculate the time it takes for the population to double itself.This time , we denote by t', would depend on the growth constant.Higher the growth constant k ,shorter the doubling time. To find this time let us reverse the procedure: take N(t) = 2n Find t for a given k : 2n = n ekt'</p> <p>Dividing by n, we get</p> <p>2 = ekt'</p> <p>Taking log on both sides, we get ln 2 = kt' or doubling time t' = ln 2/k Larger the k value or growth rate, shorter the doubling time! Now ln 2 =0.69 or take it as 0.7 Doubling time t' = 0.7/k Let us find the doubling time for Joe's model: k = 0.18 Doubling time t' = 0.7/0.18 = 3.88 or nearly 4 years.</p> <p>The population of homes in Chico town was 3000 in the year 2005. Four years later, that is in 2009, it could be 6000 homes! in the year 2013 ,it could be 12000; in the year 2017, it could be 24000. Try this problem: 1 The number of toucon birds in Amazon forest in an area of 1000 acres was 150 in the year 2004. When the bird counting was done in 2007, it was 210. 1 What is the growth constant? 2 Find the doubling time.3 Predict the toucon population in years 2010 and 2013.</p> <p>[Hint: 210 = 150 e3k</p> <p>k=0.114</p> <p>doubling time t'=</p> <p>6.2 years ]2 India's human population was 300 million in 1945. It rose to 1000 million or 1 billion in the year 1990. Find the growth rate in percent and also predict the likely population for the year 2015. Also find the doubling time. {Ans k = 0.019 or 1.9% doubling time t'= 37 years.</p> <p>Note: The population growth rate has decreased in India; so the growth constant now: k = 0.013 or 1.3% . 3 If the bacteria population in a soup triples every three hours, how long it will take for initial population of 20 bacterias to grow to 1000. [Hint tripling time t" = ln 3/k Find k since t" = 3 hours.ln3=1.1 k=0.37 t=10.5 hours] Exponential Decay There are processes in which population may decrease---say fish population in lake or deer population in a forest [because food is limited or there are predators in the environment---things conservationists and ecologists worry about.] We can model this too by writing an exponential function with (-k) t: N(t) = n e-kt</p> <p>'k' is called a "decay constant". Find k as before; "Half life" is the time it takes for the population to reduce to half...</p>