Solutions to Math 41 Final Exam December 10, 2012

Transcription

1 Solutions to Math 4 Final Exam December,. ( points) Find each of the following limits, with justification. If there is an infinite limit, then explain whether it is or. x ln(t + ) dt (a) lim x x (5 points) We are dealing with a continuous and = and ln(t + )dt =. We can apply L Hospital: lim x x ln(t + )dt ln(x + ) x x x x = indeterminate, since the numerator and denominator are x+ Note that in the first equality we used FTC: d dx (now this is again indeterminate ) (using l Hospital again) x ln(t + )dt = ln(x + ). (b) lim ( + sin x)/x x + (5 points) Note that this is an indeterminate limit of form. So, let L x +( + sin x) x. We will compute ln L: Therefore L = e ln L = e. ln L = ln lim x +( + sin x) x x + ln(( + sin x) x ) ln( + sin x) x + x x + x + = = cos x +sin x cos x + sin x (now this is indeterminate ) (by L Hospital)

2 Math 4, Autumn Solutions to Final Exam December, Page of 6. ( points) In each part below, use the method of your choice, but show the steps in your computations. (a) The curve with equation x y = y passes through (x, y) = (, ). Find the equation of the line tangent to the curve at this point. (5 points) If we view y implicitly as a function of x with the goal of finding dy dx, then we first need to take the natural log of both sides in order to differentiate x y : ln(x y ) = ln(y ) y ln x = ln(y ) We now differentiate both sides with respect to x and obtain: y ln x + y x = y y y (ln x y ) = y x y y = x(ln x y ) = (y )y x(y ) ln x At the point (x, y) = (, ) we obtain y =. This is the slope of the tangent line to that point. The equation is then: y = (x ) (b) Find h (x) if h(x) = x e t t + 3 dt e (5 points) Let F (x) be an antiderivative of x (which exists by the Fundamental Theorem of x +3 Calculus, since is continuous). We know, by the fundamental theorem of calculus that: e x x +3 h(x) = x This implies the following formula for h (x): e t t + 3 dt = F (x ) F () h (x) = d dx F (x ) d dx F () = F (x ) d dx x = xf (x ) = xex x 4 + 3

3 Math 4, Autumn Solutions to Final Exam December, Page 3 of 6 3. ( points) For this problem, we try to solve the equation x 4 4x + =. (a) Show that this equation has at least one solution between x = and x =. (State clearly what results you use and why they apply.) (5 points) Notice that f() = + = and f() = 4 + =. We will use the Intermediate Value Theorem. The function f(x) = x 4 4x + is a polynomial, therefore it is continuous. Because < <, by the IVT we obtain that there is a c (, ) such that f(c) =. This implies that the polynomial has at least one solution between x = and x =. (b) It is a fact that this equation has a unique solution x = c between x = and x = (but you do not need to prove this). Use Newton s method, with initial guess x =, to compute the successive approximations x and x 3 for c. (5 points) The formula for Newton s method is the following: x n+ = x n f(x n) f (x n ) In our case f (x) = 4x 3 4. We compute x and x 3 now taking x = : x = 4 = x 3 = 6 7 = + 56 = 9 56

4 Math 4, Autumn Solutions to Final Exam December, Page 4 of 6 4. ( points) (a) Melted ice cream drips into a large waffle-style cone of height 6 cm and radius 3 cm. (This waffle cone is shaped like a standard right circular cone.) At the moment when the depth of ice cream is 8 cm, the rate of inflow of ice cream is.5 cm 3 /sec. How fast is the level of liquid ice cream rising in the cone at this moment? (6 points) Let V be the volume of the ice-cream in the cone. Let h be the level in the cone and r the radius of the surface of ice-cream in the cone. From the assumption, we are given that dv dt =.5 cm3 /sec. We need to find out dh dt. Next, the volume of ice-cream inside the cone is also given by V = πr h 3. Using similar triangles, we see that r = 3h 6, and plugging this in the formula above we get V = π 3 ( 3 6 h) h = 3πh3 56. After taking the derivatives of this volume with respect to t, we find When h = 8, we get Since dv dt dv dt = 9π dh h 56 dt. dv dt = 9π 56 (8 ) dh dt = 9π 4 =.5 cm3 /sec, we finally get dh dt = 9π cm/sec. dh dt.

5 Math 4, Autumn Solutions to Final Exam December, Page 5 of 6 (b) (Note: ice cream imagery aside, this part is independent of any fact from part (a).) The source of the melted ice cream from part (a) is a large spherical scoop; the scoop is melting in such a way that it always maintains its spherical shape as it shrinks in size, but at any instant its rate of change of volume, dv, is equal to the product of its surface area, A, times a fixed constant: specifically, dt dv sphere = dt 5 A sphere (Here A and V are measured in square centimeters and cubic centimeters, respectively, and time is measured in seconds.) Show that the radius of the scoop is shrinking at a constant rate, and find this rate. (4 points) Let r be the radius of the sphere scoop. First we have V sphere = 4πr3 3 from the volume formula of a ball. Taking derivatives of the volume with respect to t, we get dv sphere dt = 4πr dr dt. Using the assumption dv sphere dt = 5 A sphere and area formula A sphere = 4πr, we get 4πr dr dt = 5 (4πr ). After dividing 4πr at both sides, one can obtain dr dt = 5. This implies that the radius of the scoop is shrinking at a constant rate of 5 cm/sec.

6 Math 4, Autumn Solutions to Final Exam December, Page 6 of 6 5. (9 points) Let g be the function defined on the interval { x } by: g(x) = (a) Find g( ) and g ( ). ( points) We have g( ) = x cos(πt )dt =, by the properties of definite integrals. By the Fundamental Theorem of Calculus, g (x) = cos(πx ). g ( ) = cos(π( ) ) = cos(π) =. (b) It is a fact that g() = g(); explain why this is true. cos(πt ) dt. Plugging in x =, we get (3 points) For the sake of notational convenience we set f(t) = cos(πt ), so that g(x) = x f(t)dt. Now notice that f is an even function: f( t) = cos(π( t) ) = cos(πt ) = f(t). This means that the graph of y = cos(πt ) is symmetric around the y-axis. We then conclude that f(t)dt = The signed area underneath f on [, ] = The signed area underneath f on [, ] = f(t)dt. () This fact can also be proved by substitution: u = t. du = dt. f(t)dt = f( u)( du) = f( u)du = Now g() = f(t)dt and g() = f(t)dt = f(t)dt + have g() = f(t)dt + f(t)dt = f(t)dt = g(). f(u)du = f(u)du. f(t)dt. Using equation () we (c) On what parts of the domain { x } is g increasing? decreasing? Explain completely. (4 points) Again, by FTC, g (x) = cos(πx ). Recall that g is increasing when g is positive, decreasing when g is negative. Thus we need to find out the subinterval of [, ] on which cos(πx ) stays positive or negative. We know cos θ is positive when θ π. Therefore, cos(πx ) is positive when πx < π x < < x <. Similarly, we know cos θ is negative when π < θ < 3π. Therefore, cos(πx ) is negative when π < πx < 3π < x < 3 3 < x <, or 3 < x <, However, notice that 3/ greater than and 3/ is less than, and we do not care about the range of x beyond [, ]. Hence let us put: cos(πx ) is negative when < x <, or < x <. x cos(πx ) + Therefore, g is increasing on (, ), decreasing on [, ), (, ]. Including or excluding the endpoints ±, ± is a matter of definition. We accept answers of all kind regarding in/excluding the endpoints.

7 Math 4, Autumn Solutions to Final Exam December, Page 7 of 6 (d) On what parts of this domain is the graph of g concave upward? downward? Explain completely. (4 points) By chain rule g (x) = (cos(πx )) = sin(πx )(x ) = sin(πx )πx. g is concave up if g when positive, concave down when g is negative, so we need to find out the subinterval of [, ] on which sin(πx )πx stays positive or negative. We know sin θ stays positive when < θ < π. Therefore, sin(πx ) is positive when < πx < π < x < < x <, or < x <. We then analysis the sign of g (x) = sin(πx )πx (, ) and (, ). On (, ), from the above we know sin(πx ) is negative, and πx is negative. Hence altogether g (x) = sin(πx )πx is positive on (, ). Similarly, on (, ), sin(πx ) is negative, and πx is positive. Hence g (x) = sin(πx )πx is negative on (, ). x sin(πx ) πx + + g (x) = sin(πx )πx + Therefore, g is concave up on (, ), concave down on (, ). We accept answers of all kind regarding in/excluding the endpoints ± as (c). (e) Using the information you ve found in parts (a)-(d), sketch the graph of g on the domain [, ]. Label any extreme or inflection points by their x-coordinate; also use the fact that g().37. (6 points) From (a) we know g( ) =. From (b) we know g() = g().37 =.74. From (c) we know g is decreasing near the left of x = and increasing near the right of x =. Hence by first derivative test g has a local (actually absolute, but we do not ask for it) minimum at x =. Similarly, g has a local (actually absolute) maximum at x =. Recall that a point of inflection is a point at which the function changes its concavity. From (d) we know x = is an inflection point. By this definition, x = and x = are inflection point since the function g is defined only on [, ]. However, we can still interpret x = ± as inflection points by extending g beyond [, ]. Therefore, we accept answers with or without ± as inflection points.

8 Math 4, Autumn Solutions to Final Exam December, Page 8 of 6 6. ( points) (a) Give a precise statement of the Mean Value Theorem. ( points) The Mean Value Theorem states that if f is a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) then there is some number c in (a, b) so that f f(b) f(a) (c) = b a It is very important to say that c is in the open interval from a to b! (b) Let f(x) = x e ln x. Use the Mean Value Theorem to show that if b is any number larger than e, then f(b) >. (7 points) We want to show that f(b) = b e ln b > for all b > e. To do this, note that for any number b > e, the function f(x) is continuous on the closed interval [e, b] and differentiable on the open interval (e, b). So we can use the mean value theorem with a = e, b = b, and the function f(x). From part a, the mean value theorem says that for some c between e and b, f (c) = f(b) f(e) b e Now, f(e) = e e ln(e), where ln(e) =. So f(e) =. Also, we computed that f(b) = b e ln(b).thus, we get that for some c between e and b, f (c) = b e ln(b) b e Next, f (x) = e x. Thus, f (c) = e c. We know that c is between e and b. That means, in particular, that c is bigger than e. Therefore, e c is less than, so f (c) = e c is positive. We combine the fact that f (c) is greater than with the result of the mean value theorem to get that b e ln(b) > b e We also know that b > e, so b e >. Thus we can multiply both sides by b e to get that b e ln(b) >. So, we are done. (c) Which is larger, e π or π e? Justify your answer. (Hint: you can use the result of part (b) even if you did not prove it.) (3 points) The trick to this part is to figure out what to plug in to f(x). We showed that for b > e, b e ln(b) >. That is, so raising e to both sides, b > e ln b b > ln b e e b > e ln(be) = b e Since π > e, we can set b = π and get that e π > π e.

9 Math 4, Autumn Solutions to Final Exam December, Page 9 of 6 7. ( points) A particle moving along a line has position at time t (in minutes) given by s(t) = 3 t3 3t + 4t meters. (a) Find the initial velocity and acceleration; that is, the velocity and acceleration at t =. ( points) Since the position is s(t) = 3 t3 3t + 4t, the velocity is v(t) = s (t) = t 6t + 4 and the acceleration is a(t) = s (t) = 4t 6. At time t =, the velocity is v() = 4 and the acceleration is a() = 6. (b) At what time, after t =, does the particle first come to a stop? (3 points) The particle is stopped when v(t) =. We solve t 6t + 4 = (t 3t + ) = (t )(t ) =, so v(t) = at t = and t =. The first time the particle comes to a stop is t =.

10 Math 4, Autumn Solutions to Final Exam December, Page of 6 For easy reference, the particle s position function is s(t) = 3 t3 3t + 4t. (c) Find the position of the particle at the second time the particle comes to a stop after t =. ( points) From part b, the second time the particle comes to a stop is t =. The position at that time is s() = 3 ()3 3() + 4() = 4/3. (d) What is the total distance traveled by the particle between t = and t = 3? Give all the steps in your reasoning. (5 points) The velocity v(t) = (t )(t ) is positive when t is between and, negative when t is between and, and positive when t is between and 3. The total distance traveled from time to time is s() s() = 5/3 = 5/3. The total distance traveled from time to time is s() s() = 4/3 5/3 = /3. The total distance traveled from time to time 3 is s(3) s() = 3 4/3 = 5/3. All told, from time to time 3, the particle travels a distance of 5/3 + /3 + 5/3 = /3. Note that the overall displacement (i.e., net change in position) is 5/3 /3+5/3 = 3, which is simply s(3) s(); but the correct answer to the question as written is /3.

11 Math 4, Autumn Solutions to Final Exam December, Page of 6 8. (5 points) Let f(x) = x. (a) On the axes below, sketch a graph of f over the domain [, 3], and then draw the approximating rectangles that are used to estimate the area under the curve (and above the x-axis) between x = and x = 3 according to the Midpoint Rule; use n = 4 rectangles. (3 points) (b) Write an expression involving only numbers that represents the area estimate using the rectangles described above. (You do not have to expand or simplify the expression!) (3 points) The midpoints of the intervals are x =.5,.75,.5,.75. The heights of the rectangles are f(x) at those x-values. The widths of the rectangles are all.5. The total area is.5 f(.5) +.5 f(.75) +.5 f(.5) +.5 f(.75) =.5[(.5 ) + (.75 ) + (.5 ) + (.75 )].

12 Math 4, Autumn Solutions to Final Exam December, Page of 6 (c) Find the exact area of the same region by evaluating the limit of a Riemann sum that uses the Right Endpoint Rule. (That is, do not use the Fundamental Theorem of Calculus.) Show all reasoning. (9 points) The general form for the limit of a Riemann sum using the Right Endpoint Rule is b a f(x)dx n n f(x i ) x, where x = b a n and x i = a + i x. We have a =, b = 3, and f(x) = x, meaning that x = n and x i = + i n. Now, [ 3 n ( (x )dx + i ) ] n n n i= n n ( i= n ( 8 n n [ 4i n + 4i n i= ] n ) 8 n n i + 8 n n 3 i i= i= n(n + ) + 8 ) n(n + )(n + ) n n3 6 ( 8 ( + n ) + 8 ( + n )( + n ) ) 6 = = 3.

13 Math 4, Autumn Solutions to Final Exam December, Page 3 of 6 9. (5 points) Verify the following indefinite integral expression by differentiation, showing your steps: ( + e x ) dx = x + + e x ln(ex + ) + C All we need to do for this problem is to differentiate the right hand side and show that it is equal to (+e x ). ( x + ) + e x ln(ex + ) =x + ( ( + e x ) ) (ln(e x + )) = + ( )( + e x ) (e x ) e x + (ex ) e x = ( + e x ) ex e x +. Now we combine the fractions by making the denominators common: as desired. e x ( + e x ) ex e x + = (ex + ) (e x + ) e x ( + e x ) ex (e x + ) (e x + ) = (ex + ) e x (e x + )e x (e x + ) = (ex ) + e x + e x (e x ) e x (e x + ) = (e x + ),

14 Math 4, Autumn Solutions to Final Exam December, Page 4 of 6. (9 points) Suppose r(t) is a continuous function which gives the instantaneous rate of change of North America s pigeon population, measured in thousands of pigeons per year, where t is measured in years since January, 9. (a) What does the quantity 8 r(t) dt represent? Express your answer in terms relevant to this situation, and make it understandable to someone who does not know any calculus; be sure to use any units that are appropriate, and also explain what the sign of this quantity would signify. (4 points) The number 8 r(t) dt is the net change in the North American pigeon population, in thousands of pigeons, over the period between January, 9 and January, 98. A negative quantity signifies a net population decrease, and a positive quantity signifies a net increase. Alternate solution: The number 8 ( The North American pigeon population on January, 98, in thousands r(t) dt is equal to the difference ) ( ) The North American pigeon population. on January, 9, in thousands (Note that the meaning of a negative or positive quantity is built implicitly into this formulation.) (b) Define a function Q by Q(x) = x+ x r(t) dt. In the context of this application, what does the value of Q(x) represent in terms of x? Follow the same guidelines as in part (a) in expressing your answer. ( points) Q(x) gives the net change in the North American pigeon population, in thousands of pigeons, over a ten-year period that starts exactly x years after January, 9. (Note that x need not be a positive integer; if x negative, then we begin our ten-year period exactly x years before January, 9.) If Q(x) is positive, then the population experienced a net increase over this decade; if Q(x) is negative, then the population experienced a net decrease instead. (c) Suppose Q has a critical number at x = c. Give a simple condition involving c that must hold for the function r; your answer should not involve integrals. (3 points) If Q has a critical number at c, then either Q (c) = or Q is not differentiable at c. But note that Q(x) is differentiable for all x: we have Q(x) = x r(t) dt + x+ r(t) dt = x r(t) dt + x+ r(t) dt and this is a difference of two area functions, each of which is differentiable by the Fundamental Theorem of Calculus (since we are given that r(t) is continuous). Thus, Q (c) = for any critical number c of Q. We compute Q (x) below, using the chain rule with u = x + : Q (x) = d dx = d ( x+ ) r(t) dt ( x ) r(t) dt ( x ) r(t) dt x r(t) dt ( u ) r(t) dt d dx dx ( d u ) = r(t) dt du = r(u) r(x) = r(x + ) r(x) du dx d dx Thus, c is a critical number of Q if and only if = Q (c) = r(c + ) r(c), or equivalently r(c + ) = r(c).

15 Math 4, Autumn Solutions to Final Exam December, Page 5 of 6. ( points) Evaluate each of the following integrals, showing all reasoning. ( x (a) sec x tan x + ) x x x dx (5 points) First, note that x 3 dx = x = (x )x /3 dx x /3 x /3 dx = 3 5 x x 3 + C Next, we can see by inspection that an antiderivative of = x is x x. So we get ln() (b) x 3 x + sin θ + cos θ dθ (5 points) + sec(x) tan(x) + x x dx x = 3 dx + x dx + x = 3 5 x x 3 + arcsin(x) + sec(x) x ln + C sin θ + cos θ dθ = + u du sec(x) tan(x)dx + x dx using u-substitution with u = cos θ and du = sin θdθ = arctan u + C = arctan(cos(θ)) + C

16 Math 4, Autumn Solutions to Final Exam December, Page 6 of 6 (c) t ln(t) dt (6 points) t ln(t)dt = 3 t3 ln(t) 3 t dt using integration by parts with u = ln(t), du = t dt, v = 3 t3 and dv = t dt. = 3 t3 ln(t) 9 t3 + C (d) π cos( x) dx (6 points) π cos( x)dx = π u cos(u)du using u-substitution with u = x, du = x dx. We solve for dx to get dx = xdu, so dx = udu. = u sin(u) π π sin(u)du using integration by parts with v = u, dv = du, w = sin(u) and dw = cos(u)du. Note that the limits of integration are the same. = u sin(u) π ( cos(u)) π = cos(π) cos() = 4 since sin(π) = sin() =, and cos(π) = while cos() =.