Math 170 Calculus I Final Exam Review Solutions

Transcription

1 Math 70 Calculus I Final Eam Review Solutions. Find the following its: (a (b (c (d 3 = + = = = sin( (e 0 cos( = (f 0 ln(sin( ln(tan( = ln( (g (h 0 + cot( ln( = sin(π/ = π. Find any values of (if they eist where the function f( is not continuous: (a f( = 4 +, = ± (b f( = + 3, cont everywhere (c f( = + 5, = 0, (d f( = e ln(, < 0 3. Find the average value of the function f( = + on the interval [, 3]. (Don t worry about this one for the final eam. The average value of f( over the interval [a, b] is So for this function, f ave = f ave = b a b a f(d d = [ tan ( ] 3 = tan 3 tan (π/3 (π/4 = = π ( 3 4. Use the it definition of the derivative to find the slope of the tangent line to the graph of f( = + at a general value. Then, use it to find the slope of the tangent line to the graph of f at = 4. Finally, find a value of where the function is perpendicular to the line y = 3 4. f f( + h f( ( + h + h + h ( = = = f (4 = (4 = 8 = h 0 + h = The tangent line to the function (and hence the function itself is perpendicular to the given line when its slope is 3. We have 3 = f ( = when = 3/. 5. Use the it definition of the derivative to find f ( when f( = 3 f f( + h f( + h 3 3 ( = = + h h ( + h 3 ( 3 = = + h ( + h h = ( + h = = = h 0 + h

2 6. Find f ( for each of the following: (a f( = sin 3 ( + 4 cos( = 3 sin ( cos( 4 sin( (b f( = 5 ( + 3 = 5 ( (+3 5 ( ( ( (c f( = = (d f( = = + 7 π π (e f( = tan 3 ( 4 = 3 sec 4 ( 4 (f f( = = 6 (5 + /3 / (g f( = e = e + e (h f( = + csc( + = csc( + cot( + ( ( ( ( (i f( = cos 3 = 3 cos sin + (+ + + (j f( = 6 + 3e = 8e (3e + (k f( = cos = 4 (l f( = 3 ln( + = (ln( + / Find y when 6 (a y = y, (b 3 + y + y 3 =, y = + y = y 3 + 3y (c 3y 6 + 3y 3 = 9, y = 6 3y 9y + 6y (d y + =, y = y 8. Find the derivative of y = 3 using logarithmic differentiation. + 3 ( 3 y = (

3 9. Two parallel sides of a rectangle are being lengthened at a rate of in/sec, while the other two sides are shortened in such a way that the figure remains a rectangle with area 50 in. What is the rate of change of the perimeter when the length of an increasing side is 5 in? Is the perimeter increasing or decreasing? Let l be the length of a side of the rectangle that s increasing, and let w be the other side. We have P = l+w where P is the perimeter of the rectangle, and we know that lw = 50 since the area is fied, and so w = 50/l. We can rewrite the perimeter equation as P = l + (50/l = l + 00l We re given dl/dt =. Taking derivatives (with respect to t of the perimeter equation: dp dt = dl dl 00l dt dt When l = 5, we have dp dt = ( 00(5 ( = 4 in/sec The total perimeter is decreasing since the rate is negative. 0. For each of the following functions, find (a the intervals on which the function is increasing or decreasing and (b the intervals on which the function is concave up or concave down. (Hint: Your answers may not be nice - do not epect integer values for the endpoints of your interval in either problem (a f( = ( + Decreasing on (, 3 5, ( 3+ 5, Increasing on ( 3 5, 3+ 5 Concave down on (, 0, ( 4 6, 4+ 6 Concave up on (0, 4 6, ( 4+ 6, (b f( = 3 ln( Decreasing on (, 3 e Increasing on ( 3 e, Concave down on (, 6 e 5 Concave up on ( 6 e 5,

4 . Sketch a graph of the following functions by calculating critical values, inflection points, intercepts, intervals of increasing and decreasing, intervals of concavity, asymptotes, etc. (a f( = 3 4 (c f( = + 4 y f( = + 4 f( = 3 4 (b f( = 3 9 = + 3 y (d f( = ( = 3 y f( = 3 9 f( = (

5 . A rectangular area of 300 square feet is to be fenced off. Two opposite sides will use fencing costing $ per foot, and the remaining sides will use fencing costing $ per foot. Find the dimensions of the rectangle with the lowest possible cost. If y is the length of a side that costs $ per foot, and is the length of a side costing $ per foot, then the dimensions with the lowest possible cost would be = 800 ft and y = 400 ft. The restriction equation would be 300 = y and the cost equation (that we need to find a minimum of would be C = + 4y. Replacing one of the variables using the restriction (either works: ( 300 C = + 4 for which which gives us critical values of C = 800 = 0, = ±80 Only one of these values makes sense, in terms a length: = 80. This means the dimensions that yield the lowest cost would be = 80 by y = 300/80 = 40 feet. 3. A closed bo with a square base is to be constructed using 0 m of cardboard. Assuming all material is used, determine the maimum volume the bo can have. Let be the length of one side of the square base, and let y be the height of the bo. The total surface area of the bo (four sides with area y, top and bottom each with area must be 0, so 0 = 4y + We need to find a maimum for the volume, V = y, which can be rewritten in terms of as and has derivative V = ( 0 4 = V = 5 3 = 5 3 The critical values are = ± 5/3 and again, only the positive value makes sense here. So the dimensions of the largest bo would be = 5/3, and so the largest volume of the bo would be y = 5 = 5/3 V = ( 5/3 3 So, the largest possible bo with a square base is actually a cube (this will be true for any such bo, regardless of surface area. 4. Verify that the hypotheses of Rolle s Theorem are satisfied on the interval [, 3] for the function f( = ln(4 +, and find all values of c in that interval that satisfy the conclusion of the theorem. f is continuous on the interval [, 3], and its derivative f ( = is defined on (, 3. Also, 4+ f( = f(3 = 0, so all hypothesis are met. 0 = f (c = c 4 + c c c =

6 5. Find the integrals: (a ( d = C (b d = ln( (c e d = ] 4 eu du = (e e (d + d = π (e tan(θ dθ = ln cos(θ + C (f ] / d = ln( = ln(/ / (g ( + 3 d = 48 = 3 ln(4 (h 0 3 d = 3/ 0 3 d + 3/ 3 d = 5/ (i ( + 3 d = C 6. Find F ( when F ( = sin( cos( 3 dt There is a typo in this problem! The integrand should have been entirely in terms of t: in which case F ( = F ( = sin(t 3 3 7t 6 cos(t 3 dt sin( cos( 3