Mathematical_examples.doc

Mathematical examplesExample 1. Find the natural domain for .Solution. To avoid the square root of a negative number, we must choose x so that . Thus, x must satisfy . The natural domail is therefore , which can write using interval notation as -3, 3.Example 2. Prove that .Preliminary analysis: Let be any given number. We must produce a 0 such that .Consider the inequality on the right.Now we see how to choose ; that is, =/3. Of course, any smallerwould work.Formal proof: Let0 be given. Choose=/3.Then implies thatIf you read this chain of equalities and an inequality from left to right and use transitive properties of = and , you see that.Example 3. Prove that Preliminary analysis: Our task is to find such that NowThe factor |x-3| can be made as small as we know that |x+4| will be about 7. We therefore seek an upper bound for |x+4|. To do this, we first agree to make 1. Then |x-3|0 be given. Choose=min1,/8.Then 0|x-3|0), find F(x) by the definition of the derivative.Solution: By this time you will have noticed that finding a derivative always involved taking the limit of a quotient where both numerator and denominator are approaching to zero. Our task is to simplify this quotient so that we can cancel a factor h from numerator and denominator, therefore allowing us to evaluate the limit by substitution. In the present example, this cab be accomplished by rationalizing the numerator.Thus, the derivative of F(x), is given by . Its domain is (0,).Example 5. Find (if any exist) the maximum and minimum values of on (0,1).Solution: .The only critical point is p=1/2.For every value of p in the interval (0,1) the denominator is positive; thus, the numerator determines the sign. If the p is in the interval (0,1/2), the numerator is negative; hence, G(p)0. Thus, by the First Derivative Test, G(1/2)=4 is a local minimum. Sice there are no end points or singular points to check, G(1/2)=4 is a global minmum. Example 6. Prove Comparison Property: If f and g are integral on a,b and f(x)g(x) for all x in a,b, thenIn informal but descriptive language, we day that the definite integral preserves inequalities.Proof: Let partition P: a=x0 x1 xn=b be any partition of a,b, and for each i, let be any sample point on the ith subinterval xi-1, xi. We may conclude successively thatExample7. Determine the convergence set forSolution: The nth term is , n2. Thus, We know that the series converges when 1, that is , when |x+2|3 or, equivalently, -5x1, but we must check the end points -5 and 1. At x=-5, and un converges by the Alternative Series Test.At x=1, un =ln(n)/n and un diverges by comparison with the harmonic series.We conclude that the given series converges on the interval -5, 1).Example 8. Parametric equations for appoint P moving in the plane are x=3cost and y=2sint, where t represents time.(a) Graph the path of p.(b) Find expressions for the velocity v(t), speed |v(t)|, and acceleration a(t).(c) Find the maximum and minimum values of the speed and where they occure.(d) Show that the acceleration vector based at P always points to the origin.Solution: (a) Since x2/9+ y2/4=1, the path is the ellipse, the graph is omitted.(b) The position vector is r(t)=3cost i+2sint jand so v(t)=-3sint i+2cost j, A(t)=-3cost i-2sint j.(c) Since the speed is given by , the maximum speed of 3 occurs when , that is t=/2 or 3/2. This corresponds to the points (0, 2) and (0, -2) on the ellipse. Similsrly, the minimum speed of 2 occurs when sint=0, which corresponds to the points (3, 0) and (-3, 0).(d) Note that a(t)=-r(t). Thus, if we base a(t) at P, this vector will point to and exactly reach the origin. We conclude that |a(t) is largest at (3, 0) and (-3, 0), and smallest at (0, 2) and (0, -2).Example 9. Find the general solution of the equationSolution: This equation can be written in the form and thus . Taking logarithms of both sides of this equation gives . Example 10. Find the general solution of the differential equationSolution: Here M(t,y)=3y+et, N(t,y)=3t+cosy. This equation is exact since and . Hence, there exists a function (t,y) such that (i) (ii). We will find(t,y) by each of the three methods below.First method: From (i), (t,y)=et+3ty+h(y). Differentiating this equation with respect to y and using (ii) we obtain thath(y)+3t=3t+cosy.Thus, h(y)=siny and(t,y)=et+3ty+siny. (Strictly speaking, h(y)=siny+ constant. However, we already incorporate this constant of integration into the solution when we write(t,y)=c.) The general solution of the differential equation must be left in the form et+3ty+siny=c since we can not find y explicitly as a function of t from this equation.Second method: From (ii), (t,y) =3ty+siny+k(t). Differentiating this equation with respect to t and using (i) we obtain thath(t)+3y=3y+et.Thus, k(t)= et and(t,y)=et+3ty+siny. Third method: From (i) and (ii)(t,y)=et+3ty+h(y) and (t,y) =3ty+siny+k(t).Compating these two expressions for the same function(t,y) it is obvious that h(y)=siny and k(t)= et. Hence(t,y)=et+3ty+siny. Example 11. Find the solution of the initial-value problem (*)Solution: From the first equation of (*), y=3x-x. (*)Differentiating this equation givesy=3x-x=x+y.Then, substituting for y from (*) gives3x-x=x+3x-xso that x-4x+4x=0.This implies that x(t)=(c1+ c2t)e2t (*)where c1, c2 are constants. Substituting (*) into equation (*) gives y(t) =( c1-c2+ c2t)e2t (*)The constants c1 and c2 are determined from the initial conditionsx(0)=3= c1 , y(0)=0= c1- c2.Hence, c1=3, c2=3 and x(t)=3(1+t) e2t, y(t)=3t e2tis the solution of (*).Example12. (Tossing two dice) Let an experiment consist of tossing two dice. Event A denotes double appear, event B denotes the sum is