Graded Assignment 10

Step 1: Determine Prior Parameters (Blanks A, B, C)

Concept: For a Beta distribution $\text{Beta}(\alpha, \beta)$:

1. $\text{Mean } \mu = \frac{\alpha}{\alpha + \beta}$
2. $\text{Variance } \sigma^2 = \frac{\mu(1-\mu)}{\alpha + \beta + 1}$

Calculation:

• Given $\mu = 0.4$ and $\sigma^2 = 0.02$.
• Using the variance formula: $$0.02 = \frac{0.4(1 - 0.4)}{\alpha + \beta + 1}$$ $$0.02 = \frac{0.24}{\alpha + \beta + 1}$$ $$\alpha + \beta + 1 = \frac{0.24}{0.02} = 12$$ $$\alpha + \beta = 11$$
• Using the mean formula: $$\frac{\alpha}{11} = 0.4 \implies \alpha = 4.4$$
• Finding $\beta$: $$\beta = 11 - 4.4 = 6.6$$

Matches from Option List:

• A ($\alpha$) = 4.4 [Option 12]
• B ($\beta$) = 6.6 [Option 3]
• C (Prior Distribution) = Beta(4.4, 6.6) [Option 5]

Answers:

• Q11 (A): 12
• Q12 (B): 3
• Q13 (C): 5

Step 2: Bayesian Update (Blanks D, E, F)

Concept:

• Likelihood (Binomial/Bernoulli): $L(p) \propto p^{\text{successes}}(1-p)^{\text{failures}}$
• Prior (Beta): $\pi(p) \propto p^{\alpha-1}(1-p)^{\beta-1}$

Calculation:

• Likelihood (D): $n=40, k=28$, failures $= 12$. $$L(p) = p^{28}(1-p)^{12}$$ [Matches Option 6]
• Prior Term (E): $\alpha=4.4, \beta=6.6$. $$\pi(p) = p^{4.4-1}(1-p)^{6.6-1} = p^{3.4}(1-p)^{5.6}$$ [Matches Option 13]
• Posterior Density (F): Multiply D and E. Add exponents. $$p^{28+3.4}(1-p)^{12+5.6} = p^{31.4}(1-p)^{17.6}$$ [Matches Option 1]

Answers:

• Q14 (D): 6
• Q15 (E): 13
• Q16 (F): 1

Step 3: Posterior Results (Blanks G, H)

Concept: The posterior density $p^{31.4}(1-p)^{17.6}$ corresponds to a Beta distribution with parameters $\alpha_{new} - 1 = 31.4$ and $\beta_{new} - 1 = 17.6$.

Calculation:

• Posterior Distribution (G): $\alpha_{post} = 32.4$, $\beta_{post} = 18.6$. $$\text{Beta}(32.4, 18.6)$$ [Matches Option 4]
• Posterior Mean (H): $$\text{Mean} = \frac{\alpha_{post}}{\alpha_{post} + \beta_{post}} = \frac{32.4}{32.4 + 18.6} = \frac{32.4}{51} \approx 0.635$$ Rounding to two decimal places gives 0.64. [Matches Option 2]

Answers:

• Q17 (G): 4
• Q18 (H): 2

1) Critical Points of $f(x, y) = 3x^2y + y^3 - 3x^2 - 3y^2 + 2$

Concept: Critical points occur where the gradient $\nabla f = \langle f_x, f_y \rangle$ is $\langle 0, 0 \rangle$.

Solution:

1. Partial Derivatives: $$f_x = 6xy - 6x = 6x(y - 1)$$ $$f_y = 3x^2 + 3y^2 - 6y$$
2. Solve $f_x = 0$: $6x(y-1) = 0 \implies x = 0 \text{ or } y = 1$.
3. Case 1: $x = 0$ Substitute into $f_y = 0$: $3(0)^2 + 3y^2 - 6y = 0 \implies 3y(y-2) = 0$. Points: $(0, 0)$ and $(0, 2)$.
4. Case 2: $y = 1$ Substitute into $f_y = 0$: $3x^2 + 3(1)^2 - 6(1) = 0 \implies 3x^2 - 3 = 0 \implies x = \pm 1$. Points: $(1, 1)$ and $(-1, 1)$.

Answer: The set of critical points includes $(0, 0), (0, 2), (1, 1)$.

2) Directional Derivative of $f(x, y) = \frac{xy^2}{x^2+y^4}$ at $(0,0)$

Concept: Using the limit definition of the directional derivative for a unit vector $u = (u_1, u_2)$.

Solution:

1. Substitute points: $$\frac{f(tu_1, tu_2)}{t} = \frac{1}{t} \left( \frac{(tu_1)(tu_2)^2}{(tu_1)^2 + (tu_2)^4} \right) = \frac{t^3 u_1 u_2^2}{t(t^2 u_1^2 + t^4 u_2^4)}$$ $$= \frac{t^3 u_1 u_2^2}{t^3(u_1^2 + t^2 u_2^4)} = \frac{u_1 u_2^2}{u_1^2 + t^2 u_2^4}$$
2. Take limit as $t \to 0$: $$\lim_{t \to 0} \frac{u_1 u_2^2}{u_1^2 + t^2 u_2^4} = \frac{u_1 u_2^2}{u_1^2} = \frac{u_2^2}{u_1}$$ Note: This is valid only if $u_1 \neq 0$.

Answer: The directional derivative… is $\frac{u_2^2}{u_1}$, where $u_1$ is non-zero. (Option 2)

3) Directions of no change for $f(x, y, z) = 2x^3 + y^2 - z^3$ at $(1, 1, 1)$

Concept: “No change” means the directional derivative is 0. This happens when the direction vector is orthogonal (perpendicular) to the gradient vector.

Solution:

1. Gradient $\nabla f$: $\langle 6x^2, 2y, -3z^2 \rangle$.
2. At $(1, 1, 1)$: $\nabla f = \langle 6, 2, -3 \rangle$.
3. Check orthogonality ($\nabla f \cdot \mathbf{v} = 0$) for options:
   • $\langle 1/\sqrt{5}, 0, 2/\sqrt{5} \rangle \cdot \langle 6, 2, -3 \rangle = \frac{6 - 6}{\sqrt{5}} = 0$. (Yes)
   • $\langle 0, 3/\sqrt{13}, 2/\sqrt{13} \rangle \cdot \langle 6, 2, -3 \rangle = \frac{6 - 6}{\sqrt{13}} = 0$. (Yes)
   • $\langle 2/\sqrt{17}, -3/\sqrt{17}, 2/\sqrt{17} \rangle \cdot \langle 6, 2, -3 \rangle = \frac{12 - 6 - 6}{\sqrt{17}} = 0$. (Yes)

Answer: The vectors $(1/\sqrt{5}, 0, 2/\sqrt{5})$, $(0, 3/\sqrt{13}, 2/\sqrt{13})$, and $(2/\sqrt{17}, -3/\sqrt{17}, 2/\sqrt{17})$ are correct.

4) Linear Approximation of $f(x, y) = y e^x - \frac{1}{4}(x^2+y^2)$ at $(0, 1)$

Concept: $L(x, y) = f(a, b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$.

Solution:

1. $f(0, 1) = 1 \cdot e^0 - \frac{1}{4}(0+1) = 1 - 0.25 = 0.75$.
2. $f_x = y e^x - \frac{x}{2} \implies f_x(0,1) = 1(1) - 0 = 1$. ($A=1$)
3. $f_y = e^x - \frac{y}{2} \implies f_y(0,1) = 1 - 0.5 = 0.5$. ($B=0.5$)
4. Form: $L = 0.75 + 1(x) + 0.5(y-1) = x + 0.5y + 0.25$. So $C = 0.25$.
5. Calculate $A + 2B + 4C$: $$1 + 2(0.5) + 4(0.25) = 1 + 1 + 1 = 3$$

Answer: 3

5) Cardinality of set S for $f(x, y) = 2\sqrt{x^2+4y}$

Concept: We need unit vectors $\mathbf{u}$ such that $D_u f(-2, 3) = 0$. This implies $\mathbf{u} \perp \nabla f$. In 2D, there are always exactly 2 unit vectors perpendicular to any non-zero gradient.

Solution:

1. $\nabla f = \langle \frac{2x}{\sqrt{x^2+4y}}, \frac{4}{\sqrt{x^2+4y}} \rangle$.
2. At $(-2, 3)$: Denom = $\sqrt{4+12} = 4$. $\nabla f = \langle -4/4, 4/4 \rangle = \langle -1, 1 \rangle$.
3. Perpendicular vectors to $\langle -1, 1 \rangle$ are parallel to $\langle 1, 1 \rangle$.
4. Unit vectors: $\pm (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. Total count is 2.

Answer: 2

6) Estimate $f(1.2, 0.9)$ using Linear Approximation for $f = xy e^x$

Solution:

1. Point $(1, 1)$. $f(1, 1) = e$.
2. $f_x = y(e^x + xe^x)$. At $(1, 1) \to 1(e+e) = 2e$.
3. $f_y = x e^x$. At $(1, 1) \to e$.
4. $L(x, y) = e + 2e(x-1) + e(y-1)$.
5. Input $x=1.2, y=0.9$: $$L(1.2, 0.9) = e + 2e(0.2) + e(-0.1) = e + 0.4e - 0.1e = 1.3e$$
6. Result is $\beta e$, so $\beta = 1.3$.

Answer: 1.3

Questions 9, 10, 11, 12: Max Directional Derivative

Concept: The maximum directional derivative at a point occurs in the direction of the gradient $\nabla f$. Its value is the magnitude $|\nabla f|$.

Solutions:

• Q9: For $T = e^{-(x^2+y^2+z^2)}$, $\nabla T = -2 \langle x, y, z \rangle e^{-(x^2+y^2+z^2)}$.

  • Max rate is magnitude: $2 \sqrt{x^2+y^2+z^2} e^{-(x^2+y^2+z^2)}$. (Matches first statement option).
  • Direction is $\nabla T$, which points opposite to the position vector. Normalized: $-\frac{\langle x, y, z \rangle}{\sqrt{x^2+y^2+z^2}}$. (Matches third statement option).
  • Answer: Options 1 and 3 are true.

• Q10: $f_1 = y^6 e^{5x}$. $\nabla f$ at $(0,0)$.

  • $f_x = 5y^6 e^{5x} \to 0$. $f_y = 6y^5 e^{5x} \to 0$.
  • Answer: 0

• Q11: $f_2 = 5 - 4x^2 + 2x - 6y^2$.

  • $f_x = -8x + 2 \to 2$. $f_y = -12y \to 0$.
  • Magnitude $|\langle 2, 0 \rangle| = 2$.
  • Answer: 2

• Q12: $f_3 = 6x \sin(6x) + 6y \cos(6y)$.

  • $f_x \to 0$ (since $\sin(0)=0$ and $x=0$).
  • $f_y = 6\cos(6y) - 36y\sin(6y) \to 6(1) - 0 = 6$.
  • Magnitude $|\langle 0, 6 \rangle| = 6$.
  • Answer: 6

13) Tangent Plane to $z = 5 - x^4 - y^2$ at $(1, 1, 3)$

Solution:

1. Let $F(x, y, z) = x^4 + y^2 + z - 5 = 0$.
2. Normal vector $\mathbf{n} = \langle 4x^3, 2y, 1 \rangle$.
3. At $(1, 1, 3)$: $\mathbf{n} = \langle 4, 2, 1 \rangle$.
4. Plane eq: $4(x-1) + 2(y-1) + 1(z-3) = 0$. $4x + 2y + z - 4 - 2 - 3 = 0 \implies 4x + 2y + z = 9$. $z = -4x - 2y + 9$.
5. $A = -4, B = -2, C = 9$.
6. $C - A - B = 9 - (-4) - (-2) = 9 + 4 + 2 = 15$.

Answer: 15

8) Rate of change of $T$ at $(1, 0, 0)$ toward $(8, 6, 0)$

Solution:

1. Gradient $\nabla T$ at $(1, 0, 0)$ is $\langle -2/e, 0, 0 \rangle$.
2. Vector $\mathbf{v}$ from $(1,0,0)$ to $(8,6,0)$ is $\langle 7, 6, 0 \rangle$.
3. Unit vector $\mathbf{u} = \frac{\langle 7, 6, 0 \rangle}{\sqrt{7^2+6^2}} = \frac{\langle 7, 6, 0 \rangle}{\sqrt{85}}$.
4. $A = \nabla T \cdot \mathbf{u} = \frac{-2}{e} \cdot \frac{7}{\sqrt{85}} = \frac{-14}{e\sqrt{85}}$.
5. Value asked: $10Ae = 10 \left( \frac{-14}{e\sqrt{85}} \right) e = \frac{-140}{\sqrt{85}}$.

Answer: $\frac{-140}{\sqrt{85}}$ (approx -15.18)

14) Find $f_x(1, 1)$ given tangent line info

Concept: The slope of the tangent line in the direction of the x-axis (direction $\langle 1, 0 \rangle$) is the partial derivative with respect to x, $f_x$.

Solution:

1. Tangent line passes through contact point $(1, 1, f(1,1)) = (1, 1, 3)$.
2. It also passes through $(0, 1, 5)$.
3. The change in $y$ is 0, so this line represents the slope in the x-direction.
4. Slope $m = \frac{\Delta z}{\Delta x} = \frac{5 - 3}{0 - 1} = \frac{2}{-1} = -2$.
5. Therefore, $f_x(1, 1) = -2$.

Answer: -2