M151B Practice Problems for Final Exam

Transcription

1 M5B Practice Problems for Final Eam Calculators will not be allowed on the eam. Unjustified answers will not receive credit. On the eam you will be given the following identities: n k = n(n + ) ; n k = n(n + )(n + ) ; 6 n ( n(n + ) ). k =. Compute each of the following its: a. +. b. c. d. esin( ). sin ( e ). [( + )/ / ]. e. The geometric mean of two positive real numbers a and b is defined as ab. Show that ab = ( a/ + b / ). a. Find a value for c that makes the given function continuous at all points. { sin f() =, c, =. b. Determine whether or not your function from (a) is differentiable at =. If it is differentiable at this point, compute its derivative there.. Find an equation for the line that is tangent to the graph of at the point =. f() = e +. Suppose the angle of elevation of the Sun is decreasing at a rate of.5 rad/hr. How fast is the shadow cast by a ft tall building increasing when the angle of elevation of the Sun is π 6?

2 5. Suppose f() is continuous on the interval [a, b] and differentiable on the interval (a, b). Show that if f () = for all (a, b), then there eists some value c (, ) so that 6. Let f(b) f(a) = c(b a). f() = / ( + ) /, < <. 6a. Locate the critical points of f and determine the intervals on which f is increasing and the intervals on which f is decreasing. 6b. Locate the possible inflection points for f and determine the intervals on which f is concave up and the intervals on which it is concave down. 6c. Evaluate f at the critical points and at the possible inflection points, and determine the boundary behavior of f by computing its as ±. 6d. Use your information from Parts a-c to sketch a graph of this function. 7. Find the side-lengths that maimize the area of an isosceles triangle with given perimeter P =. (An isosceles triangle is a triangle with two sidelengths equal.) 8. Find all fied points for the recursion equation a n+ = a n + a n. Sketch a graph of the function f(a) = a+, and use the method of cobwebbing to determine a whether or not one of these fied points will be achieved from the starting value a =. 9. Find all fied points for the recursion equation t+ = + t and determine whether or not each is asymptotically stable or unstable.. Suppose a function f() is continuous on the interval [, ] and that you are given the following table of values: f() /8 / /8 / 5/8 7/8 Table : Values of f() for Problem. Use an appropriate Riemann sum to approimate f()d.. Use the method of Riemann sums to evaluate + d.

3 . Evaluate the following indefinite integrals. a. e cos(e )d. b. + d b. cos d.. Evaluate the following definite integrals. a. + d. b. π sec d.. Find the area of the region bounded by the graphs of y = and y =. 5. Find the volume obtained when the region between the graphs of y = e and y = e, [, ], is rotated about the -ais. 6. Suppose the base of a certain solid is the region in the y-plane between the line y = and the parabola y =. Find the volume of the solid created if every cross section is a right isosceles triangle with hypotenuse in the y-plane perpendicular to the -ais. Solutions a. We have + = ( )( + 5) =, where we have observed that is negative for to the left of. b. We apply the Squeeze Theorem in this case, using the inequality We have and so according to the Squeeze Theorem e e sin( ) e. e = e =, esin( ) =.

4 c. We apply L Hospital s Rule twice, sin ( e ) = sin + cos ( e )( e ) = sin + cos e e = cos sin e e =. d. This it has the indeterminate form, so the first thing we do is rearrange it into an epression with the form. We have ( + )/ / = / [( + )/ ] ( + = )/ = / = ( + ) / =. / ( + ) / ( ) / e. We observe that this it has the general form, and so we can apply L Hospital s rule. We have (a/ + b / ) = e ln( a / +b / ) = e ln( a / +b / ) = e ln( a/ +b / ). In order to compute this it, we write ln(a/ + b / ) = ln( a/+b/ ) = ( a / +b / a/ (ln a)( ) + b/ (lnb)( )) = ln a + b / ln b) = (ln a + lnb), a / + b /(a/ where in this last step we have used that as. The it is e (ln a+ln b) = e ln(ab) = e ln(ab)/ = ab. a. Since sin =, we can make this function continuous at all points by choosing c =. b. Since the function is separately defined at =, we must proceed from the definition of differentiation. We compute sinh f f( + h) f() h () = = h h h h sin h h cosh sin h = = = h h h h h =,

5 where the last two steps both used L Hospital s rule. We conclude that this function is differentiable at =, and that f () =.. First, f () = ( + )(e + e ) e () ( + ) f () =, which is the slope of the tangent line. Using f() = and the general point-slope form y f(a) = f (a)( a), we conclude y =.. First, observe that what we know is dθ d =.5 rad/hr and what we want to know is, dt dt where is the length of the shadow (see the diagram). We see that the relation between θ and is tanθ =. Upon taking a derivative of this equation with respect to t, we obtain sec θ dθ dt = d dt, where we can now fi θ = π, so that 6 sec θ = cos π 6 Combining these observations, we have = =, while = tan π 6 =. d dt = dθ π dt sec 6 = ()(.5) = + ft/hr. 5. According to the Mean Value Theorem there eists some value c (a, b) so that f(b) f(a) b a = f (c). In this case f (c) = c, and so we conclude f(b) f(a) b a = c f(b) f(a) = c(b a). 5

6 6a. The derivative of f() is f () = + / ( + ) /, from which we find the critical points =,,. We see that f is increasing on (, ] [, ) and decreasing on [, ]. 6b. The second derivative of f() is f () = 5/ ( + ) /, from which we find that the possible inflection points are =,. We see that f is concave up on (, ) (, ) and concave down on (, ). 6c. Evaluating f at the critical points, possible inflection points, and at the endpoints, we have: f( ) = f( ) = / f() = / ( + ) / = / ( + ) / = +. 6d. Your plot should look something like this: 7. Let y denote the length of the sides of equal length, and let denote the length of the side between them. Then the perimeter is = y + y = 5. 6

7 By the Pythagorean Theorem, the height of such a triangle is h = y, and so the area to be maimized is A = y A() = (5 ) = 5 5, 5. (The upper it of 5 is clear both because a value of larger than this would put a negative number under the radical, and because the single side cannot be more than half the perimeter.) In order to maimize A(), we compute A () = The critical values are =, 5, where we observe that = 5 is also a boundary value. Checking A() at the critical and boundary values, we find A() = A( ) = 5 5 = 5 A(5) =. 5 We conclude that the maimum area is and the side-lengths are = and y = 5 () =. That is, an equilateral triangle. 8. The fied points solve a = a + a a = a a =. We conclude that the fied points are ±. In order to use cobwebbing, we must sketch a graph of the function f(a) = a + a. First, setting f (a) = a =, we find that the critical points are a = ±,. The function is increasing on (, ] [, ) and decreasing on [, ]. Net, f (a) = a, and so the only possible point of inflection is a =. The function is concave down on (, ) 7

8 and concave up on (, ). Finally, ( a a + a ) = f( ) = ( a + a ) = +( a + a ) = + f( ) = a ( a + a ) = The plot of this function and the cobwebbing are depicted below. We conclude a n =. n 9. In order to find the fied points, we solve which becomes (upon multiplication by ) = +, = ( )( + ) =, and the fied points are =,. In order to check for stability we set f() = +, and compute f () =. 8

9 We have f ( ) = is unstable f () = is stable.. Since no value for f() is given at either = or at =, we cannot take a Riemann sum with left or right endpoints. We see, however, that the values of are precisely the midpoints of the subintervals in the partition P = [,,,, ]. The most reasonable Riemann sum is f(c k ) k, where the c k are the interval midpoints. That is,. In this case = b a n n A n = Finally, f(c k ) k = ( + ) =. = n = n, and we use right endpoints k = + k. We have [( + k ) + ( + k ) ] n = [( + k n ) + ( + k n + k n )] n [ n = + n k + n + n k + n n n n n [ = + n(n + ) + + n(n + ) n n A n = + n = 6. a. Using the substitution u = e, for which du = e d, we find cosudu = sin u + C = sin(e ) + C. n ] k + n n(n + )(n + ) 6 ]. b. We make the substitution u = +, with du = d, and obtain u du = du = u / u / du u u = u/ / u/ / + C = ( + )/ ( + ) / + C = ( + ) / ( ) + C. 9

10 c. In this case, integrate by parts with u = cos and dv = d, for which we have du = d and v =. The integral becomes cos + d. For the remaining integral, we use fast substitution (since u has already been used) to obtain cos + C. a. We make the substitution u = + (or alternatively use fast substitution), so that du = d, and the integral becomes 8 du u = 8 u / du = u / 8 = / [ 8 ]. b. We integrate by parts, setting u = du = d dv = sec d v = tan. We obtain π sec d =tan π π = π + ln cos π tan d = π + ln( ).. First, we locate the points of intersection by solving = + =. In general, fourth order equations are difficult to solve algebraically, but this is really a second order equation in the variable, and it factors as ( )( + 5) =, so that the real roots are = ±. We observe that the upper graph is always y =, and also take advantage of symmetry to compute the area as A = ( ) d = [ 5 5 ] = [ 8 5 ] = We observe that the graph of y = e is always above the graph of y = e on [, ], and so according to the method of washers, V =π (e ) (e ) d = π e e d = π [e + e ] = π [e + e ],

11 where in this case we used fast substitution. 6. First, the base of the triangle etends from y = up to y =, so its length is. The angle opposite the base is a right angle, so if we drop a line perpendicular to the base we divide the triangle into two triangles. In this way, we see that the height of the triangle is. (Alternatively, observe that the halves of the triangle can be rearranged into a square with sidelength.) The area is The volume is A() = bh = ( ) = ( ). V = ( ) d = + d = [ ] =.