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MIT 18 01 - Lecture Notes

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18.01 Calculus Jason Starr Fall 2005 Lecture 11. October 4, 2005 Homework. Problem Set 3 Part I: (g) and (h). Practice Problems. Course Reader: 2E-4, 2E-8, 2E-9. 1. Related rates. A situation that arises often in practice is that two quantities, say x and y, depend on a third independent variable, say t. The quantities x and y are related through some constraint. Using the constraint, if the rate-of-change dx/dt is known, the rate-of-change dy/dt can be inferred. Example. For a spring displaced x units from equilibrium, Hooke’s law implies the potential energy of the spring is, 1 P = kx 2 ,2 where k is a constant with units kg/s2 . At some moment t = T , a spring is displaced 5cm from equilibrium and has velocity 5cm/s. In terms of the spring constant k, describe the rate-of-change of the potential energy at t = T .18.01 Calculus Jason Starr Fall 2005 Implicitly differentiating the equation with respect to t gives, using the chain rule, dP 1 dx dx = k(2x) = kx . dt 2 dt dt So, at time t = T , dP dx (T ) = kx(T ) (T ) = k(5)(5)cm 2/s = dt dt 252/s.kcm2. Method for solving related-rates problems. Many of thes e steps apply to any word-problem in mathematics. (i) Identify the independent variable. In the example, this is t. (ii) Label all constants. In the example, k is a constant. (iii) Label all dependent variables. In the example, x and P are dependent variables. (iv) Draw a diagram and carefully label it. (v) Write the given rate-of-change and the unknown rate-of-change. In the example, dx/dt(T ) is given as 5cm/s, and dP /dt is unknown. (vi) Using the diagram and any other information, find constraints among the dependent variables. In the example, this is the equation P = kx2/2. (vii) Implicitly differentiate the constraint equations with respect to the independent variable. In the example, this gives dP/dt = kxdx/dt. (viii) Substitute in all known quantities and solve for the unknown rate-of-change. In the example, dP/dt(T ) equals 25kcm2/s. Example. A state trooper waits a distance a from a highway for passing speeders. The speed limit is 60mph. The trooper aims her radar gun at an angle of π/4 to the road. The radar registers a passing car moving away from the trooper at a speed of 50mph. Should the trooper ticket the driver? The independent variable is time t. The constants are the distance a and the angle θ = π/4. Label a coordinate system with the trooper at the origin and the highway equal to the line y = a. Label the position of the car along the highway as x, moving in the positive direction. Denote by r the distance of the car from the trooper. Then x and r are dependent variables. The rate-of-change dr/dt(T ) is given as 50mph. The unknown rate-of-change is dx/dt(T ). The constraint is the Pythagorean theorem, 2 2 2 r = x + y .18.01 Calculus Jason Starr Fall 2005 Implicit differentiation with respect to t yields, dr dx dx 2r = 2x + 0 = 2x . dt dt dt At time t = T , x(T ) equals a, because the angle θ is π/4. Thus r(T ) equals √2a. Substituting in gives, dx 2(√2a)50 = 2(a) (T ). dt Solving gives, dx (T ) = √250 ≈ 71mph. dt So the trooper should ticket the driver. Example. A point on the x-axis moves away from the origin. There is an angle θ subtended by the point and the unit circle with equation x2 + y2 = 1. In other words, standing at the point (x, 0) and staring at the circle, θ is the angle of your field-of-vision occupied by the circle. At a moment t = T , the point is at the position (2, 0) and moving with velocity v. What is the rate-of-change of θ at t = T ? The independent variable is time t. There is no constant. The dependent variables are the x-coordinate of the point, x(t), and the angle θ(t). The rate-of-change dx/dt(T ) is given to be v. The rate-of-change dθ/dt is unknown. The constraint is somewhat tricky. There are two tangent lines to the circle containing (x, 0). These are the tangent lines to points (a, +b) and (a, −b) on the circle. Because the tangent line to the circle at (a, b) is perpendicular to the radius through (a, b), the triangle with vertices (0, 0), (a, b) and the point (x, 0) is a right triangle. The angle of the triangle at (x, 0) is θ/2. Since the radius has length 1 and the hypotenuse has length x, the constraint is, 1 sin(θ) = . x Implicit differentiation with respect to t gives, d sin(θ) dθ d(x−1) dx = dθ dt dx dt , or, cos(θ) dθ dt = −1 x2 dx dt . Since x(T ) equals 2, sin(θ(T )) = 1/2, and thus cos(θ(T )) equals √3/2. Plugging in gives, √3 dθ (T ) = −1 v = −v. 2 dt (2)2 418.01 Calculus Jason Starr Fall 2005 Solving gives, dθ (T ) = dt − (2√v/ 3). 3. Another applied max/min problem. As review for Exam 2, this is another applied max/min 2problem. A trapezoid is inscribed inside the upper unit semicircle, x2 + y = 1, y ≥ 0. The base of the trapezoid is the diameter of the semicircle lying on the x-axis. The top of the trapezoid is parallel to the x-axis joining (−x, y) to (x, y) for a point (x, y) on the unit circle in the first quadrant. What is the maximal area enclosed by such a trapezoid? The parameters are x and y. The height of the trap ezoid is y. The area of a trapezoid is the product of the height with the average of the parallel sides. Thus, (2 + 2x)A = y = (x + 1)y. 2 This is the quantity to be maximized. There is a constraint among the parameters, 2 x + y 2 = 1. Also, since (x, y) is in the first quadrant, 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. There are at least 3 ways to proceed. The most direct is to solve for y in terms of x, y = √1 − x2 . Substituting this into the equation for A gives, 2A(x) = (x + 1)√1 − x . Differentiating gives, dA −2x 1 2 = √1 − x2 + (x + 1)2√1 − x= √1 − x((1 − x 2) − (x + x)) = −(2x2 + x − 1) . 2 2 2dx √1 − xBecause the quadratic polynomial 2x2 + x − 1 factors as, 2x 2 + x − 1 = (2x − 1)(x + 1), the critical points of A are x = −1 and x = 1/2. But x = −1 does not give a point in the first quadrant. Thus A is maximized either at one of the endpoints x = 0, x = 1 or at the critical point x = 1/2. Plugging in …

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