( ) 7 ( 5x 5 + 3) 9 b) y = x x

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1 New York City College of Technology, CUNY Mathematics Department Fall 0 MAT 75 Final Eam Review Problems Revised by Professor Kostadinov, Fall 0, Fall 0, Fall 00. Evaluate the following its, if they eist: a) e 0 e b) sin(7) c) cos() + π sin() d) 0 e 7sin() e) 0 sin(). Find the horizontal asymptotes of the function y(): a) y = b) y = c) y = Find the derivatives of the following functions using the definition of derivative: a) f () = 5 b) f () = +. Find the derivatives of the following functions: a) f () = sec(5) b) f () = tan() c) f () = 5 5. Find the derivative a) y = ( +) 5 d) f () = sin(7)cos(5) e) f () = 6 dy y' = of the following functions, using logarithmic differentiation: d ( ) 7 ( ) 9 b) y = 6. Find the equation of the tangent line, in slope-intercept form, to the curve: a) f () = at (,9) b) f () = at (,) 7. Using implicit differentiation, find the equation of the tangent line to the given curve at the given point: a) y y 7 = 5 + at (, ) b) y 7y + = 9 at (0,) 8. Approimate 7. by using linearization. Use a calculator to check your answer only. 9. Wire of length 0m is divided into two pieces and the pieces are bent into a square and a circle. How should this be done in order to minimize the sum of their areas? Round your answer to the nearest hundredth. 0. Find the dimensions of a closed bo having a square base with surface area and maimal volume. Round your answer to the nearest hundredth.. If a snowball melts so its surface area decreases at a rate of cm /min, find the rate at which the diameter decreases when the diameter is 6 cm.. The radius of a sphere increases at a rate of in/sec. How fast is the volume increasing when the diameter is in?. The radius of a cone is increasing at a rate of in/sec, and the height of the cone is times the radius. Find the rate of change for the volume of that cone when the radius is 7 inches.

2 . Sand pours from a chute and forms a conical pile whose height is always equal to its base diameter. The height of the pile increases at a rate of 5 feet/hour. Find the rate of change of the volume of the sand in the conical pile, when the height of the pile is feet. 5. A cylindrical tank with radius 8 m is being filled with water at a rate of m /min. What is the rate of change of the water height in this tank? 6. A bo with a square base and an open top must have a volume of 56 cubic inches. Find the dimensions of the bo that will minimize the amount of material used (the surface area). 7. A farmer wishes to enclose a rectangular plot using 00 m of fencing material. One side of the land borders does not need fencing. What is the largest area that can be enclosed? 8. Sketch the graph of the function f () =. Determine the intervals of increase, decrease and the intervals of concavity. Calculate and label all local (relative) maima, minima and points of inflection. 9. Sketch the graphs of the following functions. Check for symmetry (even, odd), find and label any intercepts and horizontal and vertical asymptotes: a) f () = b) f () = c) f () = 0. Evaluate each the following definite integrals: a) (e + )d b) (e )d. Evaluate each the following indefinite integrals: a) 5 7 d b) 6 + d Answers to questions:. Check that all its are indefinite, of the form 0, thus one is allowed to apply l Hospital s rule. 0 a) 5 b) 7 c) 0 d) 7 e). The horizontal asymptotes one obtains by taking the its of the function as goes to +/- infinity: a) y = ± ± = ± = 8 =, so the only horizontal asymptote is y =. b) The only horizontal asymptote is y =. c) The only horizontal asymptote is y = 0.

3 New York City College of Technology, CUNY Mathematics Department Fall 0. a) f '() = b) f ( + h) f () h f ( + h) f () h ( + h) 5( + h) ( 5) h+h 5h = = = ( +h - 5) = - 5 h h ( + h) + ( + h) ( + ) = h. a) sec(5) +5 sec(5)tan(5) b) tan() 6 d) 7cos(7)cos(5) 5sin(7)sin(5) e) ( 6) h h + h = h + ( + tan ()) c) = ( h + ) = + ( ) 5. a) First take the logarithm of both sides: ln y = 5ln( +) + 7ln( ) + 9ln(5 5 + ), then take the derivative of both sides y' y = y' = y b) Follow the steps in part a) to get y' = (+ ln ) 6. a) The equation of the tangent line at = is given by y = 5. b) The equation of the tangent line at = is given by y = +. and after multiplying both sides by y, we get: 7. a) The derivative y' = 5 6y one can compute by implicit differentiation. The slope of the tangent line 6 y at the given point is y' evaluated at (,-), which gives y'() = 7 /. The equation of the tangent line is given by: y = + 7 ( ) = 7 6 b) y' = 7y +, y'(0) =, y = + ( 0) = + y First, linearize the function f () = at = 7 = by computing the tangent line to f () at (7, f (7)) : y = f (7) + f '(7)( 7), where f (7) = 7 = and f '(7) = 7, since f '() = d ( ) =. Thus, the / d equation of the tangent line is: y = + ( 7). Finally, we approimate the cube root of 7. using the 7 tangent line: 7. y(7.) = + 7 (7. 7) = + =.0. The eact answer, using a calculator, 90 is 7. = Let the first piece have a length, then the second one has a length 0. From the first piece, we can form a square with a side of length and an area. From the second piece, we can form a circle

4 with circumference 0 = πr thus having a radius r = 0 π and an area πr = π 0 π. We want to minimize the sum of the two areas f () : f () = + π 0 π. This is done by setting the ( + π ) 80 derivative to zero f '() = = 0 and solving for the critical point 0 = 80. at which the 8π + π (absolute) minimum of f () is attained (why?). This tells us that we should use about. m from the wire to make the square, in order to minimize the sum of the two areas. 0. Let the length of one side of the base square be and the height of the bo be y. The total surface area of the bo is y + = and the volume is V = y. Epress y from the first equation y = 6 and plug it into the volume: V = (6 ). Set the derivative to zero and find the critical point(s): V '() = = 0 = ± and take = since length is positive. V() attains a (local) maimum at =, which is also an absolute maimum for > 0 (why?). Therefore, the desired dimensions are: = and y = 6 =, which means that the cube maimizes the volume of the bo.. The rate of change of the diameter is dd dt = cm/min. π. The rate of change of the volume is dv dt = 78π in / sec.. dv dt = π 85. in /sec. We use V = πr h = πr since h = r.. dv dt = 0π 6.8 ft /hour. We use V = πh, since r = h. 5. V = πr h = 6πh dv dt dh = 6π dt dh dt = dv 6π dt = 6π = π m/min 6. The base is 8 and height is, thus the dimensions are: The area is maimized when one side of the rectangle is 50 m and the other is 00 m, which gives an area of 5000 m. Note: For problems #8 and #9 one may use a graphing calculator to plot the functions and confirm the analysis. Please note that students are allowed to use graphing calculators on the final eam.

5 New York City College of Technology, CUNY Mathematics Department Fall 0 The graph of y = The graph of y = The graph of y = 8. The critical points are = 0, (where y' = ( ) = 0 ). The derivative y' > 0 to the left of 0 and to the right of, thus the function is increasing there; and y' < 0 between 0 and, thus the function is decreasing there. The first derivative test tells us that the function has a local (relative) maimum at the point (0, ) (as the function is increasing to the left of 0 and decreasing to the right of 0), as well as a local (relative) minimum at the point (, 5) (as the function is decreasing to the left of and increasing to the right of ). The second derivative y'' = 6 6 becomes zero at = ; it is negative to the left of, thus the function is concave downward there, and positive to the right of, so the function is concave upward there. Since the second derivative changes sign at =, the point on the curve (, ) is the only inflection point. 9a) This is an even function since f ( ) = f (), thus the function is symmetric relative to the y-ais. The vertical asymptotes are = ± (where the denominator becomes zero) and the single horizontal asymptote y = 0 is obtained by taking the its ± = 0. Since y' =, the only critical point is = 0. When < 0, ( ) y' > 0 and when > 0, y' < 0, thus the function is increasing to the left of zero and decreasing to the right of zero, which means that y(0) = 0.5 is a local maimum (by the First Derivative Test). Sketching the function requires our knowledge of where the function is increasing or decreasing. Alternatively, the one-sided its of the function on both sides of the vertical asymptotes suffice: = +, =, =, + + = +. 9b) The y-intercept is 0, as y(0) = 0. This is an odd function since f ( ) = f (), thus the function is symmetric relative to the origin. The vertical asymptotes are = ± (where the denominator becomes zero) and the single horizontal asymptote y = 0 is obtained by taking the its ± = 0. Since y' = < 0 there are no critical ( ) points, thus no local maima and minima and the function is decreasing everywhere. Sketching the function requires our knowledge of where the function is increasing or decreasing. Alternatively, the one-sided its of the function on both sides of the vertical asymptotes suffice: + = +, =, =, + = +.

6 The graph of y = 9c) The y-intercept is 0, since y(0) = 0. This is an even function since f ( ) = f (), thus the function is symmetric relative to the y-ais. The vertical asymptotes are = ± (where the denominator becomes zero) and the single horizontal asymptote y = is obtained by taking the its ± =. Complete the solution as in part (a). First, compute the derivative and find the critical points, if any. Then, investigate the sign of the derivative around the critical points to find out where the function is increasing or decreasing. Using the first derivative test will reveal the nature of the critical points, i.e. whether local maima or minima. Finally, the knowledge of where the function is increasing or decreasing or alternatively, the iting behavior of the function on both sides of the vertical asymptotes will suffice to sketch the graph of the function. 0. a) (e + )d = e d + d = (e e ) + ( ()) = (e e ) +.05 b) (e )d = e d d = (e e ) ( ) = (e e) a) 5 7 d = (5 5 ) d = 5 5 d d = C b) 6 + d = d + d = + + ( +) + C = + C = + C