Lecture 23 Notes: Pipeline Issues: Delay Slots, Annulment, Exceptions

18.01 Calculus Jason Starr Fall 2005 Lecture 23. November 8, 2005 Homework. Problem Set 6 Part I: (i) and (j); Part II: Problem 2. Practice Problems. Course Reader: 4I­1, 4I­4, 4I­6. 1. Tangent lines to parametric curves. This short section was not explicitly discussed for general parametric curves. It was discussed for polar curves, which are a special collection of parametric curves. � � 18.01 Calculus Jason Starr Fall 2005 Given a parametric curve, � x = f (t), y = g(t), what is the slope of the tangent line at (f (a), g(a))? The relevant differentials are, dx = f � (t)dt, dy = g � (t)dt. If g � (a) is nonzero, then the slope of the tangent line is, � f � (a) dy f � (t)dt �� (a) = � = . dx g (t)dt �t=a g � (a) In particular, for a function r = r(θ), the associated polar curve is, � x = r(θ) cos(θ), y = r(θ) sin(θ) Thus the differentials are, dx = [r� (θ) cos(θ) − r(θ) sin(θ)]dθ, dy = [r� (θ) sin(θ) + r(θ) cos(θ)]dθ. Therefore the slope of the tangent line is, dy r� (θ) sin(θ) + r(θ) cos(θ) = � . dx r (θ) cos(θ) − r(θ) sin(θ) 2. Tangent lines for polar curves. Although the formula above is perfectly correct, it is a bit long to remember. There is a slightly different packaging that is much easier to remember. Define α to be the angle from the horizontal ray emanating from (x(θ), y(θ)) in the positive x­direction, and the tangent line. To be precise, there are two such angles, differing by π. The defining equation for α is, dy tan(α) = . dx And, of course, y tan(θ) = . x Define ψ to be the difference between α and θ, ψ = α − θ. The angle addition/subtraction formulas for tan(θ) are, tan(φ1 + φ2 ) = tan(φ1 ) + tan(φ2 ) tan(φ1 ) − tan(φ2 ) , tan(φ1 − φ1 ) = . 1 − tan(φ1 ) tan(φ2 ) 1 + tan(φ1 ) tan(φ2 ) 18.01 Calculus Jason Starr Fall 2005 Therefore, tan(ψ) = tan(α − θ) = tan(α) − tan(θ) . 1 + tan(α) tan(θ) Substituting in the equations for tan(θ) and tan(α) from above gives, tan(ψ) = (dy/dx) − (y/x) . 1 + (y/x)(dy/dx) To simplify this, imagine multiplying both numerator and denominator by xdx and manipulate formally, xdy − ydx tan(ψ) = . xdx + ydy The actual justification of this is a little more involved, but the formal manipulation leads to the correct equation. To compute the denominator in the expression, differentiate both sides of, r 2 = x2 + y 2 , to get, 2rdr = 2xdx + 2ydy, or equivalently, xdx + ydy = r(θ)r� (θ)dθ. To compute the numerator in the expression, differentiate both sides of, tan(θ) = y , x to get, dy ydx 1 − 2 = 2 (xdy − ydx). x x x Now substitute x = r cos(θ) in the denominator to get, sec2 (θ)dθ = sec2 (θ)dθ = 1 sec2 (θ) (xdy − ydx) = (xdy − ydx). r2 cos2 (θ) r2 Cancelling sec2 (θ) and multiplying both sides by r2 gives, xdy − ydx = r2 dθ. Thus the fraction for tan(ψ) is, tan(ψ) = r2 dθ xdy − ydx = � . xdx + ydy rr dθ 18.01 Calculus Jason Starr Fall 2005 Simplifying gives, tan(ψ) = r(θ)/r� (θ) . Example. Consider the cardioid, discussed in recitation, r(θ) = a(1 + cos(θ)). The formula for ψ is, tan(ψ) = r a(1 + cos(θ)) 1 + cos(θ) = = . � r −a sin(θ) − sin(θ) To simplify this, write θ = 2(θ/2) and use the double­angle formulas to get, 1 + cos(2(θ/2)) 1 + (cos2 (θ/2) − sin2 (θ/2)) = . − sin(2(θ/2)) −2 sin(θ/2) cos(θ/2) Replacing 1 − sin2 (θ/2) in the numerator by cos2 (θ/2), this simplfies to, 2 cos2 (θ/2) = − cot(θ/2). −2 sin(θ/2) cos(θ/2) Of course there is an identity, − cot(u) = tan(u − π/2). Altogether, this gives, tan(ψ) = − cot(θ/2) = tan(θ/2 − π/2). Therefore, ψ = (θ − π)/2. Since α equals θ + ψ, this gives, α = (3θ − π)/2. In particular, the angle of the tangent line to the cardioid at θ = π/2 is α = π/4. 3. Arc length in polar coordinates. As discussed previously, the formula for arc length of a parametric curve is, � ds = (dx/dt)2 + (dy/dt)2 dt. In the case of a parametric curve, this becomes a bit simpler. The differentials are, dx = (r� (θ) cos(θ) − r(θ) sin(θ))dθ, dy = (r� (θ) sin(θ) + r(θ) cos(θ))dθ. Squaring gives, (dx)2 = ((r� )2 cos2 (θ) − 2rr� sin(θ) cos(θ) + r2 sin2 (θ))(dθ)2 , (dy)2 = ((r� )2 sin2 (θ) + 2rr� sin(θ) cos(θ) + r2 cos2 (θ))(dθ)2 . 18.01 Calculus Jason Starr Fall 2005 Summing down columns gives, (dx)2 + (dy)2 = [(r� )2 + r2 ](dθ)2 . Taking square roots gives the differential element of arc length for a polar curve, � [r� (θ)]2 + [r(θ)]2 dθ. ds = Example. For the cardioid, r(θ) = a(1 + cos(θ)), the derivative is, r� (θ) = −a sin(θ). Thus, (r� )2 + r2 = a2 (1 + cos(θ))2 + (−a sin(θ))2 = a2 (1 + 2 cos(θ) + cos2 (θ)) + a2 sin2 (θ). This simplifies to, 2a2 (1 + cos(θ)). To simplify this further, write θ = 2(θ/2) and use the double­angle formula to get, 2a2 (1 + cos(2(θ/2))) = 2a2 (1 + cos2 (θ/2) − sin2 (θ/2)) = 2a2 (2 cos2 (θ/2)) = 4a2 cos2 (θ/2). Taking square roots gives, ds = 2a cos(θ/2). Note, this answer is only correct for −π ≤ θ ≤ π. Outside this range, we might have to take the other square root to get a positive number. In particular, the total arc length of the cardioid is, � � θ=π s = ds = 2a cos(θ/2)dθ = 2a (2 sin(θ/2)|π−π = 2a((2) − (−2)). θ=−π Simplifying, the total arc length of the cardioid is, s = 8a. Surface areas of surfaces of revolution can be computed in a similar way. This was only briefly discussed in lecture. Here is a continuation of the previous problem. Example. The top half of the cardioid, r(θ) = a(1 + cos(θ)), 0 ≤ θ ≤ π, is revolved about the x­axis to give a fairly good approximation of the surface of an apple. What is the surface area of this apple? 18.01 Calculus Jason Starr Fall 2005 Since we are revolving about the x­axis, the radius of each slice is y. Therefore the differential element of surface area is, dA = 2πyds. Substituting in y = r(θ) sin(θ) = a(1 + cos(θ)) sin(θ), and substituting in for ds gives, dA = 2π[a(1 + cos(θ)) sin(θ)](2a cos(θ/2)dθ). To simplify this, substitute both, 1 + cos(θ) = 2 cos2 (θ/2), and, sin(θ) = 2 sin(θ/2) cos(θ/2), to get, dA = 4πa2 (2 cos2 (θ/2))(2 sin(θ/2) cos(θ/2)) cos(θ/2)dθ = 16πa2 cos4 (θ/2) sin(θ/2)dθ. Thus the total surface area is, � A= � π 16πa2 cos4 (θ/2) sin(θ/2)dθ. dA = θ=0 To evaluate this integral, substitute, u = cos(θ/2) u(π) = 0, du = −(1/2) sin(θ/2)dθ, u(0) = 1 The new integral is, A = 16πa 2 � u=0 4 u (−2du) = 32πa 2 � u=1 u=1 4 u du = 32πa u=0 2 � �1 u5 �� . 5 �0 This evaluates to give the total surface area of the apple, A = 32πa2 /5. 5. Area of a region enclosed by a polar curve. What is the area of the planar region enclosed by a cardioid? By the same sort of reasoning as for volumes and arc lengths, the differential element of area of the triangular region bounded by the rays θ, θ + dθ and the curve r(θ) is, dA = r(θ)2 dθ. 2 Thus the area enclosed by a polar curve is, � � A = dA = θ=b θ=a r(θ)2 dθ. 2 18.01 Calculus Jason Starr Fall 2005 In particular, the area enclosed by the cardioid is, � 2π 2 a (1 + cos(θ))2 A= dθ. 2 0 This expands to give, a2 2 2π � 1 + 2 cos(θ) + cos(θ)2 dθ. 0 To simplify the last part of the integrand, substitute, cos(θ)2 = 1 + cos(2θ) , 2 to get, a2 2 � 0 2π 1 + cos(2θ) a2 1 + 2 cos(θ) + dθ = 2 4 � 2π 3 + 4 cos(θ) + cos(2θ)dθ. 0 Using the Fundamental Theorem of Calculus, this equals, �2π � � a2 1 3θ + 4 sin(θ) + sin(2θ)�� . 4 2 0 Evaluating gives, A = 3πa2 /2.