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MA - Calculus I for the Life Sciences Final Exam Solutions Spring -May- . Consider the funion defined implicitly near (2, 1) by x2 y 2 + 3xy = 10y. (a) [ points] Use implicit differentiation to find the derivative at (2, 1). i d d h 2 2 x y + 3xy = [10y] dx dx dy dy dy + 3y + 3x = 10 2xy 2 + 2yx2 dx dx dx dy dy dy 2yx2 + 3x − 10 = −2xy 2 − 3y dx dx dx 2xy 2 + 3y dy =− dx 2yx2 + 3x − 10 Substituting x = 2 and y = 1, we get dy 4+3 =− dx (2,1) 8 + 6 − 10 dy 7 =− dx 4 (2,1) (b) [ points] Determine the equation of the tangent line at (2, 1). 7 The equation of the tangent line is y = − (x − 2) + 1 . 4 . Let f (x) be the funion defined by f (x) = e √ x for x ≥ 0. (a) [ points] Write an equation for the line tangent to the graph of f (x) at the point where x = 1. √ d √ d √x 1 √ e =e x x = √ e x, dx dx 2 x e So the slope of the tangent line at x = 1 is f ′ (1) = . The equation for the 2 e tangent line is y = (x − 1) + e . 2 f ′ (x) = -May- Final Exam Solutions (b) [ points] Find the point where this tangent line crosses the x-axis. The line crosses the x-axis when y = 0. e (x − 1) + e = 0 2 e (x − 1) = −e 2 x − 1 = −2 x = −1 The tangent line crosses the x-axis at (−1, 0). . Let f be the funion defined by f (x) = x3 − x2 − 2x for 0 ≤ x ≤ 3. (a) [ points] Find where the graph of f crosses the x-axis. Show your work. Set f (x) = 0 and solve x3 − x2 − 2x = 0 x(x2 − x − 2) = 0 x(x + 1)(x − 2) = 0 The solutions are x = −1, x = 0, x = 2, but we are restried to [0, 3], so the only two places where f crosses the x-axis will be x = 0 and x = 2. (b) [ points] Find the intervals on which f is increasing. Find the derivative and the critical points. f ′ (x) = 3x2 − 2x − 2 Now solve 3x2 − 2x − 2 = 0 √ √ 2 ± 28 1 ± 7 x= = 6 3 √ 1+ 7 lies in our interval. We know that f (x) is decreasing if Only x = 3 √ 7 1 + f ′ (x) < 0 which happens when 0 < x < . Also, f (x) is increasing if 3 √ 1+ 7 f ′ (x) > 0 which happens when < x < 3. 3 (c) [ points] Find the absolute maximum and absolute minimum value of f . Justify your answer. f (x) is continuous on the closed interval [0, 3], so it attains its√maximum and 1+ 7 minimum. There are three points to check: x = 0, x = , and x = 3. 3 Spring MATH - Final Exam -May- Solutions From the First √ Derivative Test we can see that the absolute minimum occurs 1+ 7 at x = . Evaluating f at and , we have that f (0) = 0 and f (3) = 12, 3 so the absolute maximum occurs at x = 3. √ 20 + 14 7 The maximum value of f on [0, 3] is and the minimum value is − ≈ 27 −2.1126. . [ points] A projeile is fired toward the sea from a cliff feet above the water. If the launch angle is 45◦ and the muzzle velocity is k, then the height h of the projeile above the water when it is x feet downrange from the launch site is h(x) = −32 2 x + x + 200 feet. k2 See the figure. Determine the maximum values of h and the distance downrange where this maximum occurs if k = 2800 feet per second. h cliff x sea level 32 2 x + x + 200 k2 64 h′ (x) = − 2 x + 1 k h(x) = − Set equal to zero and solve − 64 x+1 = 0 k2 k2 x= 64 Since h′′ (x) < 0, this will be the absolute maximum. Thus, the downrange distance k2 is x = = 122, 500 feet. The height at that distance is h(122500) = 61, 450 feet 64 Spring MATH - Final Exam -May- Solutions . Find the requested quantities in each of the following. (a) [ points] Let 2 + cos(x − 1) if 0 ≤ x ≤ 1, f (x) = ax − 2 if 1 < x < 2, 2 x − 2x + b if 2 ≤ x ≤ 3. where a and b are constants. Find the values of a and b which make f (x) is continuous on [0, 3]. We need lim− f (x) = lim+ f (x). This gives the equation x→1 x→1 2 + cos(0) = a − 2 a=5 Likewise, we need lim− f (x) = lim+ f (x). This gives x→2 x→2 2a − 2 = 22 − 2(2) + b 8 = 4−4+b b=8 These values for a and b will make f (x) continuous. 6n2 − 3n + 7 . n→∞ (n + 3)2 (b) [ points] Evaluate lim 6n2 − 3n + 7 6n2 − 3n + 7 = lim n→∞ n→∞ n2 + 6n + 9 (n + 3)2 n2 n 7 6 2 −3 2 + 2 n n = lim n2 n→∞ n n 9 +6 2 + 2 n2 n n 6 = =6 1 lim . Find the following limits sin(x) . x→0 x cos(x) Use l’Hospital’s Rule since both the numerator and denominator go to . (a) [ points] lim cos(x) sin(x) = lim x→0 cos(x) − x sin(x) x→0 x cos(x) 1 = =1 1−0 lim Spring MATH - Final Exam -May- Solutions This can also be done as follows: sin(x) sin(x) 1 = lim x→0 x cos(x) x→0 x cos(x) = 1·1= 1 lim ln(xe x ) . x→∞ x Use l’Hospital’s Rule since both the numerator and denominator go to infinity. (b) [ points] lim 1 x (e + xe x ) x ln(xe x ) xe = lim lim x→∞ x→∞ x 1 e x + xe x = lim x→∞ xe x 1 = lim + 1 = 1 x→∞ x x2 − 4 (c) [ points] lim 2 . x→2 x − x + 6 l’Hospital’s Rule does not apply. 0 x2 − 4 = x→2 x 2 − x + 6 8 =0 lim . Find the following general antiderivatives. Z x2 + x + x−1 + x−2 dx (a) [ points] Z 1 1 1 x2 + x + x−1 + x−2 dx = x3 + x2 + ln |x| − + C. 3 2 x Z 2 sin(x) − 2 cos(x) + 2 sec2 (x) dx (b) [ points] Z 2 sin(x) − 2 cos(x) + 2 sec2 (x) dx = −2 cos(x) − 2 sin(x) + 2 tan(x) + C. (c) [ points] Z ! e x − e −x dx. 2 Z x −x ! Z e −e 1 1 dx = (e x − e −x )dx = (e x + e −x ) + C 2 2 2 Spring MATH - Final Exam -May- Solutions . Use the Fundamental Theorem of Calculus (part ) to evaluate the following derivatives. Zx 2 cos2 (u)e −u du then F ′ (x) equals (a) [ points] If F(x) = 1 By the Fundamental Theorem of Calculus, Part I, we know that Zx d 2 2 ′ cos2 (u)e −u du = cos2 (x)e −x . F (x) = dx 1 (b) [ points] If G(x) = Z 3x2 cos(2u) sin(2u) du then G ′ (x) equals 1 By the Fundamental Theorem of Calculus, Part I, and the Chain Rule we have Z 3x2 d ′ G (x) = cos(2u) sin(2u) du dx 1 d (3x2 ) = cos 2(3x2 ) sin 2(3x2 ) dx = 6x cos(6x2 ) sin(6x2 ) . Find the following definite integrals. Z 3 1 4x − dx (a) [ points] x 1 Z 3 1 4x − 3 1 dx = 2x2 − ln(x)1 x = (18 − ln 3) − (2 − ln1) = 16 − ln 3 ≈ 14.9014 (They may use the calculator to compute this integral, but in that case the answer is either right or wrong — no partial credit.) Spring MATH - Final Exam -May- Solutions (b) [ points] The funion h(x) is graphed below. 6 5 4 3 2 1 1 Compute Z 2 3 4 5 6 7 8 9 10 10 h(x) dx. 0 This is to be done geometrically. Z 10 h(x) dx = 0 Z 1 h(x) dx + 0 Z 2 h(x) dx + 1 Z 4 h(x) dx + 2 Z 5 h(x) dx + 4 Z 10 h(x) dx 5 1 1 1 = (0)(1) + (2 + 4)(1) + (4 + 2)(2) + (2)(1) + (1 + 5)(5) 2 2 2 = 26 . Find the requested areas. (a) [ points] Set up (but do NOT evaluate) the integral that will give the area 3π enclosed by the curves y = cos(x), y = 1+sin(2x) and the vertical line x = . 2 3 2 1 Π 2 Area = Z 3π 2 0 Π 3Π 2 [(1 + sin(x)) − cos(x)] dx 3π ≈ 6.71239. They do not have to evaluate the 2 integral, but IF they do, they must do it correly.) (The corre answer is 2 + Spring MATH - Final Exam -May- Solutions √ (b) [ points] The region R is enclosed by y = x/4 and y = x in the first quadrant. Find the area of R. 4 3 2 1 4 8 12 16 √ First, find the points of interseion by setting x = 4x . These are x = 0 and x = 16. The area is then Z 16 √ 32 x ≈ 10.6667. x − dx = AreaR = 4 3 0 (They may use the calculator to compute this integral, but they must show the integral that they are evaluating and not just the answer.) BONUS [ points each] Compute the derivatives of the following funions. You do not need to simplify. (a) f (x) = x2 + 1 . e 2x f ′ (x) = 2xe 2x − 2(x2 + 1)e 2x e 4x (b) f (x) = x5 · sin(2x). f ′ (x) = 5x4 sin(2x) + 2x5 cos(2x). (c) f (x) = cos(3x + 5x3 ). f ′ (x) = −(3 + 15x2 ) sin(3x + 5x3 ). (d) f (x) = e 2 . f ′ (x) = 0 (e) f (x) = (4x2 − 1)3 + (8x + 9)11 . f ′ (x) = 3(4x2 − 1)2 (8x) + 11(8x + 9)10 (8). Spring MATH -