Math - Final Exam Review. Find dx x + 6x +. Name: Solution: We complete the square to see if this function has a nice form. Note we have: x + 6x + (x + + dx x + 6x + dx (x + + Note that this looks a lot like the derivative of arctan(x! We do a u-substitution to make this concrete. Let u x +, so du dx. We get: dx x + 6x + dx (x + + du u + arctan(u + C arctan(x + + C. A pool initially has L of chlorine and 97 L of water in it. Fresh water is pumped in to the tank at a rate of 5 L/minute and the mixture of chlorine and water from the pool is pumped out at a rate of L/minute. The chlorine and water in the pool are well-mixed. Find an equation for Q(t, the quantity (in L of chlorine in the pool. Solution: We want to find a differential equation for Q. Note that there is no incoming chlorine, only outgoing chlorine. There are L of liquid leaving every minute. Since the pool is well-mixed, the amount of this L that is chlorine will be given by Q V, where V is the volume of the pool. Since V is initially L and has a net change of + L every minute, we get V + t. Also, our initial condition Q(, as this was stated in the problem. We get an initial value problem: dq dt Q V Q + t Q(
This is a separable differential equation. Solving we get: dq Q dt + t ln Q ln + t + C ln(q ln( + t + C Q(t e C e ln(+t/ e C ( + t / Note that we used the fact that Q, V + t are both nonnegative, so we could remove the absolute value signs. Solving for the initial condition we get: Q( e C ( / e C ( / Therefore, Q(t ( / ( + t /.. Find x x + dx. Solution: This is a rational function, so we would like to use partial fractions. In this case, the numerator has larger degree, so we first do polynomial long division: x + x x x x x Therefore, we get: x x(x x + dx + x x dx + x xdx x + dx Note that x + cannot be broken up in to linear real factors, as its solutions are all imaginary. But then, we are already in the partial fractions form, so we will just integrate. In particular, note that the numerator is the derivative of the denominator, so we use u-substitution with u x +, du xdx. We get: x x + dx x x xdx du u ln u + C x x + dx x ln x + + C
4. Determine whether n n exists. Solution: Note that the limit of the terms is, so the term test doesn t help. However, for x note that the function f(x x is positive, continuous and decreasing. We can use the integral test. We want to know if x dx exists. This will be unpleasant to integrate, so we instead will compare it to something. Note that this integral is a lot like it to exist. We bound it above by g(x as follows: x x + (x x This follows from the fact that x for x. Therefore, we get: Since 5. Find dx x x. x dx x dx dx which exists, so we expect x x dx x dx exists by the p-test, the integral exists, so the summation converges. Solution: We want to use substitution here. We could use rational substitution, but because everything is in the denominator, it is potentially easier to use a trig substitution. We use x sec(θ, θ < π/. Then dx sec(θ tan(θdθ. We get: dx x x sec(θ tan(θdθ sec(θ sec (θ tan(θdθ tan(θ dθ θ + C We know x sec(θ so cos(θ x. Therefore, θ arccos( x. Our final answer is then arccos( x + C. 6. Find t t dt. Solution: We would like to use a substitution here. Both a rational substitution and a trig substitution will work here. I will do the trig substitution, but I advise you to do the rational substitution on your own time. Because we have a t, we do the substitution t sec(θ, θ < π/. Therefore, dt sec(θ tan(θdθ. We get: t t dt sec (θ sec (θ sec(θ tan(θdθ sec 4 (θ tan (θdθ
4 This is not a simple integral, so we use trig identities to make it easier. Since (tan(θ sec (θ, we turn all but of the sec(θ in to tan(θ to get: sec 4 (θ tan (θdθ sec (θ( + tan (θ tan (θdθ (tan (θ + tan 4 (θ sec (θdθ Using the substitution u tan(θ, du sec (θ, we get: (tan (θ + tan 4 (θ sec (θdθ (u + u 4 du u + u5 5 + C tan (θ Since tan(θ sec (θ t, our final answer is: (t / + (t 5 5 + C + tan5 (θ 5 7. Find the general solution to dy dx + x y + x, where we assume x >. Solution: This is a first-order linear differential equation with a(x x, k(x + x. We first need to integrate a(x, which we do by partial fractions. Solving a(x (x (x + A x + B, we clear denominators to get A(x + + x + B(x. Therefore A + B and A B. Solving this, we get A, B. Therefore: A(x a(xdx (x (x + dx ln x ln x + ln x x + Note that since x >, we can remove the absolute value signs. Therefore, we get: m(x e A(x e ln((x /(x+ x x + + C Therefore, we get: y(x m(xk(xdx m(x x + x x + dx x x + x + x dx x ( x + (x / + C x
5 8. Let f(x x 6. Determine for what x T f(x converges. For such x, what does it converge to? Solution: In order to determine where it converges, we first need to find it. As is, the function is hard to take derivatives of, so we use partial fractions to make it easier. x 6 A x 4 + B x + 4 A(x + 4 + B(x 4 A + B, 4A 4B A 8, B 8 Therefore, we are really considering T f(x where f(x ( 8 x 4. Taking derivatives x + 4 of this, we get: f (n (x ( n n! 8 (x 4 n+ ( n n! 8 (x + 4 n+ Plugging in, we get: f( 8 4 8 4 6 f (! 8 4 8 4 f (! 8 ( 4! 8 4! 4 4 f ( (! 8 4 4! 8 4 4 f (4 ( 4! 8 ( 4 5 4! 8 4 5 4! 4 4 5. Therefore, all the odd degree terms are, and when we multiply by xn n!, the n! s cancel, leaving us with: T f(x 4 4 x 4 4 x 4 4 4 5 x 6 4 4 7 x 8 4 4 9... ( + x 4 4 + x4 4 5 + x6 4 7 +... 6 ( x n 4 n This is a geometric sum multiplied by 6. From class, we know that this will then converge for: x < 4 x < 6 x < 4
6 Moreover, we know from class what this converges to for such x. For x < 4, we have: T f(x 4 6 ( x n ( 4 x 6 n x 6 f(x Therefore, T f(x converges for x < 4 and for such x it converges to f(x. 9. Say we are given a differential equation dy F (x, y for some function F (x, y. Say we have an dx initial condition y(x x. (a Explain how to use Euler s method to approximate y(x + h for some step-size h. Include any relevant formulas. Solution: We use the formulas from Euler s method to do so. First, we start with x, y. Then, using the tangent line to approximate y(x, we get: x x + h y y + hf (x, y x x + h x + h y y + hf (x, y y + hf (x, y + hf (x, y (b Draw a picture explaining the formulas in Euler s method and include a short description.. Find dx x ln(x. Solution: This is a straightforward application of u-substitution. Note that the derivative of ln(x is x, which is present in the integral. This suggests using u ln(x, du dx x. We get: dx du x ln(x u. Determine whether the vectors u (a For all vectors (, v ln u + C ln ln(x + C (, w ( x, there are numbers a, b, c such that y Solution: Consider the two vectors u, w. -dimensional vectors as follows: a u + c w a ( satisfy the following properties: ( x a u + b v + c w. y These can easily be shown to be a basis for ( + c ( ( x y
7 We get two equations which we solve: a + c x c y a x y, b y Note that c was uniquely determined, and because c was unique, a was forced to be x y and is therefore unique. Note that if we use this a, c we get: ( ( ( x y y x a u + c w + y y Therefore, if we want to solve a u + b v + c w there are numbers a, b, c as desired. ( x y (b For all vectors, there are unique numbers a, b, c such that ( x, we can take a x y, b, c y. So, y ( x a u + b v + c w. y ( x Solution: As proved above, u, w form a basis. In particular, (x y u + y w. Once y we add in the vector v, we no longer get unique solutions though. For instance, say we wish to find a, b, c such that: a u + b v + c w ( From above, we know we can take a, b, c. However, this vector equals v, so we could also take a, b, c and this will work as well. Therefore, the numbers a, b, c are not unique.. Find e x + e 4x dx. Solution: Note that the denominator can be rewritten as + (e x. It therefore makes sense to do a trig substitution on e x. Specifically, we do e x tan(θ, so that this will simplify to sec(θ. We pick π/ < θ < π/. Solving for x, we get: x ln(tan(θ dx tan(θ sec (θdθ Therefore, our integral becomes: e x dx tan(θ + e 4x + tan (θ tan(θ sec (θdθ sec (θ sec(θ dθ sec(θdθ ln sec(θ + tan(θ + C
8 We know tan(θ e x. Since sec(θ + tan (θ, we get sec(θ + e 4x. Our final answer is then: ln + e 4x + e x + C. Let P be the plane determined by the points,,. Let v 4. Find vectors 5 u, w such that u + w v, u is normal to the plane and w lies parallel to the plane. Solution: We have not learned any way to project vectors on to a plane. What we do instead is project v on to the normal line. That will give us the perpendicular part u, which we can then subtract from v to find w v u. Note that the plane has vectors a and b computation we get: n a b. By direct To make computation a bit simpler, we will scale by to get the normal vector n to the plane (recall that scaling by a non-zero constant does not change being normal to the plane. Then, relative to the vector n, we get: 4 u v v n n n n 5 4 7 7 + 7 w v v u 4 7 5 5 4. Approximate e.9 within 5! of the actual value. (Hint: Use the fact that e < Solution: Since the Remainder Theorem has a factor of use T 4 e x. We then have: (n+!, it seems reasonable that we can T 4 e x + x + x! + x! + x4 4! e.9 + (.9 + (.9 + (.9 6 + (.94 4
9 We wish to get an upper bound on the remainder. We look at the fifth derivative f (5 (t for t [,.9]. We get: f (5 (t e t e.9 < This last inequality occurs because e.9 < e <. By the remainder theorem: e.9 ( + (.9 + (.9 + (.9 6 + (.94 (.95 4 5! < 5! This isn t quite good enough. Note that if we took a fifth-order approximation, by the same argument we will get an upper bound of 6! 5! 6 < 5!. So instead, we do: T 5 e x + x + x! + x! + x4 4! + x5 5! e.9 + (.9 + (.9 + (.9 6 + (.94 4 + (.95 We now look at the sixth derivative. This is just e t, and by the same argument as above, for t [,.9], this is less than. By the remainder theorem: R 5 f(x (.96 6! < 6! 5! 6 < 5! Therefore, a fifth-order approximation gets us within 5! of the correct value. 5. Given x >, find the solution to the initial value problem: Solution: We rearrange this to get: We have a(x ( + x arctan(x dy dx ( + x e x y y(tan( e tan( dy dx + y ( + x arctan(x e x arctan(x ( + x arctan(x, k(x e x arctan(x. To integrate a(x, we use the u-substitution u arctan(x, as then we get: dx a(xdx ( + x arctan(x du u ln u + C ln arctan(x
Since x >, arctan(x, so we can remove the absolute value signs to get A(x ln(arctan(x. Therefore, we get: m(x e A(x e ln(arctan(x arctan(x y(x m(xk(xdx m(x e x dx arctan(x ( e x + C arctan(x Using that y(tan( e tan(, we get: ( e tan( e tan( + C arctan(tan( So, C and y(x e x arctan(x. e tan( + C 6. Consider the initial value problem dy dx x y, y(. (a Draw a direction field for the differential equation and sketch the solution corresponding to the initial value above. (b Use Euler s method with step size h. to approximate y(.. (c Compare the above methods with the actual solution to y(.. Explain why, in general, Euler s method gives relatively close approximations for small step sizes. 7. Evaluate dx x + x. Solution: We first find the antiderivative. Note that partial fractions. We solve: x + x A x + B + x A( + x + Bx A + B, A A, B x+x x(x+, so it makes sense to use We then integrate: dx x + x x ( + x dx ln x ln + x + C
Now, we find dx. Note that for x, the denominator is always positive. Therefore, this x+x is improper only at. We get: b dx x + x lim dx b x + x ( b lim ln x ln + x b ( x lim ln b + x b lim ln b b + b ln( 4 Note that lim b b +b, so lim b b ln +b ln. So, the limit above is just ln( ln( 4. 8. Find the second-order Taylor polynomial of x et dt. Use this to estimate. e t dt and find an upper bound on the error in your approximation. Solution: Let f(x x et dt. Using the FTC, we will take derivatives of this function in order to find the Taylor polynomial. f( e t dt f (x e x f ( e f (x x e x f ( Therefore, T f(x x, so f(... We now wish to find an upper bound on R f(x. We look at the third derivative: f ( (t (t (t e t + 6te t (9t 4 + 6te t f ( (t (9t 4 + 6te t We wish to find the max value of this for t [,.]. However, 9t 4 +6t and e t are both increasing functions, so they take their maximum value at the maximum value of t, which is.. Therefore for t [,.]: f ( (t (9t 4 + 6te t (9(. 4 + 6(.e (. (9 + 6e (9 + 6( 45 Here, I used the fact that. < and e < to make the upper bound simpler. By the remainder theorem, we get: f(.. < 45(.! 9. For each of the following, determine if they are a meaningful expression. If so, determine whether they are a vector or a number. (a ( a b c Solution: Not meaningful.
(b ( a bc Solution: Number. (c (a b c Solution: Vector. (d a b Solution: Not meaningful. (e a b Solution: Number. (f a (( b c d Solution: Vector.. Do the vectors, 5, 9 9 Solution: We want to solve a + b form a basis? 5 + c 9 9 a + b + c x a + 5b + 9c y a b 9c z x y. We get three equations: z We subtract two times the first equation from the second and third to get rid of the a s there. We get three equations: a + b + c x b + c y x 5b 5c z x We add five times the second equation to the third to get rid of the b there. We get: a + b + c x b + c y x z x + (y x Therefore, 8x+y +z. This means that we cannot solve for any x, y, z, as they must satisfy 8x + y + z. For example, we will not be able to solve for, as this doesn t satisfy that equation. Therefore, the vectors do not form a basis. 4. (a Find a normal vector to the plane with points A,, 8. How many 4 normal vectors are there to this plane?
Solution: The plane above has vectors AB not parallel as they are not multiples of eachother. Then, we can find a normal vector n AB AC 4. 45 5 5, AC. Note that these are 4 Any scalar multiple t n for t will also be a normal vector though, as scaling does not change being perpendicular. Therefore, there are infinitely many normal vectors to the plane. (b Find a normal vector to the plane above with length equal to. How many such vectors are there? Solution: Note that if we take n from before and scale it by, it will have length. n That is: n n n n. Does π Therefore, if we then scale it again by, we will get a vector of length. So, take: m n n 4 + 4 + 45 45 This vector m has length and is normal to the plane, as it is a scalar multiple of n. There are only two such vectors, namely m and m. sec(xdx exist? Solution: First, recall that sec(xdx ln sec(x + tan(x + C. Now, in actually integrating sec(x from to π, note that it is not defined at π (as cos is here. So, the integral needs to get split up: π sec(xdx π/ sec(xdx + π π/ sec(xdx lim ln sec(x + tan(x a π/ a + lim ln sec(x + tan(x b π/ + Computing the first limit, note that as a π/, both sec(x and tan(x go to. As b π/ +, both sec(x, tan(x go to. So, the limits become: a π lim ln sec(x + tan(x a π/ + lim ln sec(x + tan(x b π/ + ln ln sec( + tan( + ln sec(π + tan(π lim ln t This gives us a case, so the limit does not exist. Note that I didn t use an equals sign in the last case, because it isn t equal. b π b
4. Evaluate (cos θ + sin θ dθ. Solution: Expanding, we get: (cos θ + sin θ dθ cos (θ + cos (θ sin(θ + cos(θ sin (θ + sin (θdθ We can deal with the middle terms with u-substitution. We use a trig identity to deal with cos, sin as follows: cos (θ + cos (θ sin(θ + cos(θ sin (θ + sin (θdθ ( sin (θ cos(θ + cos (θ sin(θ + cos(θ sin (θ + ( cos (θ sin(θdθ ( ( ( sin (θ + sin (θ cos(θdθ + ( cos (θ + cos (θ sin(θdθ For the first integral, we do the substitution u sin(θ, du cos θdθ. For the second, we do the substitution w cos(θ, dw sin(θdθ. We get: ( ( sin (θ + sin (θ cos(θdθ + cos (θ + cos (θ sin(θdθ ( u + u du + ( w + w ( dw ( + u du ( + w dw u + u w w + C sin(θ + sin (θ cos(θ cos (θ 4. Find a reduction formula for x n cos(xdx. Use this to evaluate π/4 x cos(xdx. Solution: We apply integration by parts with f(x x n, f (x nx n, g (x cos(x, g(x sin(x. We get: x n cos(xdx x n sin(x nx n sin(xdx x n sin(x n x n sin(xdx + C We apply integration by parts again to the right-hand integral with f(x x n, f (x (n x n, g (x sin(x, g(x cos(x. We get: ( x n sin(x n x n sin(xdx x n sin(x n x n cos(x + (n x n cos(xdx x n sin(x + nx n cos(x n(n x n cos(xdx Applying this formula to the case n, we get: x cos(xdx x sin(x + x cos(x cos(xdx x sin(x + x cos(x sin(x + C
5 Therefore, we get: π/4 x cos(xdx x π/4 sin(x + x cos(x sin(x ( π 6 sin(π 4 + π cos(π 4 sin(π π 6 + π ( + sin(