1. **Problem statement:** Find the linear approximation (linearization) of the functions at the given points. --- ### (a) Function: $f(x,y) = x \sqrt{y}$ at point $P(3,3)$ 2. Compute partial derivatives: $$f_x = \frac{\partial}{\partial x} (x \sqrt{y}) = \sqrt{y}$$ $$f_y = \frac{\partial}{\partial y} (x \sqrt{y}) = x \cdot \frac{1}{2\sqrt{y}} = \frac{x}{2\sqrt{y}}$$ 3. Evaluate partial derivatives at $P(3,3)$: $$f_x(3,3) = \sqrt{3}$$ $$f_y(3,3) = \frac{3}{2\sqrt{3}} = \frac{3}{2 \times \sqrt{3}} = \frac{3}{2 \times 1.732} \approx 0.866$$ 4. Evaluate function at $P(3,3)$: $$f(3,3) = 3 \times \sqrt{3} = 3 \times 1.732 = 5.196$$ 5. Write the linear approximation formula: $$L(x,y) = f(a,b) + f_x(a,b)(x - a) + f_y(a,b)(y - b)$$ Substitute $a=3$, $b=3$: $$L(x,y) = 5.196 + \sqrt{3}(x - 3) + 0.866(y - 3)$$ --- ### (b) Function: $f(x,y,z) = \sqrt{x^2 + y^2 + z^2}$ at point $P(4,2,5)$ 6. Compute partial derivatives: $$f_x = \frac{\partial}{\partial x} \sqrt{x^2 + y^2 + z^2} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}$$ $$f_y = \frac{y}{\sqrt{x^2 + y^2 + z^2}}$$ $$f_z = \frac{z}{\sqrt{x^2 + y^2 + z^2}}$$ 7. Evaluate the denominator at $P(4,2,5)$: $$\sqrt{4^2 + 2^2 + 5^2} = \sqrt{16 + 4 + 25} = \sqrt{45} = 3\sqrt{5} \approx 6.708$$ 8. Evaluate partial derivatives at $P(4,2,5)$: $$f_x(4,2,5) = \frac{4}{6.708} \approx 0.596$$ $$f_y(4,2,5) = \frac{2}{6.708} \approx 0.298$$ $$f_z(4,2,5) = \frac{5}{6.708} \approx 0.745$$ 9. Evaluate function at $P(4,2,5)$: $$f(4,2,5) = 6.708$$ 10. Write the linear approximation formula: $$L(x,y,z) = f(a,b,c) + f_x(a,b,c)(x - a) + f_y(a,b,c)(y - b) + f_z(a,b,c)(z - c)$$ Substitute $a=4$, $b=2$, $c=5$: $$L(x,y,z) = 6.708 + 0.596(x - 4) + 0.298(y - 2) + 0.745(z - 5)$$