All exercises are from Section 4.7 of the textbook. 1. Calculus with Analytic Geometry I Exam 10, Take Home Friday, November 8, 2013 Solutions. 2. Solution. The picture suggests using the angle θ as variable; θ = 0 means she rows all the way, θ = π/2, she walks all the way. There are several ways of getting all the distances; here is one. Let O be the center of the lake and consider the isosceles triangle AOB. The base angles must be equal, thus the angle AOB (angle at O) is π 2θ. The length AB is then given by AB 2 = 2 2 + 2 2 2 2 2 cos(π 2θ) = 8 + 8cos(2θ) = 8(1 + cos 2 θ sin 2 θ) = 16 cos 2 θ, so that AB = 4 cos θ. Concerning the length of the arc from B to C, its length equals the radius times the subtended angle COB. This angle is again easily seen to be 2θ; thus the length of the arc is 4θ. We now see that the time to minimize is T (θ) = 4 cos θ 2 + 4θ 4 = 2 cos θ + θ, 0 θ π 2. We find critical points, if any. We have T (θ) = 2 sin θ + 1; setting it to 0 we get sin θ = 1/2. In the interval [0, π/2], sin θ = 1/2 only for θ = π/6. The easiest way of checking is to just evaluate T at the critical point and at the endpoints of the interval. We get T (0) = 2, T (π/6) = 2 cos(π/6) + π/6 = 3 + π 6 2.26, T (π/2) = π 2 1.57 To minimize time, she should walk all the way. Here is a picture to go with the problem.
A convenient variable is x, where (x, 4 x 2 ) is the point of tangency. But we have to be careful in how we write the equation of the line, so for just a moment it may be better to denote the point of tangency by (x 0, 4 x 2 0). The value of the derivative at this point is d(4 x2 ) dx = 2x 0. The equation of the tangent x0 line is then y (4 x 2 0) = 2x 0 (x x 0 ), or y = 2x 0 x + x 2 0 + 4. From this we get at once the values of the x and y intercepts of the tangent line; they are, respectively, x0 2 + 2 x 0 and x 2 0 + 4. These are also the lengths of the legs of the triangle so that, writing again x for x 0, the area to minimize is A(x) = 1 ( x 2 2 + 2 ) (4 + x 2 ) = 1 x 4 x3 + 2x + 4 x. The domain in which we have to consider this function is (0, ). We see that A (x) = 3 4 x2 +2 4 ; setting to x2 0 and multiplying by 4x 2 (which changes nothing since x > 0) we get 3x 4 + 8x 2 16 = 0. This is a quadratic equation in x 2 with roots 4 and 4/3. Discarding the impossible root 4, we are left with x 2 = 4/3, thus x = ±2/ 3. The only critical point in the domain of relevance is x = 2/ 3. Now A (x) = 3 2 x + 8 x > 0 when 3 x > 0, so that we have a local minimum, hence an absolute minimum since it is the only critical point. The minimum area is thus ( ) 2 A 3 = 32 3. 9
3. Solution. I ll denote by AB the length of the segment AB. The length of the rope is L = P Q csc θ 1 + ST csc θ 2. We have to relate θ 2 with θ 1. We can use that tan θ 2 = ST RT RT 2 + ST so that a standard trigonometry exercise shows that csc θ 2 = 2. Now ST RT = QT QR = QT P Q cot θ 1, so writing now θ for θ 1 we get for the length of the rope L = P Q sec θ + ( QT P Q cot θ) 2 + ST 2 But this gives a really nasty derivative to minimize, so let us try to find instead the point R that minimizes the length of the rope. I ll introduce for this purpose the variable x = QR and now we can express the length L of the rope in the somewhat simpler form The domain of the variable x is 0 x QT. Then L(x) = P Q 2 + x 2 + ST 2 + ( QT x) 2. L (x) = Setting to 0, and solving for x on gets x = x P Q 2 + x QT x 2 ST 2 + ( QT x) 2 QT P Q. Is this the x that minimizes L?. We notice that P Q + ST L QT (0) = ST 2 + QT < 0, QT 2 L ( QT ) = ST 2 + QT > 0; 2 since L is only zero at x, that means L is negative to the left of the critical point, positive to the right; the critical point is a local minimum and, being the only critical point in the interval, the point at which the absolute minimum occurs. With this value of x = QR we get tan θ 1 = P Q x = P Q + ST, QT
tan θ 2 = ST QT x P Q + ST =. QT The angles are equal. NOTE: There is a clever way of solving this problem without appealing to calculus. Can you find it? 4. Solution. The amount of water will be maximum when the cross-sectional area is maximum. The cross section is a trapezoid of bases of lengths (in cm) 10 and 10 + 2 10 cos θ, height 10 sin θ. The area is thus the domain of A is 0 θ π/2. Now A(θ) = 100(1 + cos θ) sin θ. A (θ) = 100( sin 2 θ + (1 + cos θ) cos θ) = 100(cos θ + cos 2 θ sin 2 θ). To be able to solve for θ after setting A (θ)to0, it is a good idea to replace sin 2 θ by 1 cos 2 θ. With this substitution, we get A (θ) = 100(2 cos 2 θ + cos θ 1). Setting A (θ) to 0 yields a quadratic equation in cos θ that can be solved by the quadratic formula: 2 cos 2 θ + cos θ 1 = 0, so cos θ = 1 ± 3, 4 so θ = 1/2 or θ = 1. The second solution can t occur in the given domain, so cos θ = 1/2, hence θ = π/3 is the only critical point in the domain. Now The maximum value occurs when θ = π 3. A(0) = 0, A(π/3) = 75 3 129.9, A(π/2) = 100. 5. One can stop the domain at π/2 since it is obvious that a larger angle will give a worse cross-sectional area. But one can also consider a somewhat larger domain. However, one cannot take θ all the way to π because once θ = 5π/6 (150 ) the two sides of the gutter touch, the cross section is an equilateral triangle, and it doesn t make sense to keep bending the sides.
Solution. This exercise is way easier than it seems. It is a bit a lesson in anatomy. (a) A bit of trigonometry shows that the distance from B to the end of the horizontal blood vessel is b cot θ. The distance from A to B is thus a b cot θ. Trigonometry also shows that the distance from B to C is b csc θ. So assuming (as one has to, to get the required answer) that resistance is additive, the resistance from A to B is in fact ( a b cot θ R = C r 4 1 + b csc θ r 4 2 (b) I assume that the full question is: Given a major blood vessel (an artery, perhaps), a sub-vessel has to branch off to get blood to a point C at distance b from the major vessel, and it should branch off at a point at a distance less than a from the point nearest to C. So we do the usual. First of all, let us determine the domain of R as a function of θ. It should be clear that θ = 0 cannot work. In fact, the smallest value of θ has to have a tangent of b/a and the largest is π/2. The domain of R is arctan(b/a) θ π/2. Differentiating, ( +b csc R 2 θ (θ) = C 4 ). ) ( b csc θ cot θ csc θ 4 = Cb csc θ 4 cot θ ) 4 ; setting to 0, since csc θ is never 0, we get csc θ 4 cot θ 4 = 0, thus cot θ csc θ = r4 2 4 and, since cot θ/ csc θ = cos θ, we see that R (θ) = 0 if and only if cos θ = /r 4 1. 4 Is this in the domain? Since we can assume that a is fairly large compared to b, so arctan b/a is close to 0, the answer is probably yes. Justifying that this is a minimum (and not a maximum) is perhaps harder than usual, but not impossible. Here is one way. While the domain of R might not go all the way to θ = 0, the function R(θ) is define in (0, π ). If we replace cot, csc by their expressions in terms of sin, cos, the derivative becomes ( R 1 1 (θ) = Cb sin 2 θ 4 cos θ ) 4,
showing that the sign of R 1 (θ) is the same as the sign of 4 cos θ 4. Now 1 4 cos θ 4 = 1 θ=0 4 1 4 < 0 because r 2 < r 1, while 1 r 4 1 cos θ r 4 2 θ= π 2 = 1 r 4 1 > 0. The derivative goes from negative to positive, thus we have a minimum. (c) We have to find θ such that Thus θ = arccos 16 91 80. cos θ = ( 2 3 r 1) 4 r 4 1 = 16 91.