Math Problem Statement
Question
: Let
f
:
R
2
→
R
be the function
f
(
x,y
) =
(
x
2
y
2
) sin
(
1
x
2
y
2
)
,
(
x,y
)
6
= (0
,
,
0
,
(
x,y
) = (0
,
.
Prove that
f
is differentiable at the origin and compute
D
f
(0
,
.
[Hint: Consider computing the limit
lim
(
x,y
)
→
(0
,
|
f
(
x,y
)
|
‖
(
x,y
)
‖
and draw your conclusion from the result
Solution
Problem Restatement
Let be defined by
\begin{cases} (x^2 + y^2) \sin\left(\frac{1}{x^2 + y^2}\right), & \text{if } (x, y) \neq (0, 0), \\ 0, & \text{if } (x, y) = (0, 0). \end{cases}$$ We are asked to **prove that $$f$$ is differentiable at the origin** and compute $$Df(0, 0)$$. ### Step 1: Checking Continuity at the Origin To prove differentiability, we first check whether $$f$$ is continuous at the origin. We do this by considering the limit: $$\lim_{(x, y) \to (0, 0)} f(x, y).$$ We need to show that: $$\lim_{(x, y) \to (0, 0)} (x^2 + y^2) \sin\left(\frac{1}{x^2 + y^2}\right) = 0.$$ #### Approach in Polar Coordinates Convert the function into polar coordinates, where $$x = r \cos(\theta)$$ and $$y = r \sin(\theta)$$, and $$r = \sqrt{x^2 + y^2}$$. Thus, we have: $$f(x, y) = r^2 \sin\left(\frac{1}{r^2}\right).$$ We now examine the behavior of the function as $$r \to 0$$. Notice that: $$|\sin\left(\frac{1}{r^2}\right)| \leq 1.$$ Hence, the absolute value of $$f(x, y)$$ is bounded by: $$|f(x, y)| = |r^2 \sin\left(\frac{1}{r^2}\right)| \leq r^2.$$ Since $$r^2 \to 0$$ as $$r \to 0$$, we conclude that: $$\lim_{(x, y) \to (0, 0)} f(x, y) = 0.$$ Thus, $$f$$ is continuous at $$(0, 0)$$. ### Step 2: Checking Differentiability at the Origin To prove differentiability at the origin, we need to check if the following limit exists: $$\lim_{(h, k) \to (0, 0)} \frac{f(h, k) - f(0, 0) - 0}{\sqrt{h^2 + k^2}} = 0.$$ Since $$f(0, 0) = 0$$, this becomes: $$\lim_{(h, k) \to (0, 0)} \frac{f(h, k)}{\sqrt{h^2 + k^2}}.$$ Using polar coordinates again, we have: $$\frac{f(h, k)}{\sqrt{h^2 + k^2}} = \frac{r^2 \sin\left(\frac{1}{r^2}\right)}{r} = r \sin\left(\frac{1}{r^2}\right).$$ Since $$|\sin\left(\frac{1}{r^2}\right)| \leq 1$$, it follows that: $$|r \sin\left(\frac{1}{r^2}\right)| \leq r.$$ As $$r \to 0$$, $$r \to 0$$. Therefore, the limit is 0: $$\lim_{(h, k) \to (0, 0)} \frac{f(h, k)}{\sqrt{h^2 + k^2}} = 0.$$ Thus, $$f$$ is differentiable at the origin. ### Step 3: Computing the Derivative $$Df(0, 0)$$ Since $$f$$ is differentiable at the origin and the limit for differentiability is zero, we conclude that the derivative at $$(0, 0)$$ is: $$Df(0, 0) = (0, 0).$$ ### Conclusion We have proved that $$f$$ is differentiable at the origin, and the derivative at the origin is: $$Df(0, 0) = (0, 0).$$ --- If you have any further questions or need additional clarification, feel free to ask! ### Related Questions: 1. How does differentiability relate to continuity in multivariable functions? 2. What is the difference between differentiability and the existence of partial derivatives? 3. Can you prove differentiability using the definition of the derivative in multiple variables? 4. How can we use polar coordinates to simplify limits in multivariable calculus? 5. What would happen if we tried to compute $$f(x, y)$$ along different paths towards the origin? ### Tip: Always check continuity first when working with differentiability problems, as differentiability implies continuity but not the other way around!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Multivariable Calculus
Differentiability
Continuity
Polar Coordinates
Formulas
f(x, y) = r^2 sin(1/r^2)
Theorems
Differentiability criterion for multivariable functions
Continuity at a point in multivariable functions
Suitable Grade Level
College/University Level (Calculus)
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