Subjects multivariable calculus

Function Values

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Function Values


1. Find the function values for \( f(x,y) = x^2 + xy^3 \). 1.a. Calculate \( f(0,0) = 0^2 + 0 \times 0^3 = 0 \). 1.b. Calculate \( f(-1,1) = (-1)^2 + (-1) \times 1^3 = 1 - 1 = 0 \). 1.c. Calculate \( f(2,3) = 2^2 + 2 \times 3^3 = 4 + 2 \times 27 = 4 + 54 = 58 \). 1.d. Calculate \( f(-3,-2) = (-3)^2 + (-3) \times (-2)^3 = 9 + (-3) \times (-8) = 9 + 24 = 33 \). 2. Find values for \( f(x,y) = \sin(xy) \). 2.a. Calculate \( f(2, \pi/6) = \sin(2 \times \pi/6) = \sin(\pi/3) = \frac{\sqrt{3}}{2} \). 2.b. Calculate \( f(-3, \pi/12) = \sin(-3 \times \pi/12) = \sin(-\pi/4) = -\frac{\sqrt{2}}{2} \). 2.c. Calculate \( f(\pi, 1/4) = \sin(\pi \times 1/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2} \). 2.d. Calculate \( f(-\pi/2, -7) = \sin((-\pi/2) \times (-7)) = \sin(7\pi/2) = \sin(\pi/2) = 1 \) since sine is periodic with period \(2\pi\). 3. Evaluate \( f(x,y,z) = \frac{x-y}{y^2 + z^2} \). 3.a. Calculate \( f(3,-1,2) = \frac{3 - (-1)}{(-1)^2 + 2^2} = \frac{4}{1 + 4} = \frac{4}{5} = 0.8 \). 3.b. Calculate \( f(1, \frac{1}{2}, -\frac{1}{4}) = \frac{1 - \frac{1}{2}}{(\frac{1}{2})^2 + (-\frac{1}{4})^2} = \frac{\frac{1}{2}}{\frac{1}{4} + \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{5}{16}} = \frac{1}{2} \times \frac{16}{5} = \frac{8}{5} = 1.6 \). 3.c. Calculate \( f(0, -\frac{1}{3}, 0) = \frac{0 - (-\frac{1}{3})}{(-\frac{1}{3})^2 + 0^2} = \frac{\frac{1}{3}}{\frac{1}{9}} = 3 \). 3.d. Calculate \( f(2, 2, 100) = \frac{2 - 2}{2^2 + 100^2} = \frac{0}{4 + 10000} = 0 \). 4. Evaluate \( f(x,y,z) = \sqrt{49 - x^2 - y^2 - z^2} \). 4.a. Calculate \( f(0, 0, 0) = \sqrt{49 - 0 - 0 - 0} = \sqrt{49} = 7 \). 4.b. Calculate \( f(2, -3, 6) = \sqrt{49 - 2^2 - (-3)^2 - 6^2} = \sqrt{49 - 4 - 9 - 36} = \sqrt{0} = 0 \). 4.c. Calculate \( f(-1, 2, 3) = \sqrt{49 - (-1)^2 - 2^2 - 3^2} = \sqrt{49 - 1 - 4 - 9} = \sqrt{35} \approx 5.916 \). 4.d. Calculate \( f\left( \frac{4}{\sqrt{2}}, \frac{5}{\sqrt{2}}, \frac{6}{\sqrt{2}} \right) = \sqrt{49 - \left(\frac{4}{\sqrt{2}}\right)^2 - \left(\frac{5}{\sqrt{2}}\right)^2 - \left(\frac{6}{\sqrt{2}}\right)^2} \). Compute inside the root: $$ \left(\frac{4}{\sqrt{2}}\right)^2 = \frac{16}{2} = 8, \quad \left(\frac{5}{\sqrt{2}}\right)^2 = \frac{25}{2} = 12.5, \quad \left(\frac{6}{\sqrt{2}}\right)^2 = \frac{36}{2} = 18. $$ Sum these: \( 8 + 12.5 + 18 = 38.5 \), then $$ \sqrt{49 - 38.5} = \sqrt{10.5} \approx 3.240. $$ **Final answers:** 1.a \(0\), 1.b \(0\), 1.c \(58\), 1.d \(33\) 2.a \(\frac{\sqrt{3}}{2} \), 2.b \( -\frac{\sqrt{2}}{2} \), 2.c \( \frac{\sqrt{2}}{2} \), 2.d \(1\) 3.a \(0.8\), 3.b \(1.6\), 3.c \(3\), 3.d \(0\) 4.a \(7\), 4.b \(0\), 4.c \(\sqrt{35} \approx 5.916\), 4.d \( \sqrt{10.5} \approx 3.240 \).