Math 2250 Final Exam Practice Problem Solutions. f(x) = ln x x. 1 x. lim. lim. x x = lim. = lim 2

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1 Math 5 Final Eam Practice Problem Solutions. What are the domain and range of the function f() = ln? Answer: is only defined for, and ln is only defined for >. Hence, the domain of the function is >. Notice that ln =, + since + as +. Now, we can evaluate using L Hôpital s Rule; it is equal to ln = = =. f will have some maimum value; to figure out what it is, take f () = ln = ln = ln. 3/ Then f () = when = ln, meaning that ln =, or = e. Notice that f () changes sign from positive to negative at = e, so the maimum of f occurs here. Since f(e ) = ln e e = e, we see that the range of f is (, ]. e. Find the point on the graph of y = e 3 at which the tangent line passes through the origin. Answer: Let f() = e 3. Since f () = 3e 3, the tangent line to e 3 at the point = a has slope 3e 3a ; hence, using the point-slope formula, it is given by y e 3a = 3e 3a ( a) = 3e 3a 3ae 3a. In other words, the tangent line to the curve at = a is or y = 3e 3a 3ae 3a + e 3a y = e 3a (3 3a + ). This passes through the origin if we get equality when we substitute for both and y, so it must be the case that = e 3a ( 3a + ) = e 3a ( 3a).

2 Since e 3a, this means that 3a =, or a = /3. since f(/3) = e 3 /3 = e, the point whose tangent line passes through the origin is ( ) 3, e. 3. Find the equation of the tangent line to the curve at the point (3, ). y 3 y = 6 Answer: Differentiating both sides with respect to yields Thus, y 3 + 3y dy dy y d d =. dy ( 3y ) = y y 3. d dy y y3 = d 3y. Plugging in (3, ), we see that the slope of the tangent line is (3)() 3 3(3)() 3 = = 4 7. Thus, using the point-slope formula, the equation of the tangent line is or, equivalently, y = 4 7 ( 3) = 4 7 7, y = Use an appropriate linearization to approimate 96. Answer: Let f() =. Then I will approimate 96 using the linearization of f at a =. To do so, first take f () =. Then the linearization is L() = f() + f ()( ) = + ( ) = + 5 = + 5. So we approimate 96 by = f(96) L(96) = = = 98 = 9.8.

3 5. Consider the function f() = e. What is the absolute maimum of f()? Answer: Notice that f is defined for all. Also, f() = ± ± e =, (by two applications of L Hôpital s Rule) so f doesn t go off to infinity. Now, to find the critical points, compute f () = e + e ( ) = e ( 3 ), which equals zero precisely when = 3 = ( ); namely when = or = ± Thus, we just need to evaluate f at the critical points: f() = /e f() = f( ) = /e Since f its to in both directions, we see that the absolute maimum value of the function (occurring at both = and = ) is /e. 6. Water is draining from a conical tank at the rate of 8 cubic feet per minute. The tank has a height of feet and the radius at the top is 5 feet. How fast (in feet per minute) is the water level changing when the depth is 6 feet? (Note: the volume of a cone of radius r and height h is πr h 3.) Answer: If h is the height of the top of the water in the cone and r is the radius of the top of the water, then r 5 = h, so r = h/. Now, the volume of water in the tank is V = 3 πr h = 3 π(h/) h = π h3. In turn, this means that dv dt = π dh 3h dt = π dh h 4 dt. Since dv dt = 8, this means that 8 = π dh h 4 dt, or dh dt = 7 πh. Thus, when h = 6, the water level is changing at the rate dh dt = 7 36π = π. 7. The function f() = is concave down for what values of? Answer: To determine concavity, we need to compute the second derivative. Now, so f () = 4 3 8, f () = 36 = ( 3). Notice that f () < precisely when < < 3, so the function f is concave down on the interval (, 3). 3

4 8. Evaluate the it ( 6)/. Answer: Let f() ( 6) /. Taking the natural log of f yields Now, by L Hôpital s Rule, ln(( 6) / ) = ln( 6) ln( 6) =. ln( 6) ln(f()) = = since this is the it of ln(f()), we know that 9. Let f() = cos. What is f (π/)? 6 6 f() = e 6. 6 = 6 = 6. Answer: I will use logarithmic differentiation to find f (). To that end, let y = f() = cos. Then Differentiating both sides, Hence, ln y = ln( cos ) = cos ln. dy y d = cos cos sin ln = sin ln. f () = dy d = y ( cos sin ln ) = cos ( cos sin ln ). f (π/) = (π/) cos π/ ( cos π/ π/ ) sin π/ ln(π/) = (π/) ( ln(π/)) = ln(π/).. For t 5, a particle moves in a horizontal line with acceleration a(t) = t 4 and initial velocity v() = 3. (a) When is the particle moving to the left? Answer: The particle will be moving to the left when its velocity is negative. To determine the velocity, note that a(t)dt = (t 4)dt = t 4t + C. Hence, v(t) = t 4t + C for some C, which we can determine by plugging in t = : 3 = v() = 4() + C = C, so v(t) = t 4t + 3 = (t 3)(t ). Notice that this function is negative when < t < 3, so the particle is moving to the left between t = and t = 3. (b) When is the particle speeding up? Answer: The particle is speeding up when its acceleration is positive, which is to say when so the particle is speeding up when t >. < a(t) = t 4, 4

5 (c) What is the position of the particle at time t if the initial position of the particle is 6? Answer: Since v(t)dt = (t 4t + 3)dt = t3 3 t + 3t + D, we know that s(t) = t3 3 t + 3t + D for some real number D, which we can solve for by plugging in t = : 6 = s() = 3 3 () + 3() + D = D, so the position of the particle at time t is s(t) = t3 3 t + 3t If 6 f()d = and f()d = 7, find 6 4 f()d. Answer: Notice that 6 4 f()d = 6 f()d f()d = 7 = 3.. Evaluate the definite integral π/4 π/6 sin tdt. Answer: Since cos t is an antiderivative of sin t, the Fundamental Theorem of Calculus tells us that π/4 [ ] π/4 3 3 sin tdt = cos t = cos(π/4) ( cos(π/6)) = π/6 + =. π/6 3. Evaluate the integral t 3 dt. Answer: Since t 3 looks vaguely like t, we should epect that the natural log comes into play. In fact, ln(t 3) is an antiderivative of t 3, so dt = ln(t 3) + C. t 3 4. Evaluate the definite integral Answer: Re-write the integral as ( Now, + 4 d ) d = d + 4 d = d = [ / / d = / ] 4 d + = [ 4 ] 4 = 8 4 = 4. 4d. 5

6 On the other hand, + 4 d = 5. Suppose the velocity of a particle is given by 4d = [ ] 4 = 3 = 3. d + v(t) = 6t 4t. What is the displacement of the particle from to? Answer: The displacement is given by s() s(). Since s (t) = v(t), the Fundamental Theorem tells us that s() s() = s (t)dt = v(t)dt = the displacement is 8 units. 6. Suppose that What is f()? Answer: Let g() = +. Then, ( ) g () = d f(t)dt d where u =, using the Chain Rule. 4d = = 34. (6t 4t)dt = [ t 3 t ] f(t)dt = +. by the first part of the Fundamental Theorem, In other words, Now, we know that g() = +, so = d ( u ) du f(t)dt du d g () = f(u) = f( ). f( ) = g (). = (6 8) ( ) = 8. Hence, g () = + () = +. f() = g ( ) = 3 = 3. 6

7 7. Evaluate the integral 3e tan sec d. Answer: Let u = tan. Then du = sec d, so we can re-write the above integral as 3e u du = 3e u + C. Now, substituting back in for u, this yields the answer 3e tan + C. 8. Evaluate the definite integral π/6 8 tan(4) d. Answer: Notice that tan(4) = sin(4) cos(4), so the above integral is equal to π/6 Let u = cos(4). Then du = 4 sin(4) d. 8 sin(4) cos(4) d. 8 sin(4) cos(4) d = ( 4 sin(4)d) = cos(4) u du. We want to replace the given integral with an integral in terms of u, but that means we also need to change the its of integration: u() = cos(4 ) = cos() = and u(π/6) = cos(4 π/6) = cos(π/4) =. the integral we were given can be re-written as / u du = [ ln u]/ = ln(/ ) + ln() = ln(/ ( ) = ln /). But now, from the properties of logarithms, ( ln /) ( ( = ln /) ) = ln ( ) = ln. Hence, we conclude that π/6 8 tan(4) d = ln. 9. What is the area of the red region in the figure? The blue curve is given by y = ln green lines are the lines = e and = e. Answer: First, notice that the curve y = ln crosses the -ais when ln =, and the vertical 7

8 which only occurs when ln =, meaning when =. Between = /e and =, the -ais (i.e. the curve y = ) is above the blue curve, whereas the blue curve is above the -ais when < e. the red area is equal to /e ( ln ) d + e ( ln ) d = ln /e d + For both of these integrals, let u = ln. Then du = d. Moreover, u(/e) = ln(/e) = ln(e ) = u() = ln() = ln(e ) = u(e) = ln(e) = ln(e ) =. e ln d. the sum of integrals above is equal to so the red region has area. u du +. What are the domain and range of the function u du = [ u ] + [ u ] = ( () + ( ) ) + ( ) =, f() = + e e? Answer: The function is well-defined everywhere ecept when the denominator is zero, which happens when = e, or, equivalently, e =. This only happens when =, so we see that the domain of f is the set of all real numbers. As for the range of f, notice that + e e = + 8

9 and Now, and + e + e =. + e e = + e + e = e + e = by L Hôpital s Rule, so f has horizontal asymptotes at and. Finally, since f () = ( e )e ( + e )( e ) ( e ) = e e + e + e e ( e ) = ( e ) is never zero, f has no critical points and, thus, no maima or minima. Putting this all together, the range of f consists of those real numbers such that < < or < < +.. What is the equation of the line tangent to the graph of y y + 3 = 4 at the point (, )? Answer: The goal is to determine y by using implicit differentiation. Differentiating both sides yields: or, equivalently, Hence, 3y y + 6y + 3 y y + 6 =, y (3y + 6 y) + 6y + 6 =. y = 6y 6 ) 3y + 6 y. the slope of the tangent line at (, ) will be 6()( ) 6() 3( ) + 6() ( ) = 3 = 4. Hence, using the point-slope formula, the tangent line at (, ) will be y + = 4( ) = 4 4, or, equivalently, y = Evaluate the it cos +. Answer: Notice that both numerator and denominator go to zero as. Hence, we can apply L Hopital s Rule: cos + = sin + =, since sin() =. 9

10 3. The volume of a cube is increasing at a rate of cm 3 /min. How fast is the surface area increasing when the length of an edge is cm? Answer: If V (t) is the volume of the cube after t minutes, A(t) is the surface area, and l(t) is the length of an edge, then we know that V (t) = (s(t)) 3 A(t) = 6(s(t)) V (t) = for all t s(t ) = where t isthemomentofinterest and we re asked to figure out A (t ). Now, we know that so A (t) = 6 s(t) s (t) = s(t)s (t), A (t ) = s(t )s (t ) = ()s (t ) = s (t ), so all we need to do is determine s (t ). To do so, let s differentiate V : Then, plugging in t = t, we have V (t) = 3(s(t)) s (t). = V (t ) = 3(s(t )) s (t ) = 3() s (t ) = 3s (t ), meaning that s (t ) = 3 = 3. we know that A (t ) = s (t ) = so the surface area is increasing at a rate of 4 cm /sec. 3 = 4, 4. Find the maimum and minimum values, inflection points and asymptotes of y = ln( + ) and use this information to sketch the graph. Answer: Notice that and, by the Quotient Rule, y = + = + y = ( + )() () ( + ) = + 4 ( + ) = ( + ). Now, the critical points occur when y =, which is to say when + =. The only happens when =, so is the only critical point. Notice that y () =, which is greater than zero, so the second derivative test implies that is a local minimum. y = when =, meaning when = ±, so there are inflection points at = ±. Finally, so there are no horizontal asymptotes. ln( + ) = = + ln( + ), Putting all this together, we see that y has a minimum at and is concave up between and and concave down everywhere else and has no asymptotes, meaning that the graph looks something like this:

11 (-, ln ) (, ln ) Use an appropriate linearization to approimate e /. Answer: Consider the function f() = e. Then I want to approimate f(/) = e / using the linearization of f at =. This linearization is given by L() = f() + f ()( ) = f() + f (). Now f() = e = and f () = e, so f () = e =. Hence, the linearization is given by L() = +. f(/) L(/) = + = =.. 6. What is the absolute maimum value of f() = / for >? Answer: Taking the natural log of both sides,. Now differentiating, we see that so ln f() = ln( / ) = ln f () f() = ln = ( ln ), f () = f() / ( ln ) = ( ln ). Since / is never zero for >, f () = only when ln =, meaning that ln =. This only happens when = e, so e is the only critical point of f. Notice that f () changes sign from positive to negative at = e, so the first derivative test implies that f has a local maimum at e. However, since this is the only critical point and there are no endpoints, this must, in fact, be the global maimum of f.

12 Hence, the absolute maimum value of f() for > is 7. Suppose the velocity of a particle is given by f(e) = e /e. v(t) = 3 cos t + 4 sin t. If the particle starts (at time ) at a position 7 units to the right of the origin, what is the position of the particle at time t? Answer: Let s(t) be the position of the particle at time t. Then we know that s (t) = v(t) and that s() = 7. Now, v(t)dt = (3 cos t + 4 sin t)dt = 3 sin t 4 cos t + C. since s(t) is an antiderivative of v(t) = s (t), we know that s(t) = 3 sin t 4 cos t + C for some real number C. To solve for C, plug in t = : so we see that C =. the position of the particle is given by 8. Evaluate the definite integral Answer: Note that 7 = s() = 3 sin() 4 cos() + C = 4 + C, + cos 3 θ cos θ π/6 s(t) = 3 sin t 4 cos t +. π/6 + cos 3 θ cos dθ. θ = cos θ + cos θ = sec θ + cos θ. + cos 3 θ cos dθ = θ π/6 By the Fundamental Theorem of Calculus, this is equal to ( sec θ + cos θ)dθ. [ tan θ + sin θ] π/6 = ( tan(π/6) + sin(π/6)) ( tan() + sin()) = Evaluate csc r cot r dr. Answer: Recall that the derivative of csc r is csc r cot r, so csc r cot r dr = csc r + C.

13 3. Let g() = What is the derivative of g? Answer: Let h(u) = u sin t t dt and let f() =. Then so we can compute g () using the Chain Rule: Now, and, by the Fundamental Theorem of Calculus, sin t t dt. g() = h(f()), g () = h (f())f (). f () = h (u) = sin u u. Hence, g () = h (f())f () = h ( ) = sin( ) = sin( ) = sin( ). 3. If f() = ln, find f (e 3 ). Answer: Using the quotient rule, f () = ln d d () d d (ln ) (ln ) = ln (ln ) = ln (ln ). f (e 3 ) = ln(e3 ) (ln(e 3 )) = 3 3 = At what value(s) of (if any) is the function f() defined below discontinuous? if < f() = if + if > For <, f() = , which is a polynomial and, hence, continuous. Also, for < <, f() =, which is also a polynomial and hence continuous. Finally, for >, f() = +, which is continuous since it s the composition of continuous functions (specifically, f() = g h() where g(u) = + u and h() = ). the only possible points of discontinuity for f are at = and =. At =, f() will be continuous if f() = f( ) = =. Clearly, + f() = + =. On the other hand, f() = ( ) = ( ) + 4( ) + 5 = =. 3

14 since + f() = and f() =, we can see that f() is not continuous at =. Turning to =, f() will be continuous if f() = f() = =. Now f() = =. On the other hand ( ) f() = + + =. + Since the its from the left and the right are both equal to, we conclude that so f() is continuous at =. f() =, f() is continuous at all real numbers ecept =. 33. Find the equation of the tangent line to the curve at the point (, π 3 ). cos y = Answer: We will find the slope of the tangent line using implicit differentiation. Differentiating both sides yields cos y + ( sin y y ) = or and so we have that cos y y sin y =. cos y = y sin y y = cos y sin y. at the point (, π 3 ) the slope of the tangent line is y = cos(π/3) sin(π/3) = / 3 By the point-slope formula, the equation of the tangent line is or y π 3 = ( ) 3 y π 3 = the equation of the tangent line is y = π 3 3. = 3. 4

15 34. A particle moves so that its position at time t is given by s(t) = 3t + 4. If v(t) is the velocity of the particle at time t, what is t + v(t)? Answer: First, we know that the velocity v(t) is just the derivative of the position function. In other words, v(t) = s (t) = (3t + 4) / 3t 6t = 3t + 4. v(t) = t + t + 3t 3t + 4. Now, notice that the leading terms of both numerator and denominator are essentially t (since the t in the denominator is under the square root). we should simplify things by multiplying both numerator and denominator by t : t + 3t 3t + 4 = t + = t + = t + = 3 3 = 3. we can conclude that t + v(t) = 3. 3t 3t + 4 t 3t t t t (3t + 4) t 35. A technical writer is producing a book which must have -inch side margins and -inch top and bottom margins. If the area of the page can be at most 5 in, what dimensions give the most printed area per page? Answer: If we call the width of the page and the height of the page y, then consists of two -inch margins plus the width of the tet, so the actual width of the tet is. Likewise, y consists of two -inch margins plus the height of the tet, so the height of the tet is y 4. Now, we know that the area of the page is 5 in, meaning that y = 5. Solving for y, we see that y = 5. The quantity we want to maimize is the area of the printed tet, which, by the discussion above is ( ) 5 ( )(y 4) = ( ) 4 = = Hence, we re looking to maimize the function A() = Of course, it must be the case that >, and in fact this is the only constraint (since >, y = 5 is automatically bigger than zero). 5

16 To find the maimum, let s determine the critical points: A () = when meaning that and so A () = 4. 4 =, = 4, = 4. Hence = 5 and so = ±5. Clearly 5 is not greater than zero, so the only critical point we care about will be = 5. Since A () = 3, which is negative for all >, we see that = 5 is a local maimum. Since it s the only critical point in the interval we re interested, that means it must be the absolute maimum. the printed area will be maimized when the width of the page is = 5 and when the height of the page is y = 5 = 5 5 =. 36. Suppose f(t) = t + cos t. Evaluate the definite integral π/ π/ f(t)dt. Answer: Since t + sin t is an antiderivative of f(t), the Fundamental Theorem of Calculus tells us that π/ π/ (t + cos t)dt = [ t + sin t ] π/ ( (π = = π π/ ) + sin(π/) ) 4 + π 4 ( ) =. ( ( π ) ) + sin( π/) 37. Evaluate the it Answer: Notice that and ln(3 + e 3 ). 6 ln(3 + e3 ) = 6 =. we can apply L Hôpital s Rule, so the above it is equal to 3+e e = 6 e 3 3+e 3 e 3 = 3 + e 3. 6

17 Again, both numerator and denominator are going to, so we can apply L ôpital s Rule again to get we conclude that e 3 3 e 3 3 = =. ln(3 + e 3 ) = As a spherical raindrop falls, it evaporates (i.e. loses volume) at a rate proportional to its surface area. Show that the radius of the raindrop decreases at a constant rate. (Hint: the volume of a sphere of radius r is 4 3 πr3, and the surface area is 4πr ) Answer: First, let s record what we know. We know that the volume is given by V (t) = 4 3 π(r(t))3 and that the surface area is given by A(t) = 4π(r(t)). Moreover, since the volume changes at a rate proportional to the surface area, we know that for some constant C. V (t) = C A(t) Now, what we re trying to show is that the radius decreases at a constant rate; in other words, that r (t) is constant. To get at r (t), let s differentiate V (t): V (t) = 4 3 π 3(r(t)) r (t) = 4π(r(t)) r (t) by the Chain Rule. On the other hand, we know that V (t) = C A(t), so we have that Dividing both sides by 4π(r(t)) yields that 4π(r(t)) r (t) = C A(t) 4π(r(t)) r (t) = C 4π(r(t)). r (t) = C, so indeed the rate of change of the radius is constant. 39. Suppose f (t) = 3e t sec t tan t and that f() = 4. Give a formula for f(t). Answer: First, let s take the indefinite integral of f (t): f (t)dt = (3e t sec t tan t ) dt = 3e t sec t + C for some constant C. In other words, f(t) = 3e t sec t + C for some C. To determine C, we want to plug in the initial value: f() = 3e sec() + C 4 = 3 + C 4 = + C, so we see that C = 3. the formula for f(t) is f(t) = 3e t sec t