Solutions to Math 41 Eam November 10, 011 1. (1 points) Find each of the following its, with justification. If the it does not eist, eplain why. If there is an infinite it, then eplain whether it is or. (a) (4 marks) ln( π/) π/ + tan() When π +, tan(). The it is infinite because sin(π/) is a nonzero constant and cos(π/) = 0. The sign is since cos() < 0 and sin() > 0 for values of which are close to π/ but greater than π/. On the other hand, when π +, π 0+, so ln( π/) = ln(y) =. π + y 0 + Since both the numerator and denominator approach, we see that the it in question is an indeterminate form. By L Hôpital s rule, it follows that π + ln( π ) tan() 1/( π = ) π + sec () = π + cos () π. The latter it is visibly a 0 indeterminate form, so again using L Hôpital s rule we 0 find that it is equal to π + sin() cos() 1 = sin( π) cos( π ) = 1 0 = 0, since this it can be computed by direct substitution. [Note that mindlessly applying L Hôpital s rule to π + 1/( π/) sec () without first simplifying will result in another indeterminate form. If you don t simplify that and just apply L Hôpital again, you will get another indeterminate form. And so on, and so on, and so on.] (b) (4 marks) 0 + 1/ We first consider the it of the logarithm. Using laws of logarithms we find 0 + ln(1/ ) = 0 + ln(). Now ln() = and = 0, so this is of the form, which is not an 0 + 0 + 0 indeterminate form, and so L Hôpital s rule does not apply! Rather, the it is known to be infinite, and we must determine the sign. Since ln(), the sign of the numerator is negative. Since 0 +, the sign of the denominator is positive. Since = neg. it follows that neg. pos. Finally we compute using this that ln(1/ ) =. 0 + 1/ = eln(1/) = 0 + 0 + y ey = 0.
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page of 13 [Note that it is not correct to invoke the continuity of the function f(t) = e t to write eln(1/) = e ln(1/ ) 0 + = e = 0, 0 + since is not a number, so e does not make sense. The mathematically correct way to make sense of e = 0 is to write y ey = 0.] Alternative solution When 0 +, the base goes to 0 and the eponent goes to +. 0 is not an indetermination, it is 0. (c) (4 marks) 1 ( )tan( π ) As in (b), we first consider the it of the logarithm of ( ) tan( π ). We have [ ln ( ) tan( π )] = tan( π ) ln( ) 1 1 by the laws of logarithms. Since tan( π ) = +, and ln( ) = ln(1) = 0, the 1 1 it displayed above is a 0 indeterminate form. To handle this we write the it as ) ln( ) tan( sin( π π ) ln( ) = 1 1 cos( π), which is now a 0 indeterminate form. By L Hôpital s rule, plus the product rule and 0 the chain rule, it follows that this it equals ( ) 1 sin( π ) + π cos( π ) ln( ) 1 π sin( π) = ( 1) 1 sin( π) + π cos( π ) ln( 1) π sin( π) = = 1 + 0 π = π We can use direct substitution, since the quantity of which we are taking the it is actually continuous at = 1 and the denominator does not become zero. Since e is a continuous function, we therefore have 1 ( )tan( π ) tan( = π [( eln[( ) ) ] ln ) tan( π )] = e 1 = e /π. 1
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 3 of 13. (11 points) A coastguard is sitting on an elevated chair on the beach 50 m from the shoreline. He is observing a shark in the sea coming in a straight line towards him. When the shark is 60 m away from the base of the chair, it is approaching the shoreline at a speed of 4 m/sec. At that same moment, his chair is sinking in the sand at a rate of 0. m/sec and his eyes are m above the floor. How fast is the angle α in the picture changing at that moment? (Assume that the shark is just a point on the surface of the sea.) Let be the distance from the base of the chair to the shark, and y the height of the coastguard s eyes, as indicated in the picture. We are given d dy = 4 m/sec and = 0. m/sec when = 60 and y =. Both rates of dt dt change are negative because and y are decreasing. We are looking for dα dt. Note that tan(α) = y ( y ), so α = arctan. Taking derivative with respect to time t, we obtain: dα dt = 1 1 + (y/) Now, plugging in the given values we have: dα dt = 1 ( 0.)60 ( 4) 1 + 4/3600 3600 dy dt d dt y = 3600 4 3604 3600 = 4 3604 = 1 901 So the angle α is decreasing at a rate of 1 901 Alternative solution radians per second. One can differentiate with respect to t directly on the equation tan(α) = y sec (α) dα dy dt = d dt dt y We can now find z = 3600 + 4 = 3604 using the Pythagorean theorem, so sec(α) = and so sec (α) = 3604 3600. 3604 60
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 4 of 13 So dα dt = 4 3604 = 1 901 radians per second. 3604 dα 3600 dt = 4 3600
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 5 of 13 3. (5 points) A function f with domain [ 5, 3] satisfies f( 5) = and f () = 0 for all 5 3. Show that f is a constant function. We will show that f() = for all 5 3, and thus that f is a constant function. Note that f is differentiable on its domain since we re given that f () = 0 for all. Thus the Mean Value Theorem applies. We already know that f( 5) =. Let b be any number that satisfies 5 < b 3. By the Mean Value Theorem applied to f on the interval [ 5, b], there is a c between 5 and b such that f(b) f( 5) = f (c). b ( 5) We know that f (c) = 0 and f( 5) =, so we get f(b) = 0, or f(b) =. Since b was arbitrary, this shows that f() = for all 5 3.
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 6 of 13 4. (11 points) A farmer has a straight fence with a length of 10 yards that he wants to break into two pieces. Then he will bend the first piece to make a round pen for horses and bend the second piece to make a square pen for pigs. What is the minimum combined area of the two pens? (Note: He will build two pens, building only one pen is not an option.) Let be the length of fence used to make the round pen. Its circumference is = πr, where r is its radius, so r = /(π), and the area of the round pen is Area of round pen = π(r) = 4π Then 10 yards of fencing will be used to make the square pen. If s is the length of the side of the square, then s = (10 )/4 and the area of the square pen is Area of square pen = (10 ) 16 The total area enclosed in both pens is A() = (10 ) + 4π 16 0 < < 10. We wish to minimize A(), so first we find the derivative: A () = π 10 ( 1 = 8 π + 1 ) 5 8 4 ( ) 4 + π = 5 8π 4 A () equals zero only at the point = 5 4 8π 4 + π = 10π 4 + π. This is the only critical point of A() in our domain because A () is a polynomial, and so is defined everywhere in (0, 10). Furthermore A () = 4+π > 0 for all in (0, 10), so by the 8π second derivative test for global etrema, A(10π/(4 + π)) is a global minimum in (0, 10). Thus the minimum possible area is ( ) ( 10π (10π/(4 + π)) A = + 4 + π 4π ) (10 10π/(4 + π)) 16 yards
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 7 of 13 5. (1 points) Let f() = ln() (a) (1 mark) Find the domain of f(). Recall that ln() is defined only for > 0. However, since we are dividing by ln() we have to be careful not to divide by zero, which happens when = 1 as ln(1) = 0. Therefore, the domain is: (0, 1) (1, ). (b) (6 marks) Determine if f has any asymptotes, with complete reasoning. If there is any vertical asymptote, compute both corresponding one-sided its. Horizontal asymptotes: ln, has the form / ; hence, by L Hospital s rule ln = 1 1 = =. Note that we don t look at what happens when as this is not part of the domain. We conclude that there are no horizontal asymptotes. Vertical asymptotes: To find VA we look at what happens to f() as leaves the domain. There are tree its that we need to compute. First, 0 + ln = 0, because the numerator goes to 0 and the denominator goes to. So there is no vertical asymptote at = 0. Second, 1 + ln, has the form 1/0 and so the it involves infinity. When > 1 and is near 1, then ln() > 0 and > 0 so 1 + ln =. On the other hand, the third it of interest gives 1 ln =, as < 1 implies that ln() < 0. occurs when = 1. We conclude that the only vertical asymptote
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 8 of 13 For your convenience, here is f() = ln() again. (c) (5 marks) On what interval(s) is f() increasing? Decreasing? Justify your answer and find the and y-coordinates of all the local maima and minima (if any). We first compute the derivative of f() which is f() = ln 1 [ln()]. Note that on the domain ln() 0; therefore the only critical number correspond to ln() 1 = 0 (when the numerator is equal to zero). This happens when = e. We also need to include = 1 in our considerations to study the signs because it is not in the domain. ln 1 (ln ) f () f() (0, 1) + decreasing (1, e) + decreasing (e, ) + + + increasing By the first derivative test since f() is going from decreasing to increasing at = e then (e, e) is a local minimum. (d) (4 marks) It is known that f () = ln(). On what interval(s) is f() concave up? ln() 3 Concave down? Justify your answer and find the and y-coordinates of all inflection points (if any). Note that the denominator is never zero in the domain. Hence, we only worry about f () = 0 which happens when ln = 0 or equivalently when = e. We need to include = 1 again in our consideration to study the signs. ln ln 3 () f () f() (0, 1) + concave down (1, e ) + + + concave up (e, ) + concave down There is a change of concavity when = 1 and = e. However, since = 1 is not in the domain the only inflection point is at = e and f(e ) = 1 e.
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 9 of 13 (e) (5 marks) Using the information you have found, sketch the graph of f() below. Label your and y-aes and clearly label the location of any local maima (LMAX), local minimum (LMIN) and inflection points (IP), if any.
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 10 of 13 6. (13 points) Find the derivative, using any method you like (that works). You do not need to simplify your answers. (a) (4 marks) f() = e sin(5 3 ) Use the quotient rule and the chain rule, to find f () = e sin(5 3 ) 15 cos(5 3 )e sin. (5 3 ) Alternate solution 1: Rewrite f() = e csc(5 3 ) or f() = e (sin(5 3 )) 1 and use the product rule. You will obtain an epression equivalent to the above. Alternate solution : Use logarithmic differentiation: ln(f()) = ln(e ) ln(sin(5 3 )) = ln(sin(5 3 )), by the laws of logarithms, so differentiating implicitly we find f () f() = 15 cos(5 3 )(sin(5 3 )) 1, and multiplying this by f() yields an epression for f () equivalent to the one above. (b) (4 marks) Z() = arctan( 1 + ) The key here is to apply the chain rule correctly. It is crucial to epress your answer with parentheses in the correct place! One finds using the chain rule and the formula d arctan(t) = 1 that d t 1+t Z () = 1 1+ 1 + ( 1 + ).
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 11 of 13 (c) (5 marks) g(t) = ln(t ) t5 +3t Note that g(t) (t 5 + 3t) ln(t ) = ln((t ) t5 +3t ). As usual, for a function f such as ln, the notation f() a means (f()) a not f( a ). To solve the problem use logarithmic differentiation. Writing y = g(t) we have [ ] ln y = ln ln(t ) t5 +3t = (t 5 + 3t) ln[ln(t )] by the laws of logarithms. So y y = 1 (t5 + 3t) ln(t ) 1 t t + (5t4 + 3) ln[ln(t )] by the product rule and the chain rule (applied repeatedly). Multiplying by y we find ( g (t) = ln(t ) t5 +3t (t 5 1 + 3t) ln(t ) 1 ) t t + (5t4 + 3) ln[ln(t )]
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 1 of 13 7. (10 points) Consider the curve 3y = 4y (a) (6 marks) Find an equation for the tangent line to the curve at the point (1, 1). Using implicit derivative on the curve, and thus (3 + 8y) dy d = 1 4y, or 1 3 dy d = d d (4) d y + 4 d (y ) = 4y + 8y dy d The slope of the tangent line at (1, -1) is dy d = (,y)=(1, 1) dy d = 1 4y 3 + 8y. 1 4( 1) 3 + 8 1 ( 1) = 3 5 and the tangent line has to be the line with slope 3/5 and pass through (1, -1). Therefore the desired equation is y = 3 5 ( 1) 1 = 3 5 8 5 (b) (4 marks) Show that this curve does not have any horizontal tangent line. The curve has the horizontal tangent line if and only if dy = 0 at some point on the curve. d If dy = 0, we have 1 d 4y = 0 and thus y = ± 1. When plugging in y = 1 to the original curve, 3 1 ( ) 1 = 4 3 = which is impossible. Similarly, when y = 1, we have + 3 = on the curve, which is also impossible. Therefore there is no point on the curve that has y-coordinate equal to ± 1, and hence there is no point on the curve at which the tangent line is horizontal.
Math 41, Autumn 011 Solutions to Eam November 10, 011 Page 13 of 13 8. (9 points) (a) (6 marks) Use linear approimation or differentials only once to estimate the 63 value of 3. Your answer should be given in the form c, where c and d are integers. 63 d (Hint: 6 = 64) 63 We want to approimate 3 by linear approimation. We will use the function f() = 63 3, so we want to approimate f(63). The linear approimation formula is L() = f(a) + f (a)( a). so that f() f(a) + f (a)( a) Notice that f() = 1/ 1/3 = 1/6. So f () = 1 6 5/6. Now we plug in, using the fact that 64 = 6. L(63) = f(64) + f (64)(63 64) = ( 6 ) 1/6 + 1 6 (6 ) 5/6 ( 1) = 1 6 = 383 19 1 3 So our estimate is 63 3 63 383 19. (b) (3 marks) Is your estimate in part (a) smaller or larger than 63 3 63? Justify your answer. To see whether we made an under- or over-estimate, we need to check the concavity of f(). So we take the second derivative. Since f () = 1 6 5/6, f () = 1 6 5 6 11/6 = 5 36 11/6. Since f () will be negative whenever you plug in a positive number, we know that f (64) < 0. So f() is concave down around = 64 meaning the tangent line to f() will lie over the curve. Therefore our estimate is an overestimate.