Math 4 Final Sample A Solutions Tyrone Crisp (B) Direct substitution gives, so the limit is infinite. When is close to, but greater than,, the numerator is negative while the denominator is positive. So the limit is negative infinity. (E) Calculate the one-sided limits separately. As +, we have =, so the function is equal to and therefore has limit. As, we have =, so the function is equal to and therefore has limit. The one-sided limits are different, so the limit doesn t eist. (C) Direct substitution! 4 (D) Looking only at leading terms gives lim 6 9 = lim 6 = 6. 5 (A) Use the quotient rule and the identity cos + sin = : ( ) d sin ( cos )(cos ) sin (sin ) = = cos d cos ( cos ) (cos ) = cos. 6 (A) Write f() = ( cos(() / ) ) /, and use the chain rule (four times!): f () = ( cos(() / ) ) / ( sin(() / )) ( () / ) = sin cos. 7 (E) We need to know the slope of the tangent, i.e. y. Use implicit differentiation: take derivatives of both sides with respect to, using the product rule on 4y: yy +4(y +y)+ =. Now set = and y = and solve for y. You find y + 8y + 8 = so y = 4. We could 5 now use the point-slope formula, or just notice that answer E is the only one with the correct slope. 8 (A) Let r be the radius, A the area. Given: dr relating A and r: A = πr. Differentiate both sides with respect to t: da knowns and solve for the unknown: da π m /s. = π = /. Find: da when r =. Equation dr = πr. Substitute the = π. So the area is increasing at a rate of 9 (E) Possible local etrema occur at critical numbers. To find the critical numbers, first take the derivative: f () = /, and then find those in the domain of f for which f () is zero or doesn t eist. Now, the domain of f contains all real numbers, so no need to worry about that. We have f () = when / =, i.e. = 8. On the other hand, f () doesn t eist when = (because of the negative power of ). So the critical numbers are and 8. Now, apply the first-derivative test to each critical number. (You could also use the secondderivative test at 8, but not at because the second derivative is not defined at ). When is a little bigger than, the quantity / is a huge positive number, so f () >. When is a little smaller than, the quantity / is a huge negative, so f () <. So, by the first-derivative test, f has a local minimum at. Similarly, we find that f has a local maimum at 8. (A) Vertical asymptotes: First, we must CANCEL COMMON FACTORS. So cancel the ( 6) from top and bottom. There is now only one factor in the denominator, and hence just one vertical asymptote (namely = ).
Horizontal asymptotes: since the degree of the numerator is greater than that of the denominator, this function will have infinite limits at ±, and hence no horizontal asymptote. Slant asymptote: the degree of the numerator is eactly one greater than that of the denominator, so the graph will have a slant asymptote. (C) Antidifferentiate a(t), to get v(t) = sin t + C. To find C, use the fact that when t = π/, v =. So = sin(π/) + C, giving C =. So v(t) = sin t +. Now antidifferentiate v(t), to get s(t) = cos(t) + t + D. To find D, use the fact that when t = π/, s = π/. So π = cos π + π + D, giving D =. So s(t) = cos t + t +. (A)Let u = 4, so du = d. Inserting appropriate factors of and, the integral becomes ( ) 4 d = u / du. To get rid of the, notice that = 4 u by our choice of u. So the integral is equal to (4 u)u / du = 4u / u 4/ du = u 4/ 4u / du. (E) Let u = + /, so du = d. Since this is a definite integral, we must also translate the limits of integration. When =, u =. When = 4, u =. So inserting appropriate factors of and, we get 4 ( + ) d = u du = [ u ] [ = ( )] =. 4 (B) Since the denominator consists of a single term, we can simplify the integrand by writing it as a sum: 4 + = 4 + = +. We then have + d = [ ] = [4 4 ] ( ) = 5 4. 5 Terrible, terrible question. They want you to use the FTC and the chain rule. BUT the FTC is only applicable at those for which the integrand is continuous on the interval between sin and π. Since sin < π for all, this interval always contains the number π/, where the secant function has an infinite discontinuity. The upshot is, that it s not clear whether the function you re supposed to be differentiating is even defined for any value of at all! (It turns out that this function is actually defined for all, because the positive and negative infinite limits of the secant function on either side of π/ cancel each other out in some sense. But in this course, you re not assumed to know about this.) 6 (A) To integrate an absolute value, break the integral up into two pieces. Since { if = if <,
we should split the integral up at. We get + d = + ( ) d + + d = d = [ ] = 4. 7 (D) First choose which method to use: the disks/washers method will tell us to integrate with respect to (i.e. parallel to the ais of rotation), and this is good since the boundary curve is given in the form y = f(). So use the disks/washers method. We re integrating with respect to, so the limits of integration will be the smallest and largest values of in the solid. These are and 5. So the volume will be V = 5 A() d, where A() represents the area of the vertical cross-section through. This cross-section will be a circle of radius, so we have A() = π( ) = π( ). The volume is therefore 5 [ ] 5 [ ( )] 5 4 V = π( ) d = π = π 5 = 5π. 8 (A) First notice that the graph will have vertical asymptotes at = ±. We can eliminate (c), which has an etraneous vertical asymptote at =. Now look at the behaviour of f near the vertical asymptotes. For eample, we have lim f() = and lim f() =, + which rules out (b), (d) and (e). The only option left is (a). (Note: this is certainly not the only way to solve this problem.) 9 (B:False) For f to be continuous at a, we require also that f(a) be defined and equal to the limit. (B:False) Just about any function f you can think of will give a counter eample. The point is that the integral behaves nicely with respect to additive operations (like adding two functions together), but NOT with respect to multiplicative operations (like multiplying two functions, taking the square, etc.). (A:True) The thing inside the parentheses doesn t depend on, so it is constant with respect to and its derivative is therefore zero. The slope of the tangent line is dy = d 5 4. The problem is to find the value(s) of which maimise this quantity: i.e., we re trying to maimise the function f() = dy = d 5 4. So, we look at the derivative of f (which will be the second derivative of y): f () = d y d = 4 6. Possible maima occur when f () =, or f () doesn t eist. Since f () eists for all, there are no critical numbers of the second kind. We can factor f () = 6(4 ), so the critical numbers are, and. Apply either the first- or the second-derivative test to each of these. I ll use the second-derivative test, which involves looking at (Note that this is now the third derivative of y.) f () = 4 8.
4 When =, f () = 4 >, so f has a local minimum at. When = ±, f () < and so f has local maima at = ±. Now, f is an even function, so f() = f( ). We now want to argue that this common local maimum value is actually the global maimum of f. But f is a polynomial, with lim f() = lim f() =. This means that f has a global maimum value, which must then occur at a local maimum. We found that the only local maima are at = ±, and that f has the same value at these two points, so these values of give the global maimum. So, the slope of the tangent line to the given graph is steepest at = ±. Draw a picture. The curves intersect at = and =. Since neither curve is on top of the other throughout the whole interval [, ], we must compute the area as a sum of two integrals. On the interval [, ], the line is on top of the parabola, so the area of that piece is found by integrating. On the interval [, ], the parabola is above the line, so the area of that piece is found by integrating. So the area is A = d+ [ d = ] + [ ] = ( )+ 8 4 + =. 4 (a) I would first just sketch y = 8/ and y =. They intersect at one point, and in order to make our picture accurate we need to know if the line = goes to the left or to the right of this point. So, find the coordinates of the intersection point, by solving 8 =. The solution is =. So the intersection point is (, 4), and the line = therefore lies to the left of the intersection point. Strictly speaking, there are now two more intersection points: the point (, 8) where the line intersects y = 8/, and the point (, ) where the line intersects the parabola. You should label these as well. 4 (b) Choose which method to use: For disks/washers I d need to integrate parallel to the ais of rotation, i.e. with respect to. For cylindrical shells I d need to integrate perpendicular to the ais of rotation, i.e. with respect to y. My boundary curves are both given in the form y = f() i.e. is set up as the independent variable so I prefer to integrate with respect to. So I ll use disks/washers. Integrating with respect to, the limits of integration will be the smallest and largest -values in the solid. In part (a), we found that these are respectively and. So the volume is A() d. A() is the area of the annulus obtained by taking a vertical slice through. This annulus has outer radius R = 8 (the distance from the ais of rotation up to the top curve), and inner radius r = (the distance from the ais of rotation to the bottom curve). So ( ) 8 A() = π π( ) = π ( 64 4). The volume is therefore given by V = π(64 4 ) d. 4 (c) Choose the method: The boundary curves are still in the form y = f(), so I still want to integrate with respect to. Since we re now rotating around a vertical ais, it is now the cylindrical shells method that involves integrating with respect to. So I use cylindrical shells.
Again integrating with respect to, the limits of integration are still the largest and smallest values of in the region, which are still and respectively. So the volume is S() d, where S() is the surface area of the cylinder through. This cylinder has radius (the distance from the ais of rotation out to the edge of the cylinder), and it has height 8 (top y-coordinate minus bottom y-coordinate). So ( ) 8 S() = π = π(8 ), and the volume is therefore given by V = π(8 ) d. 5