= 1 3. r in. dr in. 6 dt = 1 2 A in. dt = 3 ds

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1 . B. Consider the octagon slit u into eight isosceles triangles with vertex angle o and base angles o /. We want to calculate the aothem using the tangent half-angle formula and a right triangle with base /=. a tan a cos a sin asn 8 A. C. Given: da =,A = s,at s = 6 da = s ds = (6) ds Û ds = r in = dr in s = ds 6 = A in = r in da in dr in = ( /) = = r in. D. Since we want the total area, we give all areas a ositive sign. For each eriod of the sine wave, the area is, so. eriods gives a total area of 6.. C. V = R H = (/) (8) = 6. C. S = R +Rl, where l is the lateral length, calculated by Pythagorean theorem: l = R +H =. S = 6 +6 = 6 6. C. The volume of a frustum is given by V = ( R H -r h), where R and H are the fixed lengths of the full cone, and r and h are the radius and height of the deteriorating cone. Through similar triangles, we have the relation r h = R H Û r = R h, which can be substituted into the volume H equation to give V = æ R H - R ç è H h ö since we know how h changes. Taking the time derivative ø and assuming R and H are constant gives dv R dh h R h dh. At seconds, or H H / minute, the height has droed /6 units, giving us a volume decrease of

2 6 units /min dv, and then taking the oosite answer since we asked for negative rate of change B. The two curves change at x=, so we need to make two integrations. Using the shell/cylinder method: V = t( cos t - sin ò t) + t sin t - cos t ò ( ) V = ò tcost - ò tcost é V = t sint + cost ù ú ë é - t sint + cost ù ú ë é V = 8 - ù ë ú - é - ù ë 8 ú = 8. B. We set u the integration and solve for B(a,b). ò B(a,b) x - (- x) - dx = ò x (- x) dx = B(a,b) Using integration by arts... x (- x) - ò - x (- x)dx = B(a,b) + é x - ù 6 x6 ú =B(a,b) ë B(a,b) = 6. A. We need to calculate the first derivative of the df, set equal to and solve for x. f X (x;,6) = 6x(- x) - 8x (- x) = x(- x) [6(- x)- 8x]= x = = 7

3 . B. Set u the integral and evaluate: m = ò 68x (- x) dx m = ò 68x (- x +x -x + x - x )dx m = ò 68x - 8x +68x -68x 6 + 8x 7-68x 8 dx m = m = =. C. The first three terms of sin x are P (x)= x - x! + x.the integral is then! 6 x x x dx x x x.!!! 6! 7 7. A. We need to do a chain rule rate of change on the integral: da da dx x x x x e x sinx x e x sinx dx x e 7sin. E. If we define one of the vertices of the rectangle as (x,y), then the erimeter of the rectangle is P = x + y. From the equation of the ellise, we know that differentiating wrt x x y dy dy b x gives. We want to maximize P, so dp = + dy. Substituting gives a b dx dx a y dx dx dp b x. Solving for x gives dx a y a x b y. We can substitute this back into the equation of b a ab the ellise to get that y = and also x =. The area is then A xy. a + b a + b a b. C. The area enclosed by a chord and a arabola is given by the formula A = bh, essentially / of the area formed by the arallelogram. The height is the focal length and the base is the length of the latus rectum. Converting the equation of the arabola to standard form gives (y -) =(x +). The focal length is /=, the height must be and wih, /(6)=.. A. Simson s rule alied to sec x on the given subinterval gives the formula

4 /-(- /) A = é ë sec(- )+ sec(- )+sec+ 8 sec( )+ 8 sec( )ù = é + - ( ) ++ ( - ) ë + ù ú = é ù ( ) ú ë 6. D. If the bases are x and the legs are x, then the height of the triangles must be given by h = ( x ) - æ è ç x ö ø = x. The integral becomes ( ) x V = æ ö ò x è ç ø dx = ò x dx 8 V = 8 7. C. First we need to determine the intersections of the two curves: x +6x - = x - 7 x +x - = (x + )(x - ) = x = {-,} ò x x -6x +dx = 6 8. B. With the rojection into the xy-lane, we know that a= and b=. Keeing the x- x z comonent, we know that the xz-rojection must be and that c>. Given the c eccentricity, we can calculate c to be e = c - c = \c = Þc =. The volume of an ellisoid is given by V = abc = ()()( )=6. D. Shoelace theorem.. B. The centroid is calculated by finding the intersection of the line segments connecting the centroids of the four triangles made by the vertices of the quadrilateral choosing three at a time for the vertices of those triangles. The oosite centroids for two triangles are (,-) and (/,/), while the other set of oosite centroids are (7/,7/) and (/,). Connecting

5 these two airs of oints and finding the intersection yields the oint (67/,77/), so y 77.. D. Paus centroid theorem allows us to calculate the centroid easily. We know that the volume of a revolved solid is V =ra, where r is the distance from the axis to the centroid. The area of the lamina is r and the volume of the shere is r. Equating and solving for r gives r. V = ò / - / / (secx +) -(tanx +) dx V = ò sec x +secx +- tan x -tanx -dx - /. A. / V = ò +secx -tanxdx by trigonometric identities - /. A. The right endoint Riemann sum aroximates the integral to ln lnln ln ln. The true value of the integral is lnxdx ò = xlnx - x = ln--ln+=ln -. The right Riemann sum will overestimate the e e ln ln ln ln integral, giving us an absolute difference of 6.. D. We need to calculate the area of the arallelograms formed by each combination of vectors, multilied by to account for the arallel faces. For each air, the magnitude is [,,] [-,,] = (-) + + =, [,,] [,,] = 8 +(-) + = 7, and [-,,] [,,] = + +(-8) =. The sum of each times two is the answer.. C. To determine the bounds for the inner loo, we set the equation r= and solve for angles: = + cosq Û - = cosq Þq = ì, ü í ý. Setting u the integral on olar area: î þ / / A = ò (+ cosq) dq = / ò +8cosq +8cos q dq = -6. / Dx i = n = b-a i Þ b - a = + n n = a+idx Þ a =,b = i æ + i ö 6. D. è ç n ø = x \ ( x +)dx = x + xù = - = 8 ò

6 If V = 7. C. R, then dv = dr R.Likewise, S = R Þ ds dr = 8R. Setting u the ratio dv / R ds dv R gives 8 R ds /. 8. A. The formula for the annular sector is given by A R r R r da d dr d d dr d R ( r) r R r r dr / d t t r t da / t / () t t t da t t 7 t t / 6. sec 7 V = ò. C. x dx = - = x. B. A acos t d( asin t) asin t d( acos t) A a cos t sin t a sin t cos t a A sin t cos t sin t cos t a cost cost A a a cost A cos t 8 8 A a t sint 8 a 8