Calculus with Analytic Geometry I Exam 10 Take Home part

Calculus with Analytic Geometry I Exam 10 Take Home part Textbook, Section 47, Exercises #22, 30, 32, 38, 48, 56, 70, 76 1 # 22) Find, correct to two decimal places, the coordinates of the point on the curve y = sin x that is closest to the point 4, 2) Solution We find the point x, y) on the curve that minimizes the square of the distance D = x 4) 2 + y 2) 2 The constraint on x, y) is that y = sin x, so as a function of x, Dx) = x 4) 2 + sin x 2) 2 The domain in which we have to consider this function is all of the real line; x could be anywhere We begin finding the critical points of D; we compute d x) and set it to 0 D x) = 2x 4) + 2sin x 2) cos x = 0 This equation can be rewritten as 2 sin x) cos x = x 4 This is not an easy equation to solve, which is why the book has the graphic calculator or computer symbol attached to it Those of you who get stymied, will you bother to ask for help? But thanks to the magic of Wolfram alpha, it gets to be quite easy to complete the exercise First I will draw a graph to get an idea of how many solutions there might be Notice that the left hand side is always between 2 and 3 because 2 sin x is always 1 and 3, so with 1 cos x 1, the worst case scenario is that cosx = 1 and 2 sin x = 2, so the product is 2, or cos x = 1 and 2 sin x = 3 so the product is 3), so that we must have 2 x 4 3, or 2 x 7 So if there is a solution, it will be in the interval [2, 7] This can help us with the graph What I will have Wolfram alpha do is plot the two curves y = 2 sin x) cos x and y = x 4 in the interval [2, 7] and see how many times they intersect; if any at all The result is We see clearly that there are three solutions So next we ask our friend W alpha to find them, by inputting solve 2 sin x) cos x = x 4 From the graph we already have a good idea of what the solutions should be, close to 265, 5 and 6 W alpha s answer is x 265119, 509625, 612841 We need to decide which is a minimum, if any Let s be logical about it Use our head Think of x as being far away to the left, very negative,and now it begins to move to the right The distance between x, sin x) and 4, 2) has to be decreasing This means that 265119 is either a local minimum or an inflection Let us check this using the second derivative test, keeping in mind that the x-value is only approximate We have D x) = 21 + cos 2 x sin 2 x + 2 sin x); D 265119) = 49966 D cam out sufficiently positive that we can rule out an inflection point; we must have a local minimum After this, D increases; this makes 509625 either a local max, or an inflection Computing, D 509625) = 314789, Well, one has to be careful The curve could have a number of little oscillations that are too small to show up on the graph If this were a life and death project, one might want to run a few more tests

2 so we have a local maximum This forces the third critical point to be a local minimum, since D must increase as x It is now quite evident that whichever one of D265119), D612841) is smaller, is the minimum value We have D265119) = 4157, D612841) = 017 The conclusion is that the first critical point is the one we are looking for We could also have graphed D itself; here is how the graph looks This confirms our calculations; the critical point at approximately 265 is where the absolute minimum occurs Now sin265119) = 047098 Rounding off to the required two decimal places, the answer is 265, 047) 2 #30) A right circular cylinder is inscribed in a cone with height h and base radius r Find the largest possible volume of such a cylinder Solution We must assume that inscribed means the the base of the cylinder is contained in the base of the cone and geometrical intuition shows that to get the largest cylinder we should have the center of the bases of cylinder and cone be the same Let x be the radius of the cylinder, y its height If we cut by a vertical plane through the common axis, we get the following picture

3 The volume of the cylinder is V = πx 2 y We need to relate x and y, which is easily done by similar triangles In the profile pictured, the triangle above the green rectangle is similar to the big triangle We get h y)/h = x/r Solving, y = h hx/r) Then V x) = πx 2 h hx ) ) = πh x 2 x3 r r The domain of this function is 0 x r Now V x) = πh 2 2x 3x2 r ) ; setting to 0 we have two solutions, x = 0 an end-point) and x = 2r/3 Because of the Extreme Value Theorem, there is a maximum Because interior maxima are achieved at critical points, the maximum value of the volume occurs for x = 0, 2r/3, or x = h Now V 0) = 0, V 2r/3) = 4πhr 2 /27, V h) = 0 The maximum value of the volume is V = 4πhr2 27 3 #32) A Norman window has the shape of a rectangle surmounted by a semicircle Thus the diameter of the semicircle is equal to the width of the rectangle) If the perimeter of the window is 30 ft, find the dimensions of the window so that the greatest possible amount of light is admitted Solution Drawing a picture is indicated With the variables from the picture, the area of the window is A = xy + 1 x ) 2 2 π πx 2 = xy + The constraint 2 8 is 2y + x + πx/2 = 30 Solving for y, substituting into the area, simplifying a bit, Ax) = 15x π + 4 x 2 8 The domain of this function is a bit messy; clearly x 0 We also must have y = 15 π + 2 x 0, which 4 forces x 60 60 So 0 x π + 2 π + 2 Differentiating A we get A x) = 15 π + 4 x Setting A x) = 0, there 4 is a single cortical point, namely x = 60/π + 4) To justify it is the maximum, for once we can compute A ; A x) = π + 4)/4 < 0 Because 60/π + 4) is the ONLY critical point in the interval, and it is a relative max, it is where the max occurs The dimensions are x = 60 π + 4, y = 30 π + 4

4 4 # 38) A fence 8 ft tall runs parallel to a tall building a ta distance of 4 ft from the building What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? Solution Here s a picture The ladder is depicted in red The square of its length is S = x 2 + y 2 ; I ll minimize the square of the length rather than the length From similar triangles, y/x = 8/x 4); thus y = 8x/x 4) and S = x 2 + 64x2 x 4) 2 The domain of this function is 0 x < Differentiating and setting to 0 we get S x) = 2x 512x x 4) 3 = 0; solving, x = 0 and x = 41 + 2 2/3 ) The only relevant solution is the second one; x = 0 is not in the domain This time I ll use the first derivative test to justify that we found a minimum Notice that lim x 4+ S x) =, lim x S x) = For this to happen, S must be negative close and the the right of 4, positive for large values of x This means, S goes from negative to positive at the critical point, thus it is a local minimum Being the ONLY critical point, there ia an absolute minimum at that point Now, at this value of x, Sx) = x 2 1 + ) 64 x 4) 2 = 161 + 2 2/3 ) 2 1 + ) 64 = 16 1 + 2 2/3) 3 32 2 1/3 The length of the ladder is the square of this quantity, thus the length of the shortest ladder is L = 4 1 + 2 2/3) 3/2 1665 ft 5 #48) A woman at a point A on the shore of a circular lake with radius 2 mi wants to arrive at the point C diametrically opposite A on the other side of the lake in the shortest possible time see figure) She can walk at the rate of 4 mi/h and row a boat at 2 mi/h How should she proceed? Solution Lets add some data to the picture:

5 Using a bit of high school geometry or, what should be high school geometry) the central angle subtending the arc BC is twice the angle θ By the law of cosines, and high school trigonometry BC 2 = r 2 + r 2 2r r cosπ 2θ) = 2r 2 1 cosπ 2θ)) = 2r 2 1 + cos 2θ) = 4r 2 cos 2 θ, where r = 2 is the radius of the circular lake That is BC 2 = 16 cos 2 θ, thus BC = 4 cos θ The length of the arc BC is 2θr = 4θ The total time the woman would take if she rows to B and then walks, is T θ) = 4 cos θ 2 + 4θ θ = 2 cos θ + θ The domain of T is the interval [0, π/2] Differentiating and setting to 0 gives T θ) = 2 sin θ + 1 = 0, thus sin θ = 1/2, hence θ = π/6 The minimum and the maximum) value of T has to occur at one of 0, π/6, π/2 We see that T 0) = 2, T π/6) = 3 + π 6 226, T π 2 ) = π 2 157 The minimum occurs for θ = π/2, which means that the woman should walk all the way 6 #56) What is the smallest possible area of the triangle that is cut off by the first quadrant and whose hypotenuse is tangent to the parabola y = 4 x 2 at some point? Solution The picture shows one such triangle The variables to be used are the coordinates of the point of tangency, namely a, b) Since y = 2x, the slope of the tangent line at a, b) is 2a The equation of the tangent line is y b = 2ax a) But b = 4 a 2, so it is y 4 + a 2 = 2ax a), simplifying y = 2ax + 4 + a 2 The y and x intercepts of this line are, respectively, 4 + a 2 and 4 + a 2 )/2a The x-intercept is the base, the y-intercept the height, of the triangle, the area is thus Aa) = 4 + a2 ) 2 4a = 4 a + 2a + a3 4

6 The domain is 0 < a 2 b = 4 a 2 0 forces a 2 4, so a 2) Now setting A a) = 0: 0 = A a) = 4 a 2 + 2 + 3 4 a2 = 3a4 + 8a 2 16 4a 2 Thus 3a 2 + 8a 2 16 = 0, a quadratic equation in a 2 ; with the solutions a 2 = 8 ± 16)/6 a 2 cannot be negative, so the only valid solution is a 2 = 8/6 = 4/3, so a = 2/ 3, which is in the domain Now A a) is seen to be negative near 0, positive for large values of a, thus we have a minimum single critical point in the interval criterion!) The corresponding area is A = 32 3 9 7 #70) A steel pipe is carried down a hallway 9 ft wide At the end of the hall there is a right angled turn into a narrower hallway 6 ft wide What is the length of the longest pipe that can be carried horizontally around the corner? Hint: Suppose L is the length of the slanted line in the picture, the one forming an angle of θ with the first hallway A bit of trigonometry might allow you to relate this length to θ The longest the pipe could be might perhaps be the largest possible value of L? You might also notice that 0 θ π/2 Solution The so called hint has a little mistake in it The longest the pipe can be is the shortest of all the lengths L of the type in the picture, because if it is to go through the bend, it can t be longer than this shortest length Let us call x the length of the part of L in the wider corridor, y the part in the narrower corridor, so that L = x + y Now x = 9/ sin θ, y = 6/ cos θ see picture) Thus Lθ) = 9 sin θ + 6 = 9 csc θ + 6 sec θ the domain being 0 < θ < π/2 Now cos θ L θ) = 9 csc θ cot θ + 6 sec θ tan θ = 9 cos θ sin 2 θ + 6 sin θ cos 2 θ = 6 sin3 θ 9 cos 3 θ cos 2 θ sin 2 θ Setting L θ) = 0, we get 6 sin 3 θ = 9 cos 3 θ, thus tan θ = 3 9/6 = 3 15 Is the value at this θ a minimum? The best way of seeing that it is, is using an intuitively clear fact, which is also an easy consequence of the Extreme Value Theorem, and the fact that interior extrema are achieved at critical points Here is the intuitively clear fact: Let f be defined and continuous on an interval a, b) that is fx) is defined for all x such that a < x < b, and f is continuous at all points of a, b)) If lim x a + fx) = and lim x a fx) =, then f assumes a

7 minimum value in a, b); that is, there is a point x 0 in a, b) such that fx 0 ) is the absolute minimum value of f If f is differentiable at that point, then f x 0 ) = 0 The previous result is true also if a =, in which case lim x a + fx) is replaced by lim x fx) and/or if b = ; then lim x b + fx) becomes lim x fx) Noticing that in our case lim θ 0 + Lθ) = lim θ π/2) Lθ) =, the principle just mentioned guarantees a point in 0, π/2) where an absolute minimum occurs; it also guarantees that at that point L is zero Well, there is only one point where L is zero; L at that point must be the absolute minimum Since we are asked for the length of the longest pipe which coincides with the smallest value of L) we have to calculate L 3 15) This means computing the cosine and sine of this 3 15 and I will avoid using approximations and calculators until it becomes absolutely necessary So we set up the standard triangle The tangent of the pictured angle θ is 3 15 We see that cos θ = the shortest L, and the longest pipe, will have length 1, sin θ = 15 1/3, 1 + 15 2/3 1 + 15 2/3 L = 9 1 + 15 2/3 ) 15 + 6 1 + 15 15 2/3 = 6 1 + 15 2/3 1/3 15 + 1 1/3 ) 3/2 = 6 1 + 15 2/3 15 2/3 + 1 = 6 15 2/3 + 1) And the answer is 3/2 6 15 2/3 + 1) 2107 ft 8 #76) The blood vascular system consists of blood vessels arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart This system should work so as to minimize the energy expended by the heart in pumping the blood In particular, this energy is reduced when the resistance of the blood is lowered One of Poiseuille s laws gives us the resistance R of the blood as R = C L r 4 where L is the length of the blood vessel, r is the radius, and C is a positive constant determined by the viscosity of the blood Poiseuille established this law experimentally, but it also follows from equation 2 in section 4 of Chapter 8) The figure shows a main blood vessel with radius r 1 branching at an angle θ into a smaller vessel with radius r 2

8 a) Use Poiseuille s law to show that the total resistance of the blood along the path ABC is a b cot θ R = C r1 4 + b csc θ ) r2 4 where a and b are the distances shown in the figure Solution We have to assume that resistance is additive Consider the triangle with vertices at B, C, and the third vertex at the right end of the main blood vessel The length from point B to that third vertex is b/ tan θ = b cot θ The distance from A to B is thus a b cot θ, making the resistance for that portion of the main vessel equal to Ca b cot θ)/r 4 1 The length from B to C is b/ sin θ = b csc θ, so the resistance in that secondary vessel will be R 2 = Cb csc θ/r 4 2 Adding up R 1 + R 2, we get precisely the R given above b) Prove this resistance is minimized when cos θ = r4 2 r1 4 Solution The assumption here is that a, b are constants; otherwise we d have no idea how to proceed This means that the smallest θ can be is when its tangent is b/a; the largest is π/2 That is, the domain of R is arctanb/a) θ π/2 We find the critical points of R Differentiating the C in front is a nuisance, by the way) R θ) = C 1 r1 4 b csc 2 θ 1 ) r2 4 b csc θ cot θ = Cb sin 2 θ 1 r 4 1 cos θ r 4 2 Setting to cos θ = r2/r 4 1 4 follows Is it a minimum? I took a point away for not justifying that a minimum or maximum was achieved In the final, I will probably penalize you more for lack of justification! The answer to the question, is it a minimum is actually not necessarily Looking at the expression for R θ) it is clear that R θ) > 0 if θ is to the right of the point; for example when θ = π/2, R π/2) = Cb/r1 4 > 0 It is also easy to see that R θ) < 0 to the left of our presumed minimum point The big problem is: Is θ such that cos θ = r2/r 4 1 4 in the domain of R? Since cos decreases in [0, π/2] we must have r2/r 4 1 4 than the value of cos on the left end point of the domain; that is, r2 4 cosarctan b a ) = a a2 + b 2 r 4 1 Nothing is said about the relation between a, b and the radii of the vessels Presumably for the same radii r 2, r 1 we might have a very large value of b, making a/ a 2 + b 2 very small, so small that the inequality r 4 2/r 4 1 a/ 1 2 + b 2 does not hold In that case the minimum is achieved for θ = arctanb/a) Our beloved author does not seem to have considered this possibility or explained why it can t happen) )

9 c) Find the optimal branching angle correct to the nearest degree) when the radius of the smaller blood vessel is two-thirds the radius of the larger vessel Solution Assuming that r2/r 4 1 4 a/ a 2 + b 2, the minimum occurs for cos θ = r2/r 4 1, 4 which in this case equals 2r 1 /3) 4 /r1 4 = 16/81 In degrees the angle is θ = arccos 16 81 7861 In the text there is now a picture of actual blood vessels)