MAT 272 Calculus and Analytic Geometry III Sample solutions for test 2 1. For f (x, y) = x3 − 3x2 + y 3 − 3y, calcufxx (x, y) = 6x − 6 late the 1st partials fx (x, y) = 3x2 − 6x, and 2 fyy (x, y) = 6y fy (x, y) = 3y − 3. Their common zeros are the fxy (x, y) = 0 four critical points (2, ±1) and (0, ±1). Calcund late the 2 partial derivatives, and the discrim- D(x, y) = 36(x − 1)y inant, and evaluate at all critical points. (0, −1) −6 −6 0 + loc max 10/11/2011 (0, 1) −6 6 0 − saddle (2, −1) 6 −6 0 − saddle (2, 1) 6 6 0 + loc min 2. Write the constraint as the zero-level set of the function g(x, y) = x + y − 5. Form the Lagrangian L(λ, x, y) = f (x, y) + λg(x, y) and calculate the first partials Lλ = x + y − 5, Lx = 2(x − 2) + λ,Ly = 2(y − 1) + λ. and equate all to zero. The resulting system of three linear equations has the unique solution (λc , xc , yc ) = (−2, 3, 2). Since the gradient of g is never zero, and f is convex, this critical point must be the unique global minimum. At any local minimum, the constraint curve must be tangent to the level sets of f critical, which means that the gradients of f and g must be scalar multiples of each other. 3. f (a+∆x,b+∆y)−f (a+∆x,b) is the slope of the secant lines DL. ∆y ∂f ∂y (a + ∆x, b) is the slope of the tangent line DJ. ∂f ∂f ∂x (a, b) · ∆x + ∂y (a, b) · ∆y is the length of the line segment (a,b) √ The slope of the line segment AL is f (a+∆x,b+∆y)−f . (∆x)2 +(∆y)2 ~ (a, b) · √ h∆x,∆yi . The slope of the line segment AI is ∇f (∆x)2 +(∆y)2 HI. 4. For a twice differentiable function z = f (r, θ) and polar coordinates the partial derivatives simplify to −y p 2 ∂r ∂ ∂θ ∂ θ θ x2 + y 2 = √ 2x2 2 = −r cos = cos θ and ∂x = ∂x arctan xy = x y 2 = x2−y = −rrsin = sinr θ . 2 ∂x = ∂x r +y 2 2 x +y 1+( x ) 2 ∂ ∂θ Using these results the chain rule yields zx = zr · cos θ − zθ sinr θ . and ∂x (cos θ) = − sin θ · ∂x = sinr θ . 5. At P the contour of f is parallel to the y-axis, and hence the gradient of f is parallel to the x-axis. Since it points up, at P it points East. From above ∂f ∂y (P ) = 0. Using a central difference quotient to estimate the other f (175,350)−f (100,350) partial yields ∂f = 17,000−13,000 ≈ 53.33. ∂x (P ) ≈ 175−100 75 The average slope between T and B is approximately m = 2,000−27,000 ≈ −76.72 725−400 (meters per kilometer), i.e., an downhill slope of about 8%. This occurs fre quently on freeways, but typically warrants a warning sign (check brakes, etc.). The cross-section from T to B does not confirm the statement that the “the flanks [exhibit] a concave upward profile”. as stated, e.g., on http://en.wikipedia.org/wiki/Olympus_Mons.