College #3 - Questions and Answers

Question 1 We know that a 350 kg sailfish has anacceleration of 0.62 m/s^2 at an angle of 64 degrees north ofeast.find the magnitude and direction of the net force that acts onthe sailfish. Answer a 350 kg sailboat has anacceleration of 0.62 m/s² at an angle of 64 degrees north ofeast.find the magnitude and direction of the net force that acts onthe sailboat. mass m =350 kg acceleration a = 0.62 m/s² angle θ = 64 degrees north of east the magnitude anddirection of the net force that acts on the sailboat. = ? F = ma = 350kg × 0.62 m/s² = 217 N direction of the netforce that acts on the sailboat will be in the directio ofthe acceleration i.e 64degrees ntowards north of east Question 2 Can u find the point on the plane x + 2y + 3z = 13 that is closest to the point (1,1,1). You need to minimize the function Answer 1 Apply Lagrange Multipliers: D = √((x-1)² + (y-1)² + (z-1)²) Now, we can use Constraint Surface: x + 2y + 3z = 13 to g(x, y, z) = x + 2y + 3z - 13 = 0 Formulate Condition (for constant scalar multiplier "λ"): ∇D = λ * ∇g ∇D = [2(x-1)i + 2(y-1)j + 2(z-1)k] ∇g = [1i + 2j + 3k] (1/2) * ∇D = λ * ∇g x - 1 = 1λ y - 1 = 2λ z - 1 = 3λ x - 1 = (y - 1)/2 x - 1 = (z - 1)/3 => y = 2x - 1 and z = 3x - 2 x + 2y + 3z = 13 x + 2(2x - 1) + 3(3x - 2) = 13 x + 4x - 2 + 9x - 6 = 13 14x = 21 x = 3/2 x = 3/2 y = 2(3/2) - 1 = 2 z = 3(3/2) - 2 = 5/2 Answer 2 Given x+2y+3z=13 Point (1,1,1) The normal vector to the plane is <1,2,3> The point you seek would have to be multiple of this vector added to (1,1,1) ..P = (1, 1, 1) + c < 1,2,3 > = (1+c,1+2c,1+3c) But this point has to satisfy the plane's equation. (1+c)+2(1+2c)+3(1+3c)=13 1+c+2+4c+3+9c=13 14c=13-6 c = 7/14 = 1/2 .. the point P is (1 + 1/2, 1 + 21/2, 1 + 31/2) = (3/2, 2, 5/2) Question 3 The most common form of drag experienced by vehicles (bikes, cars, planes, etc.) at operating speeds is called form drag. Itincreases quadratically with velocity (essentially because theamount of air you run into increase with v and so does the amount of force you must exert on each small volume of air). Thus F_drag = CdAv² where A is the cross-sectional area of the vehicle and Ca is called the coefficient of drag. Part A: Think about a car moving at a constant speed. v. Find the power dissipated by form drag. Express your answer in terms of Ca, A, and speed v. Part B: A certain car has an engine that provides a maximum power Po. Suppose that the maximum speed of thee car, vo, is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power P₁ is 10 percent greater than the original power (P₁ = 110%Po). Assume the following: The top speed is limited by air drag. The magnitude of the force of air drag at these speeds is proportional to the square of the speed. By what percentage, (v1 - vo) / vo, is the top speed of the car increased? Express the percent increase in top speed numerically to two significant figures. Answer Part A Power, P = F_drag * v Since F_drag and v are opposite therefore cos(180°) = -1 P = - F_drag * v = -(Cd * A * v²) * v P = -Cd * A * v³ Part B In magnitude P₀ = C_d * A * v₀³ P₁ = 110% of P₀ P₁ = 1.10 * P₀ Also P₁ = C_d * A * v₁³ [in magnitude] C_d * A * v₁³ = C_d * A * v₀³ * (110/100) v₁³ = (110/100) * v₀³ v₁ = (11/10)^(1/3) * v₀ v₁ ≈ 1.0323 * v₀ (v₁ - v₀) / v₀ = (1.0323 * v₀ - v₀) / v₀ = 0.0323 % increase = 3.23 % ≈ 3.2 % [in two significant figures] Question 4 The nth Taylor polynomial Tn(x) for f(x) = ln(1-x) is the base at zero? Find the value of n such that Taylor's inequality guarantees the lowest value |ln(x) – ln(1 - x) | < 0.01 for all x in the interval I = [-1/3 , 1/3] Answer Solution: Given: f(x) = ln(1-x) About b=0: the maclaurins series is: f(x) = ln(1-x) = (-x - x²/2 - x³/3 - x⁴/4 ... ∞) = (-1) * Σ(from n=1 to ∞) xⁿ/n Over [-1/3, 1/3] by the Taylor's inequality: |Rₙ| ≤ M*|x-b|ⁿ⁺¹ /(n+1)! Now: |x - b| = 1/3 and fⁿ⁺¹(x) = -1/(1-x)ⁿ⁺¹ So: fⁿ⁺¹(x) over [-1/3, 1/3] is maximised (n+1)! > +∞ => n > 99 Question 5 Let W be the subspace spanned by the u's, and write y as the sum of a vector in W and a vector orthogonal to W. y = [-1, 1, 3], u1 = [1, 1, 1], u2 = [-1, 1, -2] Answer y_hat = (yu1/u1u1) * u1 + (yu2/u2u2) * u2 yu1 = (-1)(1) + (1)(1) + (3)(1) = -1 + 1 + 3 = 3 u1u1 = (1)(1) + (1)(1) + (1)(1) = 3 yu2 = (-1)(-1) + (1)(1) + (3)(-2) = 1+1-6=-4 u2u2 = (-1)(-1) + (1)(1) + (-2)(-2) = 1+1+4=6 y_hat = (3/3) * [1, 1, 1] + (-4/6) * [-1, 1, -2] = [1, 1, 1] + (-2/3) * [-1, 1, -2] = [1, 1, 1] + [2/3, -2/3, 4/3] = [5/3, 1/3, 7/3] y - y_hat = [-1, 1, 3] - [5/3, 1/3, 7/3] = [-8/3, 2/3, 2/3] Answer: y = [5/3, 1/3, 7/3] + [-8/3, 2/3, 2/3]