MATH 18.01, FALL PROBLEM SET #5 SOLUTIONS (PART II)

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1 MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II (Oct ; Antiderivatives; points Recall that in pset 3A, you showed that (d/dx tanh x x Here, tanh (x denotes the inverse to the hyperbolic tangent function a Find the two constants A and B such that x A + x + B x This is called the method of partial fractions We will discuss this method in much more detail later in the course Note, A + x + B A( x + B( + x x ( + x( x (A + B + (B Ax x So we can choose A B and get A + x + B x x b Use part a to compute x dx Simplify your answer as much as possible Note, dx ln + x + c + x dx ln x + c x Therefore, x dx + x + x dx ln + x ln x + c c Explain the connection between the function tanh (x and your answer to part b We know that d dx tanh (x x

2 MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II Therefore tanh (x and our answer in part b are both the antiderivative to the function x Antiderivatives differ by a constant term so we must have, tanh (x ln( + x ln( x + c for some constant c Here we were allowed to remove the absolute value since < x < To find the constant we can substitute x We have tanh(, so tanh ( Additionally ln( Therefore we must have c, and we conclude that tanh (x ln( + x ln( x (Oct ; Antiderivatives; 6 points Suppose that E(x e x dx That is, suppose that E(x is a function such that E (x e x Remark: E(x cannot be expressed in terms of simpler functions that you already know In this way, you should think of E(x as a new function Express the antiderivative xe x x dx in terms of the function E and polynomials Note, e x x e (x + We make the substitution u x + We have, so, for a constant c du xdx xe x x dx (xe (x + dx e u du E(u + c E(x + + c, 3 (Oct 3; Differential equations and separation of variables; points Section 5:, Section 85: 8, pg 9 problem 5ab; find the blow-up time t in part b Solution to 5 - : We consider P as a function of time t measured in weeks since the curse was uttered Then the problem tells us that P ( 676 and dp dt P We are interested in know for what t > we have P (t (ie when are all the people dead Separating variables gives us Integrating we get P / dp dt P / t + c for some constant c If we put t we get c (676 / 5

3 MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II 3 We have P precisely when P /, so this is precisely when t c In other words, P (t when t 5 So it takes 5 weeks (roughly a year before all the people are dead Solution to 85-8: Let M(t be the fool s money in dollars at time t where t is the time since he started gambling measured in hours We are told, dm dt 3 M(t Separating variables we get M dm 3 dt Integrating we get Setting t we get Therefore, ln(m t 3 + c ln(m( c M e t 3 +c e t 3 M( We want to know for what t we have M(t M( This is equivalent to finding the t such that e t 3 The solution to this is t 3 ln(/ 3 ln( Therefore the fool will have lost half his money after 3 ln( hours Solution to pg 9 problem 5ab : Separating variables in the doomsday equation we get N (+ɛ dn k dt Integrating we get N ɛ kt + c ɛ Setting t we get c N( ɛ ɛ ɛn( ɛ So we get N(t ( ɛ(kt + c ɛ ( ɛ( ɛn( ɛ /ɛ kt We notice that this is undefined at t kɛn(, and as the denominator approaches ɛ from the right as t t we have, lim N(t t t (Oct 3; Differential equations; points In this problem, you will learn about terminal velocity Consider an object of mass m dropped (straight down off of a very tall cliff Its downward velocity is initially Let v(t denote its downward velocity at time t (ie v > means the object is falling towards the earth In a very simple model of air resistance, the air provides a drag force that

4 MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II is proportional to v That is, the air exerts an upward force of magnitude v on the object, where > is a constant (which has dimensions of mass Let g > length denote the downward acceleration of gravity For the purposes of this problem, we can assume that g is a constant (approximately equal to 98 meter/sec Under these assumptions, Newton s laws of motion imply that ( d dt v m v + g a Use separation of variables to solve the differential equation with the initial condition v( Hint: A modified version of the method of partial fractions from problem might be useful ( Separating the variables we get m v dv dt To be able to integrate the left hand side we follow the hint and try to find A and B such that Since, A v we can just choose because then Thus we have, m v + (A B m v B v + A v + A ( v + (A B B v + ( + B v v v + (A + Bv B A, and A m g + (A + Bv ( m g v m g + v m + v +

5 MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II 5 Thus integrating our separated ODE ( we get t + c m g v + v + dv ( m ln g v + ln v ( ( ( m ln g + v ln v ( ( ( m ln + v ln v g ( m g tanh v where the last equality comes from tanh (x (ln( + x ln( x From this we get g tanh (v (t + c m (3 v ( g tanh (t + c m This is the general solution to the ODE We need to impose the condition v( Substituting t in the solution we get ( g tanh c m which implies c Therefore our solution to the differential equation ( with initial condition v( is ( g v tanh t m b Does lim t v(t exist? If so, explain what happens when you plug the limiting velocity into the right-hand side of the differential equation ( and why the result makes sense so We have lim v(t lim t t lim tanh(t t ( g tanh t m When we plug this limiting value into our ODE ( we get d dt v ( + g m d v g + g dt

6 6 MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II In other words in the limit the derivative of velocity goes to (in other words acceleration is in the limit This is because the closer we get to terminal velocity the acceleration approaches The velocity never actually reaches terminal velocity, but it will get arbitrarily close to it c Explain how your answer to part b changes if the object is thrown off the cliff (straight down with initial velocity v > Using the general solution (3 we still get the same limit in (b ( g tanh (t + c m lim v(t lim t t Therefore the answer doesn t change Our solution assumes that v is less than the terminal velocity If not, then our solution to (a needs to be modified since we used ( ln v ln v The answer to (b would still come out the same though 5 (Oct 8; Definite integrals; points First do Section 63: 9 Then use this problem to help you calculate / x dx by using right Riemann sums By a right Riemann sum, we mean that the right endpoint of the approximating rectangles lies on the graph of y x Solution to 93-9(a: Set m, θ x/, n k in the product formula to get sin x kx [ ( ( ] k + k sin x + sin x which can be rewritten as sin x ( kx sin (k + x + sin ((k x Solution to 93-9(b: We have, ( n sin x kx So, k k k sin [ (( sin k + (( x sin k ] x (( sin k + (( x sin k x k (( n + x sin ( x kx sin (( ( n + x sin x sin ( x k

7 MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II 7 Solution to 93-9(c: The product formula with θ x/ gives us ( nx ( (n + x sin sin (( ( n + x sin x Thus, kx k k sin ( nx ( (n+x sin ( x Solution to main problem: If we split the interval [, /] into n equal intervals, then the associated right Riemann sum of x is ( k n n Setting x n in our solution to 93-9(c we get ( k sin ( ( (n+ n sin ( We can now calculate / k (x dx lim n k lim n ( sin sin ( k n sin ( ( ( sin n n ( (n+ sin ( n ( (n + lim ( (n + lim n ( lim n lim n n sin ( sin ( 6 (Oct 5; First fundamental theorem; properties of integrals; points Section 67: We can draw a rectangle that goes from (, to (a, a n We can calculate the area of this rectangle in two ways First we can use the basic formula which tells us that the rectangle must have area a n+ Secondly we can split the rectangle into two parts using the graph of y x n and then calculate each part separately The part below the curve is a xn dx The area above the curve is equivalent to the area below the curve of y x /n from x to x a n since we can go from one to the other by reflecting in the diagonal line y x Therefore the area of the second piece of our rectangle is a n y /n dy Therefore we must have a a n x n dx + y /n dy a n+

8 8 MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II Let us now check that this equation actually holds From the first fundamental theorem of calculus we know b x k dx bk+ a k+ k + So, a a n a x n dx an+ n+ n + y /n (an /n+ /n+ /n + an+ n + nan+ n + Thus, a a n ( x n dx + y /n dy n + + n a n+ a n+ n + which proves our equation holds as we knew from geometric arguments

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