Math, Exam III November 6, 7 The Honor Code is in effect for this examination. All work is to be your own. No calculators. The exam lasts for hour and min. Be sure that your name is on every page in case pages become detached. Be sure that you have all pages of the test. PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!. (a) (b) (c) (d) (e). (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 4. (a) (b) (c) (d) (e). (a) (b) (c) (d) (e) 6. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e) 8. (a) (b) (c) (d) (e) 9. (a) (b) (c) (d) (e). (a) (b) (c) (d) (e) Please do NOT write in this box. Multiple Choice.. 3. 4. Total
.(6 pts.) Compute lim x Name: Multiple Choice 9x + x +. x (a) 3/ (b) (c) 3/ (d) / (e) / Solution: Dividing top and bottom by x, we obtain 9x + x + 9 + 9x + x + lim = lim x x + x x x x = lim x = 9 = 3 x x Here we must use the minus sign out front because as x, we have that x is negative, so we must choose the negative value when we take the square root of the inside..(6 pts.) Find all asymptotes of the curve y = 4x + x. (a) (b) (c) (d) horizontal asymptotes y = 4, slant asymptote y = 4x, no vertical asymptotes. slant asymptote y = 4x +, vertical asymptote x =, no horizontal asymptotes. horizontal asymptotes y = 4, vertical asymptote x =, no slant asymptotes. slant asymptote y = 4x + 8, vertical asymptote x =, no horizontal asymptotes. (e) vertical asymptote x =, no other asymptotes. Solution: Since the degree of the numerator is one greater than the degree of the denominator, this will have no horizontal asymptote and a slant asymptote. There will be a vertical asymptote at x =. If we do the long division, we get 4x + 8 7 x, so the slant asymptote is at y = 4x + 8.
3.(6 pts.) Mr. McDonald (the one who had a farm) wants to use ft of fence to build a rectangular corral along the side of his goat barn, as shown in the figure. Of course, being frugal, he wants to enclose as much area as he can with his ft of fence. What is the maximal area he can enclose in the corral? (a) ft (b) ft (c) ft (d) 6 ft (e) none of the above. Solution: If McDonald has ft of fence, then we have that w + h =, where w is the width and h is the height. (Notice that the goat barn takes up one of the sides of the rectangle, so we only need one contribution from h.) Therefore, we have h = w. Thus the area of the enclosure is A(w) = w( w) = w w. Then to maximize this, we take the derivative and set it equal to zero. The derivative is 4w, which is zero when w =. Since the derivative is positive if w < and negative if w >, the first derivative test tells us that this is the maximum. Therefore A() = ( ) =. 4.(6 pts.) Solving the equation x 3 + x = using Newton s method with initial guess x =, what is x? (a) 4 (b) 6 (c) 4 (d) 6 (e) Solution: Recall that Newton s method is given by x n+ = x n f(x n) f (x n ). Since x =, we have x = f() f () = f (). Now f (x) = 3x + x, so f () =. Therefore x = = 4. 3
+ x.(6 pts.) Calculate the following indefinite integral dx. x3/ (a) x / + x / 4 + C (b) (x / x / ) + C (c) x / + 3 x 3/ + C (d) x x / + C (e) x / + x 3/ 3 + C Solution: + x dx = x3/ dx + x3/ dx = x/ x 3/ dx + x / dx = x / + x/ + C 6.(6 pts.) Find the left endpoint approximation to the definite integral 3 3x dx using four approximating rectangles of equal base width. (a) 38 (b) 4 (c) 6 (d) 4 (e) Solution: Since we are using four approximating rectangles with equal width onm the interval (, 3), each rectangle will have width. Since we are using a left endpoint approximation, the value will be (f(x ) + f(x ) + f(x 3 ) + f(x 4 )) = (f( ) + f() + f() + f()) = ( + ( ) + + ) = + + = 4 4
7.(6 pts.) The graph of g(x) shown below consists of two straight lines and a semicircle. Use it to calculate the integral g(x)dx. (a) + π (b) π (c) 3 4π (d) 3 π (e) 3 + 4π Solution: We can calculate this integral using g(x)dx = g(x)dx + g(x)dx + g(x)dx. Since the value of the integral is exactly the signed area between the curve and the x-axis, we can use facts of geometry to compute the area. g(x)dx = by the formula for the area of a rectangle. g(x)dx = by the formula for the area of a triangle. Since the area of a circle with radius is 4π but we only have half of a circle, g(x)dx = π. Notice we have a negative sign since the area is below the x-axis. Therefore = + π = π. 8.(6 pts.) Let f(x) = x + sin(t ) dt. Find f (x). (a) x cos(x ) + sin(x 4 ) (b) x + sin(x ) (c) x + sin(x 4 ) (d) + sin(x4 ) (e) x cos(x ) + sin(x4 )
Solution: By the fundamental theorem of calculus, f (x) is the derivative of x + sin(t )dt, which by the chain rule is f (x) = + sin(u ) du dx, where u = x. Therefore du dx = x and u = x 4. Thus f (x) = x + sin(x 4 ). 6
9.(6 pts.) Find (a) π/4 sin(θ) + cos(θ) dθ. 3 (b) + (c) + (d) (e) Solution: We know that sin(θ)dθ = cos(θ) + C. By letting u = θ, we have that cos(θ)dθ = cos(u) du = sin(u) + C = sin(θ) + C. Therefore π/4 sin(θ) + cos(θ)dθ = ( cos(θ) + sin(θ)) π/4 = ( cos(π/4) + sin(π/)) ( cos() + sin()) = + + = 3.(6 pts.) Find x + x + x + dx (a) (b) (c) (d) (e) Solution: Notice that (x+) = x +x+, so x + x + = (x+) on the interval x + x + x + dx = (, ). Therefore x + x + dx = dx = (x) = =. 7
Partial Credit You must show your work on the partial credit problems to receive credit!.( pts.) If cm of material is available to make a box with a square base and an open top, find the largest possible volume of the box. Show all work, and make sure you justify that your answer is a maximum. Solution: Since the box has an open top and a square base, then the formula for surface area is the sum of the area of the bottom plus the sum of the area of the sides. If b represents the side length of the base and h represents the height, then we have SA = b + 4bh. Since the maximum volume must occur when the surface area is maximized, we can assume we use all cm material, so = b + 4bh. Therefore, Thus the volume is V = b b h = b 4b derivative and set it equal to zero. The derivative is = b 4b = h. b b3. To maximize the volume, take the 4 3b. Therefore we solve for b: 4 3b =, so 3b =. Therefore b = 4, so b = ±. Since a side length cannot 4 be negative, this means b =. The first derivative test ensures this is a maximum since for b < the derivative is positive and for b > the derivative is negative. Therefore the maximum possible volume is V = b b b3 h = = 4 8 = 4 cm 3. 4 4 8
.(3 pts.) Evaluate the definite integral shown below using right endpoint approximations and the limit definition of the definite integral Hint: + + 3 + + n = n i= i = x dx n(n + )(n + ) 6 Solution: Firstly, devide the closed interval [, ] into n equal pieces, so the length of each piece is n, and the end points of the pieces are, n,, n,. Then, with n respect to this partition, the approximate value of the integral is n i= n f(i n ) = n n n (i 8i n ) = n = 8 n i = 8 n(n + )(n + ) 4(n + )(n + ) =. To get 3 n 3 n3 6 3n i= i= i= the exact integral value, we need to take the limit with respect to n. Thus x dx = 4(n + )(n + ) 4( + lim n = lim n )( + n ) = 4 = 8 3n n 3 3 3. (b) Verify your result using the fundamental theorem of calculus. Solution: One of the antiderivatives of the function x can be taken as x3 (I say 3 one of the antiderivatives, because there are many other antiderivatives, which differ by a constant between each other). According to the fundamental theorem of calculus, x dx = x3 3 = 3 3 3 3 = 8. We get the same result as in part (a). 3 9
3.(3 pts.) The velocity of a particle (in meters per second) is given by v(t) = t 4t + 3. (a) What is the displacement of the particle over the interval [, ]? Solution: Denote the displacement function by s(t), so s (t) = v(t), i.e. s(t) is an antiderivative of v(t). By the fundamental theorem of calculus, s() s() = v(t) dt = t 3 3 t + 3t = ( 3 3 + 3 ) ( 3 3 + 3 ) = 3. (b) What is the distance traveled by the particle over the interval [, 3]? (Hint: Not the same answer as in (a).) Solution: We know that if v(t) >, then the particle is moving forward, and if v(t) <, then the particle is moving backwards. The solutions for the equation v(t) = are,3, and both of them are in the interval [, 3]. From the properties of quadratic functions, we know that v(t) > on (, ) and v(t) < on (, 3). So the particle moves forward from to, and moves backward from to 3. On [, ], the travel distance should be s() s(), and on [, 3], the travel distance should be s(3) s(). Again by the fundamental theorem of calculus, s() s() = v(t) dt = t3 3 t + 3t ( 3 3 +3 ) ( 3 3 +3 ) = 4 3, and s(3) s() = 3 v(t) dt = t3 3 t + 3t ( 33 3 3 + 3 3) ( 3 3 + 3 ) = 4 3. Thus, the total distance is s() s() + s(3) s() = 4 3 + 4 3 = 8 3. 3 = =
4.( pts.) You will earn points if your instructor can read your name easily on the front page of the exam and you mark the answer boxes with an X (as opposed to a circle or any other mark). You may use the space below for rough work.
Math, Exam III November 6, 7 ANSWERS The Honor Code is in effect for this examination. All work is to be your own. No calculators. The exam lasts for hour and min. Be sure that your name is on every page in case pages become detached. Be sure that you have all pages of the test. PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!. ( ) (b) (c) (d) (e). (a) (b) (c) ( ) (e) 3. (a) ( ) (c) (d) (e) 4. ( ) (b) (c) (d) (e). (a) ( ) (c) (d) (e) 6. (a) (b) (c) ( ) (e) 7. (a) ( ) (c) (d) (e) 8. (a) (b) ( ) (d) (e) 9. ( ) (b) (c) (d) (e). (a) (b) (c) (d) ( ) Please do NOT write in this box. Multiple Choice.. 3. 4. Total