2 (1 + 2 ) cos 2 (ln(1 + 2 )) (ln 2) cos 2 y + sin y. = 2sin y. cos. = lim. (c) Apply l'h^opital's rule since the limit leads to the I.F.

Transcription

1 . (a) f 0 () = cos sin (b) g 0 () = cos (ln( + )) (c) h 0 (y) = (ln y cos )sin y + sin y sin y cos y (d) f 0 () = cos + sin (e) g 0 (z) = ze arctan z + ( + z )e arctan z Solutions to Math 05a Eam Review + = ( + ) cos (ln( + )) = sin y (ln ) cos y + sin y cos y + z = (z + )earctan z. (a) Apply l'h^opital's rule since the it leads to the indeterminate form (I.F.) 0=0. ln(cos ) ( sin ) sin cos = =!0 sin!0 cos!0 cos = 0 (b) Method : Rewrite the epression using sine and cosine:!(=) sec + tan =!(=) cos + sin cos =!(=) cos + sin = Method : Since the original it leads to the I.F. = we can apply l'h^opital's rule.!(=) sec + tan =!(=) sin cos cos =!(=) sin = (c) Apply l'h^opital's rule since the it leads to the I.F. =:! ln p =! p = p = 0! (d) The original it leads to the I.F. 0, in order to apply l'h^opital's rule (twice in this case), it is necessary to rewrite the epression so that the it yields the I.F. =. (e)! e =! e =! e =! +! + = e = 0. Consider = y. Implicitly dierentiate both sides with respect to. y (y ) y y y y = = (y ) (y ) Therefore, = ) (y y. (a) The local linearization of f() = p + near = 0 is given by f() f(0) + f 0 (0)( 0). Note: f(0) = and f 0 () = Therefore, f() +. p ( + ) so f 0 (0) =. (b) To approimate p :95, let = 0:05, then p :95 = f( 0:05) + ( 0:05) = 0:98.

2 (c) The graph of f is concave down for >, so the tangent line at = 0 will sit above the graph of f over that same interval. Therefore, the local linearization at = 0:05 will be an p overestimate for the actual value of : Consider the curve given by ln y + y = + ln. (a) Implicitly dierentiate both sides with respect to, solve for, and then substitute in the and y-values at (; ). ln y + y + y = =) = ln y + y y At (; ), = ln + = () (b) The local linearization of the curve near = is the equation of the tangent line passing through (; ). The slope of the tangent at (; ) is. Therefore, the local linearization is y + ( ). (c) When = :, y + (: ) = : An eample of a slanted ellipse is given by the equation + y + y =. (a) Implicitly dierentiate both sides with respect to, solve for, and then substitute in the and y-values at ( p ; 0). + y + + y = 0 =) = y + y At ( p ; 0), = p 0 p = + 0 (b) A line is normal to a curve at a point if it is perpendicular to the curve's tangent at that point. For a given line of slope m 6= 0, the slope of a line perpendicular line it is =m. Therefore, at the point ( p ; 0) the slope of any line perpendicular to the tangent is =. Therefore, the equation of the normal line is y = p. (c) = 0 for any points on the ellipse with a horizontal tangent. = 0 when y = 0 =) y =.

3 Substitute y = into the original equation for the ellipse and solve for. + ( ) + ( ) = () = () = Substitute the -values into y = to nd the corresponding values for y. The ordered pairs of the points with horizontal tangents are ( ; ) and (; ). 7. A stone dropped into a still pond sends out a circular ripple. The radius is increasing at a rate of ft/sec. After ten seconds, what is the rate of change of the area? We need to nd da. Area A =, dierentiating with respect to r t gives da t = 0 seconds, the radius is r = 0() = 0 feet. Therefore, da dr = r. At time = (0)() = 80 ft /sec. 8. A 5 foot ladder is leaning against a house when its base starts to slide away. By the time the base is feet from the house, the base is moving at a rate of 7 ft/sec. (a) We are given = 7 ft/sec, but need to nd. Using the Pythagorean Theorem we can establish a relationship between the sides of the right triangle. + y = 5 Dierentiating both sides with respect to t and solving for gives: + y = 0 =) = y p When = then y = 5 = 9. Therefore, = (7) 9: ft/sec. 9 In other words, the top of the ladder is sliding down the wall at 9. ft/sec. (b) We are given d = 7 ft/sec, but need to nd. Using trigonometry we can establish a relationship between the base of the right triangle and the angle. sin = =) = arcsin 5 5 Dierentiating both sides with respect to t gives: d = q 5 5

4 When = then y = 9, therefore, q d = = 7 = 0:78 radians/sec. 9 In other words, the angle between the ladder and the wall is changing at a rate of 0.78 radians/sec. 9. Let f() = + a + b + c, where a, b, and c are some constants. Then f 0 () = + a + b and f 00 () = 6 + a. (a) Since the graph of f has an inection point at =, then f 00 ( ) = 0 6( ) + a = 0 =) a = 9 Therefore, a = 9, but there are no restrictions on the values of b and c. (b) From (a), an inection point at = implies a = 9 therefore, f 0 () = b. Since the graph of f has a local maimum at =, then f 0 ( ) = 0. ( ) + 8( ) + b = 0 =) b = Therefore, a = 9 and b =, but there is no restriction on the value of c. (c) Since a = 9 and b =, f() = +9 ++c, f 0 () = +8+, and f 00 () = 6+8. To nd critical points: f 0 () = = ( + )( + ) = 0 =) = ; Use the Second Derivative Test to nd local etrema: f 00 ( ) < 0, therefore the graph of f has a local maimum at =. f 00 ( ) > 0, therefore the graph of f has a local minimum at =. g() 0. (a) A function f() dominates g() as! if! f() = 0. (b) ln(00) =! =! = =! = = 0 Therefore, g() = = dominates f() = ln(00) as!.. Consider f() = = ( 5). Then f 0 () = (a) f 0 () is undened when = = 0 =) = 0. f 0 () = 0 when 5( ) = 0 =) =. 5( ) and f 00 0( + ) () =. = 9 = (b) We can use a sign chart to see that f 0 () < 0 for 0 < <, and f 0 () > 0 for < 0 and >. Therefore, f is increasing on ( ; 0) and (; ). The graph of f is decreasing on (0; ). (c) Inection points occur when the function changes concavity. f 00 () is undened when 9 = = 0 =) = 0. f 0 () = 0 when 0( + ) = 0 =) =. Therefore, f has possible inection points at = 0 and =. We can use a sign chart to see that f 00 () < 0 for <, and f 00 () > 0 for < < 0 and > 0. Therefore, the graph of f changes from concave down to concave up at =, and f does not change concavity at = 0. In other words, f has an inection point at =. (d) Using the information in part(b), the graph of f has a local maimum at = 0 and a local minimum at =

5 (e) The graph of f is provided below. (f) To nd the global etrema of f we need to evaluate f at the critical points and the endpoints of the interval [ ; ]. f( ) = 6 f() = p 9 f(0) = 0 f() = p Therefore, on the interval [ ; ], f has a global minimum of 6 at =, and f has a global maimum of 0 at = 0. (g) From above, we know that the maimum value of f on the interval [ ; ] is 0, while the minimum value of f is 6. Therefore, 6 f() 0 on [ ; ].. The US Postal Service will accept a bo for domestic shipment only if the sum of its length and girth (distance around) does not eceed 08 inches. What dimensions will give a bo with a square end the largest possible volume? We want to maimize the volume of the bo. Let represent the length of each side of the square end. Let y represent the length of the bo. We need to maimize V = y. The girth given by + y is ed at 08 inches. Therefore, y = 08, substituting this epression for y into the formula for volume gives V = (08 ) = 08. dv = 6 = ( 8) = 0 when = 0; 8. Given the contet of the problem, = 8 is the only appropriate critical value. At = 8, d V = 6 (8) < 0, therefore the volume of the bo is maimized when = 8 inches.. A silo (base not included) is to be constructed in the form of a cylinder with a hemisphere attached to the top. The cost of construction per square inch of surface area is twice as great for the hemisphere as it is for the cylindrical sidewall, for which the cost is $5 per square foot. Determine the dimensions to be used if the volume of the silo is ed at 5000 cubic feet and the cost of construction is to be kept to a minimum. Useful formulas: Volume: cylinder V = r h sphere V = r Surface Area: cylinder A = rh sphere A = r 5

6 We want to minimize cost C = 5(rh) + 50(r ). The volume of the silo is given by V = r h + is ed at 5000 cubic feet. r Therefore, 5000 = r h + =) r h = 5000 r = 5000 r r r. Substituting this epression for h into the formula for cost gives C = 50r Solve dv 5000 r r + 00r = r r = 0 to nd critical points. dv = r r r 7500 r = = 0 when r = r. The second derivative d V = > 0 for all values of r > 0. Therefore, cost is r r 7500 minimized when r = 8:. The dimensions of the silo are r 8: feet and h 6:8 feet.. (a) Mean Value Theorem: if f is continuous on [a; b] and dierentiable at (a; b), then there eists a number c, with a < c < b, such that f 0 (c) = f(b) f(a). b a (b) Below is a graph to providing a geometric interpretation of the Mean Value Theorem for the function described in c. The slope of the tangent line at = c is equal to the slope of the line through points A and B. (c) Since f is continuous on [0; ] and dierentiable on (0; ), f satises the conditions of the Mean Value Theorem. Therefore, there eists a number c in the open interval (0; ) such that f 0 (c) = f() f(0) = 5 0 = 5. In particular, since f 0 () = +, c + = 5 =) c = p : 6