Graded Assignment 9

Questions 1 - 4: Matching Functions to Graphs

Concept: To match the functions, we analyze their shapes and symmetries:

• Linear functions ($ax + by$) create flat planes.
• Quadratic functions ($x^2 + y^2$) create curved surfaces like paraboloids (bowls).
• Oscillating functions ($\sin$, $\cos$) create waves or ripples.
• Decaying functions ($e^{-x^2}$) approach zero as you move away from the origin.

Analysis:

1. Function i) $f(x, y) = 2x + 3y$

   • This is a linear equation of the form $z = ax + by$.
   • It represents a flat surface (a plane) passing through the origin.
   • Match: Graph 2) (The only flat plane).

2. Function ii) $f(x, y) = x^2 + y^2$

   • This describes a circular paraboloid.
   • At the origin $(0,0)$, $z=0$. As $x$ or $y$ increases, $z$ increases rapidly.
   • It looks like a bowl opening upwards.
   • Match: Graph 1) (The bowl shape).

3. Function iii) $f(x, y) = \frac{\sin(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}$

   • This is the classic “Sombrero” or “Sinc” function.
   • It depends only on the distance from the origin ($r = \sqrt{x^2+y^2}$).
   • At $r=0$, the limit is 1 (the central peak). As $r$ increases, it oscillates and decays (ripples).
   • Match: Graph 4) (The central peak with ripples).

4. Function iv) $f(x, y) = xe^{-(x^2+y^2)}$ [Assumed from shape, likely typo in image label or just complex]

   • This function is odd with respect to $x$ (if you swap $x$ for $-x$, the sign flips).
   • This creates a “peak” on one side and a “valley” on the other side of the axis.
   • Match: Graph 3) (Shows a peak and a valley antisymmetric across the center).

Answers:

1. The graph of the function i) has the serial number 2.
2. The graph of the function ii) has the serial number 1.
3. The graph of the function iii) has the serial number 4.
4. The graph of the function iv) has the serial number 3.

Question 5: Partial Integration

Problem: Find $f(x, y)$ such that $\frac{\partial f}{\partial x} = y \cos(xy)$.

Solution: To find $f(x, y)$, we integrate the partial derivative with respect to $x$. When integrating with respect to $x$, we treat $y$ as a constant.

Using u-substitution where $u = xy$ (so $du = y , dx$):

Important: In partial integration, the “constant of integration” $C$ can be any function that depends only on the variable held constant (in this case, $y$).

Evaluating the Options:

• $\sin(xy)$: Valid (where $C(y) = 0$).
• $xy \cos(xy)$: Invalid.
• $\sin(xy) + xy$: Invalid ($xy$ depends on $x$).
• $\sin(xy) + \cos(y)$: Valid ($C(y) = \cos(y)$, depends only on $y$).
• $\sin(xy) + \phi(x)$: Invalid (cannot depend on $x$).
• $\sin(xy) + \phi(y)$: Valid ($C(y) = \phi(y)$, depends only on $y$).

Correct Options:

• $f(x, y) = \sin(xy)$
• $f(x, y) = \sin(xy) + \cos(y)$
• $f(x, y) = \sin(xy) + \phi(y)$, for some non-constant function $\phi(y)$.

Question 6: Differentiability of the Cone

Problem: $f(x, y) = \sqrt{x^2 + y^2}$. Analyze partial derivatives at $(0,0)$ and $(1,1)$. Analysis: The graph of this function is a cone with its sharp vertex at the origin $(0,0)$. Sharp points usually imply derivatives do not exist.

1. Check $f_x(0,0)$: Definition: $\lim_{h \to 0} \frac{f(0+h, 0) - f(0,0)}{h} = \lim_{h \to 0} \frac{\sqrt{h^2} - 0}{h} = \lim_{h \to 0} \frac{|h|}{h}$

   • If $h > 0$, limit is $1$.
   • If $h < 0$, limit is $-1$.
   • Since limits don’t match, $f_x(0,0)$ does not exist.

2. Check $f_y(0,0)$: By the same logic, $f_y(0,0)$ does not exist.

3. Check $f_x(1,1)$: For points other than the origin, we can use standard derivative rules.

   $$\frac{\partial}{\partial x} (x^2 + y^2)^{1/2} = \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot (2x) = \frac{x}{\sqrt{x^2+y^2}}$$

   Substitute $(1,1)$:

   $$f_x(1,1) = \frac{1}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$$

Correct Options:

• $f_x(0, 0)$ does not exist.
• $f_y(0, 0)$ does not exist.
• $f_x(1, 1) = \frac{1}{\sqrt{2}}$.

Question 7: Geometry of the Graph G(f)

Problem: Analyze the graph $G(f)$ as a subset of $\mathbb{R}^3$. $z = f(x, y)$.

Analysis:

1. Intersection with axes:

   • x-axis ($y=0, z=0$): Intersection requires $f(x,0) = 0$. This depends on the specific function. It might have no solutions or many. (False)
   • y-axis ($x=0, z=0$): Intersection requires $f(0,y) = 0$. Same issue. (False)
   • z-axis ($x=0, y=0$): Intersection requires $z = f(0,0)$. Since $f$ is a function, it yields exactly one output for the input $(0,0)$. Thus, there is exactly one point $(0, 0, f(0,0))$ on the graph intersecting the z-axis. (True)

2. Intersection with planes parallel to xy-plane ($z = c$):

   • The statement “Every plane… has non-empty intersection” means for every height $c$, there is some point on the graph.
   • Mathematically: For all $c \in \mathbb{R}$, there exists $(x,y)$ such that $f(x,y) = c$.
   • This is the exact definition of a Surjective (or “onto”) function.
   • It does not imply Injectivity (one-to-one), because a plane could slice the graph in a circle (like the paraboloid) or a line, meaning many inputs give the same output.

Correct Options:

• The intersection of the z-axis with $G(f)$ contains a single element.
• If every plane parallel to the xy-plane has non-empty intersection with $G(f)$, then $f$ is surjective.

Question 8: Continuity at the Origin

Problem Statement: Determine the number of correct statements regarding the continuity of functions $f$ and $g$ at the origin.

1. $f(x,y) = \frac{x^3}{3x^2y}$ if $(x,y) \ne (0,0)$, else $0$.
2. $g(x,y) = x^4 + x^3y + xy^3 + y^4$ if $(x,y) \ne (0,0)$, else $0$.

Analysis of $f(x,y)$: To check continuity at the origin, the limit as $(x,y) \to (0,0)$ must equal the function value ($0$). Let’s check different paths.

• Path 1 ($y = x$): $$\lim_{x \to 0} \frac{x^3}{3x^2(x)} = \lim_{x \to 0} \frac{x^3}{3x^3} = \frac{1}{3}$$
• Path 2 ($y = 2x$): $$\lim_{x \to 0} \frac{x^3}{3x^2(2x)} = \lim_{x \to 0} \frac{x^3}{6x^3} = \frac{1}{6}$$ Since the limits are different ($1/3 \ne 1/6$), the limit does not exist. Statement 1 is False.

Analysis of $g(x,y)$: The function $g(x,y) = x^4 + x^3y + xy^3 + y^4$ is a polynomial. Polynomial functions are continuous everywhere on their domain. The limit as $(x,y) \to (0,0)$ is simply $0 + 0 + 0 + 0 = 0$, which matches the defined value. Statement 2 is True.

Conclusion: Only Statement 2 is correct.

Answer: 1 (Because there is 1 correct statement)

Question 9: Difference of Limits

Problem Statement: For $f(x,y) = \frac{xy}{x^2+y^2}$, find $l_2 - l_1$.

• $l_1$: Limit along line $y=x$.
• $l_2$: Limit along parabola $y=x^2$.

Solution:

1. Calculate $l_1$ (Path $y=x$):

   $$\lim_{x \to 0} \frac{x(x)}{x^2 + x^2} = \lim_{x \to 0} \frac{x^2}{2x^2} = \frac{1}{2} = 0.5$$

2. Calculate $l_2$ (Path $y=x^2$):

   $$\lim_{x \to 0} \frac{x(x^2)}{x^2 + (x^2)^2} = \lim_{x \to 0} \frac{x^3}{x^2 + x^4} = \lim_{x \to 0} \frac{x^3}{x^2(1+x^2)} = \lim_{x \to 0} \frac{x}{1+x^2} = \frac{0}{1} = 0$$

3. Calculate Difference:

   $$l_2 - l_1 = 0 - 0.5 = -0.5$$

Answer: -0.5

Question 10: Limit Value

Problem Statement: Find the value to which $f(x,y) = \frac{xy}{2x^2+y^2}$ approaches along the line $y=2x$.

Solution: Substitute $y=2x$ into the function:

Answer: 0.33 (rounded to two decimal places)

Question 11: Jacobian Determinant

Problem Statement: Find the determinant of the Jacobian matrix $A$ for $f(x,y) = (f_1, f_2)$ at $(4, -3)$.

• $f_1(x,y) = xy + y^2$
• $f_2(x,y) = x + xy + 1$

Solution:

1. Find Partial Derivatives:

   • $\frac{\partial f_1}{\partial x} = y$
   • $\frac{\partial f_1}{\partial y} = x + 2y$
   • $\frac{\partial f_2}{\partial x} = 1 + y$
   • $\frac{\partial f_2}{\partial y} = x$

2. Evaluate at $(4, -3)$:

   • $\frac{\partial f_1}{\partial x} = -3$
   • $\frac{\partial f_1}{\partial y} = 4 + 2(-3) = 4 - 6 = -2$
   • $\frac{\partial f_2}{\partial x} = 1 + (-3) = -2$
   • $\frac{\partial f_2}{\partial y} = 4$

3. Construct Matrix A:

   $$A = \begin{bmatrix} -3 & -2 \\ -2 & 4 \end{bmatrix}$$

4. Calculate Determinant:

   $$\det(A) = (-3)(4) - (-2)(-2) = -12 - 4 = -16$$

Answer: -16

Question 12: Directional Derivative

Problem Statement: Find the directional derivative of $f(x,y) = x^2y^3$ at $(3,4)$ in the direction of vector $\vec{v} = (0, 3)$.

Concept: The directional derivative is given by $D_{\vec{u}}f = \nabla f \cdot \vec{u}$, where $\vec{u}$ is the unit vector in the direction of $\vec{v}$.

Solution:

1. Calculate Gradient $\nabla f$:

   $$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2xy^3, 3x^2y^2)$$

2. Evaluate Gradient at $(3,4)$:

   • $f_x = 2(3)(4^3) = 6(64) = 384$
   • $f_y = 3(3^2)(4^2) = 27(16) = 432$
   • $\nabla f(3,4) = (384, 432)$

3. Find Unit Vector $\vec{u}$: Magnitude of $\vec{v} = \sqrt{0^2 + 3^2} = 3$.

   $$\vec{u} = \frac{(0,3)}{3} = (0, 1)$$

4. Calculate Dot Product:

   $$D_{\vec{u}}f = (384, 432) \cdot (0, 1) = 384(0) + 432(1) = 432$$

Answer: 432

Question 13: Limit Calculation

Problem Statement: Find $\lim_{(x,y)\to(0,0)} (4x+1)\sin y$.

Solution: Since the function is a product of polynomial and trigonometric functions, it is continuous at $(0,0)$. We can use direct substitution.

Answer: 0

Comprehension Set: Questions 14, 15, 16

Context: Price function $f(x,y) = x^2 + xy + y^2$. $x$: Price of raw materials. $y$: Price of transportation.

Question 14: Rate of Change Equality We want the rate of change wrt $x$ to equal the rate of change wrt $y$.

This implies the price of raw materials is the same as the price of transportation. Answer: Statement 3 (Enter 3)

Question 15: True Statements

• Linear? No, degree is 2.
• Homogeneous? Yes, $f(cx, cy) = (cx)^2 + (cx)(cy) + (cy)^2 = c^2(x^2+xy+y^2) = c^2f(x,y)$. (True)
• Continuous? Yes, it is a polynomial. (True)
• Limit check: If $y \to 0$ and $x \to 5$: $f(x,y) \to 5^2 + 5(0) + 0^2 = 25$. The statement saying it approaches 30 is False. The statement saying it approaches 25 is (True). Answer: Select the checkboxes for Homogeneity, Continuity, and Price approaches 25.

Question 16: Directional Derivative Value Given: Rate of change along $(1, m)$ is $\frac{1}{\sqrt{1+m^2}}[ka + lb]$ at point $(a, b)$. Find $2k - l$.

1. Calculate Directional Derivative: Unit vector $\vec{u} = \frac{1}{\sqrt{1+m^2}}(1, m)$. Gradient at $(a, b)$ is $(2a+b, a+2b)$.

   $$D_{\vec{u}}f = \nabla f \cdot \vec{u} = \frac{1}{\sqrt{1+m^2}} [1(2a+b) + m(a+2b)]$$

2. Group terms by $a$ and $b$:

   $$= \frac{1}{\sqrt{1+m^2}} [2a + b + ma + 2mb]$$$$= \frac{1}{\sqrt{1+m^2}} [a(2+m) + b(1+2m)]$$

3. Compare with given form: Given form: $\frac{1}{\sqrt{1+m^2}} [ka + lb]$ Therefore:

   $$k = 2+m$$$$l = 1+2m$$

4. Compute $2k - l$:

   $$2k - l = 2(2+m) - (1+2m)$$$$= 4 + 2m - 1 - 2m$$$$= 3$$

Answer: 3