Calculus III Exam Key

KEY Math 225: Exam the Second November 8, 2017 You have two hours to complete this closed-book, closed-note, closed colleague exam. You may use a calculator for arithmetic only (trig functions and exponentials are okay, but no plotting and no calculus).   Let (f(x,y)=x^{2}+2x+\sqrt{y}+ye^{x}) (a) Find the tangent plane to (f(x, y)) at the point (0,4). (f(0,4)=0^{2}+2(0)+\sqrt{4}+4e^{0}=6) (f_{x}(0,4)=2x+2+0+ye^{x}|_{(0,4)}=2(0)+2+0+4e^{0}=6.) (f_{y}(0,4)=0+0+\frac{1}{2\sqrt{y}}+e^{x}|_{(0,4)}=\frac{1}{4}+e^{0}=\frac{5}{4}.) Plane: (z=6+6(x-0)+\frac{5}{4}(y-4)) (b) Use the tangent plane to approximate (f(0.04, 3.98)). (f(0.04,3.98)\approx6+6(0.04-0)+\frac{5}{4}(3.98-4)) (=6+.24+\frac{5}{4}(-.02)) (=6.215) 2. Let (f(x,y)=\sqrt{x^{2}+y^{2}}) (a) Find (\nabla f) at the point (-5, 12). (f_x=\frac{1}{2}(x^2+y^2)^{-\frac{1}{2}}(2x)=\frac{x}{\sqrt{x^2+y^2}}) (f_y=\frac{1}{2}(x^2+y^2)^{-\frac{1}{2}}(2y)=\frac{y}{\sqrt{x^2+y^2}}) (\nabla f=\langle\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\rangle|_{(-5,12)}=\langle\frac{-5}{13},\frac{12}{13}\rangle) (b) Find (D_uf) as we move from (-5, 12) towards (-3, 15) (Please leave your answer as a radical). (-5, 12) -> (-3, 15) = <2, 3> (\vec{u}=\langle\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}}\rangle) (\langle\frac{-5}{13},\frac{12}{13}\rangle\cdot\langle\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}}\rangle=\frac{-10+36}{13\sqrt{13}}=\frac{2}{\sqrt{13}}) (c) Argue that, at any point except the origin, (\nabla f) is a unit vector. For a bonus, use the graph of f to explain why. (|\langle\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\rangle|) so long as ((x,y)\neq(0,0)) (=\sqrt{(\frac{x^2}{x^2+y^2})+(\frac{y^2}{x^2+y^2})}=\sqrt{\frac{x^2+y^2}{x^2+y^2}}=1) 3. Let f(x, y) = 6xy - x² - y³. Find and classify the critical points of f. fx = 6y - 2x = 0 --> x = 3y fy = 6x - 3y² = 0 6(3y) - 3y² = 0 18y - 3y² = 0 3y(6 - y) = 0 y = 0 y = 6 (y = 0, D = fxx * fyy - (fxy)² 6 x = 0 x = 18 (18,6) = (-2)(-6y) - (6)² Points (0,0) D = 20 D > 0 12y - 36 Saddle [fxx < 0, max 4. Student learning learning is dependent on two factors: Sleep (S, hours per night) and Caffeine (C, doses per day). The equation L(S,C) = 3S² + 2C² + 4C is bound by the constraint that S² + C² = 40 How much sleep and how much caffeine should a student get in order to maximize learning. L = 3S² + 2c² + 14C S² + c² = 40 Ls = 6S = 2λS 5:0 λ = 3 Lc = 4c + 4 = 2λc C = √40 L = 0 + 80 + 8√10 ≈ 104. 4c + 4 = 6c λ = 3 2c = 4 c = 2 s ⇒ S² · 4 = 40 s = 6 L = 108 + 8 + 8 = 124 may max or min 5. We can define the normal line to a surface of the form F(x, y, z) = k at the point (a, b, c) to be the line through (a, b, c) in the direction of ∇F. (a) Find the normal line to the surface (xy + xz + yz = 14) at the point (2, 1, 4). ∇F = |_(2,1,4) = <5, 6, 3> (\vec{l}(t) = <2, 1, 4> + t<5, 6, 3>) (b) Prove that the normal line at any point on a sphere goes through the origin. Sphere: (x^2 + y^2 + z^2 = a^2) ∇F = <2x, 2y, 2z>|_(x₀, y₀, z₀) = <2x₀, 2y₀, 2z₀> (\vec{l}(t) = + t<2x₀, 2y₀, 2z₀>) when t = -1/2 l = <0, 0, 0> 6. A function f(x, y) is harmonic if fxx + fyy = 0. Prove that f(x, y) = e^x sin(y) is harmonic. f(x, y) = e^x sin(y) fx = e^x sin(y) fy = e^x cos(y) fxx = e^x sin(y) fyy = -e^x sin(y) fxx + fyy = e^x sin(y) + (-e^x sin(y)) = 0. 7. Give an overestimate of the integral ∫∫(x²-2x+y²) dy dx using 4 subdivisions. rectangle divided into 4 subrectangles with the following numbers labeled inside each subrectangle: 4, 12, 9, 12 Overestimate: 1·2·4 + 2·2·12 + 2·2·9 + 12·2·12 = 26·2 = 52 8. Determine [\iint_{R}y\sqrt{xy+1}dA] where (R=[0,4]\times[0,6]) [\int_{0}^{6}\int_{0}^{4}y\sqrt{xy+1}dxdy] = [\int_{0}^{6}\frac{2}{3}[\frac{2}{5}(4y+1)^{5/2}-y]dy] = [\frac{2}{15}[(4y+1)^{5/2}-y^2]|_{0}^{6}] = [\frac{2}{15}(25^{5/2}-4)-\frac{2}{3}] = 2127/3 9. Consider [\iint_{R}xy dA] where R the region in the first quadrant bound by (y=x^2) and (y=4.) Set this integral up using both orders of integration, and verify that your answers are the same. [\int_{0}^{4}\int_{0}^{\sqrt{y}}xydxdy] = [\int_{0}^{4}\frac{x^2y}{2}|_{0}^{\sqrt{y}}dy] = [\int_{0}^{4}\frac{y^2}{2}+\frac{y^{3/2}}{2}dy] = [\frac{y^3}{6}+\frac{y^{5/2}}{5}|_{0}^{4}] = 24 + 64/5 = 16.8 [\int_{0}^{2}\int_{x^2}^{4}xydydx] = [\int_{0}^{2}\frac{xy^2}{2}|_{x^2}^{4}dx] = [\int_{0}^{2}8x-x^5/2dx] = [4x^2-x^6/6|_{0}^{2}] = 24 - 7.2 = 16.8