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1 February egional 8 ) f x x x f x x f ' ' 9 6 ) ) ) 6x x 6 7 N x x x N ' x N ' 6 6 x x x x x x V r h V r r 6 y taking the derivative of the volume with respect to time, we can find the rate at which the radius is changing when the radius of the conical well is 6. r r dv dr dr dr dt dt dt dt 8 r 6 (positive due to rate of decreasing) ) I is false because if t represents the position of the boat, the rate of change of the boat s distance will never be. II is true because ' t and '' t for t. III is false because when t, the boat is feet from shore. ut it initially was foot away from the shore, so it moved feet. IV is true via the MVT for derivatives. Therefore, the answer is 6) There are quite a few different methods to solve this problem. The solution below approaches the overall scenario using shells. lso, remember that everything is with respect to the y-axis. y y dy y y dy Continued on the next page: Split this integral into two: y y dy y dy Take a look at the second integral. ewrite it as: y dy. Essentially, this is just the area of a semicircle with radius multiplied by. The value of this integral becomes. (trig substitution can also be used to solve if you d like!) Now, the first integral. ecognizing that this is an integrable function that is odd with respect to the point, leads to the conclusion that the value of the integral is. lternatively, if one did not

2 February egional 8 u y. The recognize that the function is odd, use the following substitution to help: integral then becomes: u du. ecause the limits of integration are both, the value of the integral is. Therefore, the volume obtained when the region bound by the graph of axis is when revolved around the x-axis. x y and the y- 7) Let the height of the hexagonal prism be H and let x be one of the side lengths for the hexagonal base. Since we are looking for the hexagonal prism with maximum volume, the bases of the prism are equilateral. Therefore, x is also the distance from the center of a hexagonal base to one of the vertices. 6 H radius x H x The volume of the prism inscribed in the sphere: 6 6 V prism x x x x dvprism x 7x x x x 6 x 6 x 6 x To find the side length that yields maximum volume: 7x x x 6 x 7 6 x x x x Plugging this value in to find volume: V prism ) The answer is C, Leibniz. It comes from the book The Early Mathematical Manuscripts of Leibniz, translated from the latin texts published by Carl Immanual Gerhardt with critical and historical notes. 9) f x cos xcosx sinxsinx cosx f ' x sin x f '' x cosx ) To maximize a triangle under a symmetrical, non-circular graph, it would need to be isosceles. This will occur at the axis of symmetry for the parabola, which occurs at the point,.

3 February egional 8 ) Let the total length of the pipe be L. While it is turning the corner, let L be the length that is still in the smaller hallway and L be the length that is in the wider hallway. Using this, create a formula to represent the length of the entire pipe: L L L csc 6sec L' csc cot 6sec tan L' : To find optimum θ, set csc cot 6sec tan sec tan sin sin tan tan 6 csc cot cos cos 6 Using a right triangle, it becomes known that csc csc 6sec L L L L 6 and sec ) f x decreases where f ' x. x x f x f ' x x x x ecause the numerator is negative at all x except the location of the vertical asymptotes, f x is decreasing on,,,. C ) It is given that x 8. Therefore, x Using some rearranging: x x x ) ecause f ' x is not defined where x, x x lim NE x x ) To find this limit, rewrite the function: x x x x x x x x x x x x x x x x x x x lim a b ab x x x E

4 February egional 8 9 x F F x x. For this integral, use u-sub: u x udu 6) 9 u u u du u u u. For this integral, use another substitution: vu v dv v dv v v v v v ln ln 7) M x can be represented by the determinant of the matrix. M x x 6x x M ' x x x M " x x The x-coordinate of the point of inflection of M x is where the second derivative changes signs, which is the case at x. 8) Linear pproximation allows us to utilize tangent lines to guess values for a function that would otherwise be difficult to compute by hand. For this approximation, the tangent line at x will be 6 used to approximate the value of sin : 7 y mx b f f x f ' x x f x 7 8 9) I is not necessarily true, consider f x x. II is not necessarily true, it depends on whether f x is differentiable. III is true due to properties of integrals IV is true because this is the definition of a maximum of a twice differentiable function. The answer is ) T av ) ecause we do not know the temperatures in-between the interval times given,,, and C all may or may not be true. However, is true due to the MVT for derivatives N the fact that there is a maximum at some time between the hours of :am and :pm.

5 February egional 8 ) This is the limit definition of the derivative for differentiation: x y x x ln y ln x ln ln x y ' x ln x y x x y x at x. So, using logarithmic and implicit x y' x x ln x y' ln ln x y x 8 x 6x x ) Take the derivative and set it equal to to give the optimum value for x and y: d 6x x 8x 8x x 6 and y xy 6 86 Finally, ) Find the velocities and positions of osita and Spencer using initial conditions determined by the problem. Then, set the two position functions equal to one another to find the moment in time when Spencer surpasses osita. a t a t 6 ) S v t t v t 6t p t t p t t t p t S 7 S p t S t t t 7 t t 7 t 7 C d x x x C (See the hint at the end of the test) E 6) Since the side length of the cube is meters, the radius of the inscribed sphere is meters. V r V 8 7) epresent point P as k, k equation of The x-intercept of g x in slope-intercept form: g x is. Using this, find the slope of g x and use the y-intercept to write the k, k g x k x 8 k d x-int dk k

6 February egional 8 egin to work with point P via the distance formula to relate P to k: 9 6 x y k k k k Since we now see how P relates to k, find the relationship between k and time using what we know about how P changes with time: d k dp dk k t k t dt dt dt d x-int d x-int dk dt dk dt k When k = : 6 C 8) First, make a substitution: sin cos x d x x sin cos sin cos d d sin d 9 9x x sin cos Now expand out sin and perform the integration after simplifying: cos sin sin cos cos d d d d = cos sin cos d sin 8 8 C 9) rc Length ' x x ln cos x ' x tan x tan x sec x sec x ln sec x tan x ln ln ln.

7 February egional 8 ) This scenario is easier visualized with the use of a picture. Let the height and radius of the cylinder be h and r, respectively. Let the radius of the sphere be. First, work with the sphere: VS 6 dvs d d d 7 dt dt dt dt elate the height and radius of the cylinder to the radius of the sphere: 9 h r h r Plug this relationship in to the volume for the cylinder: c 9 9 V r h r r r r 8 dvc r r r r 9 r dt 9r 9r To find the cylindrical radius that yields maximum volume: 8r r 8r r r 8 9 r 6 r Use this value of r to find h: dh dr h r h r dt dt earrange the relationship of h and r to to relate h to directly: h h h h r r h To find the rate at which the height of the inscribed cylinder is changing, differentiate, and use the value for d that was found early in the problem: dt d dh dh dh h dt dt dt dt ) ) ) ) ) 6) 7) 8) C 9) ) ) ) C ) ) E ) 6) 7) 8) 9) ) ) ) ) ) C ) E 6) 7) C 8) 9) )