\[ f(x,y)= xy; \space c(t) = (\cos(t), \sin(t)); \space 0 \leq t \leq 2 \pi \]

\[ f(x,y) = x^2 + y^2; \space c(t)= (\cos(t), 8 \sin(t)); \space 0 \leq t \leq 2 \pi \]

This problem refers to **calculus** and aims to **understand** that over a **closed** and **bounded** interval, **the continuous** function of one **variable** always reaches the **maximum** and **minimum** values. The weights of the **range** of the function are always **finite.**

In this **problem,** we are given a **function** and path that the function is being **estimated** along. We have to calculate the **maximum** and **minimum** associated with the function along the path.

## Expert Answer

**Part a:**

Given that, $f(x,y)= xy$ and $c(t) = (\cos(t), \sin(t));$ for $0 \leq t \leq 2 \pi$.

\[ f(x,y)= xy \]

\[ f(x,y)= \cos(t). \sin(t) \]

Using the **trigonometric** formula $ \sin(2x)= 2 \sin(x)\cos (x)$:

$\sin(x) \cos(x)$ is equals to $\dfrac{\sin(2x)}{2}$.

Inserting $\sin(x) \cos(x)$ in $f(x,y)$:

\[f(x,y)= \dfrac{\sin(2x)}{2} \]

We know that the range of **sine function** is always between $-1$ to $1$, that is:

\[ -1 \leq \sin(2x) \leq 1 \]

\[ \dfrac{-1}{2} \leq \dfrac{ \sin(2x)}{2} \leq \dfrac{1}{2} \]

\[ \dfrac{-1}{2} \leq f(x,y) \leq \dfrac{1}{2} \]

**Part b:**

Given that $f(x,y)= x^2+y^2$ and $c(t) = ( \cos(t), 8\sin (t));$ for $0 \leq t \leq 2 \pi$.

\[ f(x,y)= x^2 + y^2 \]

\[ f(x,y)= (\cos (t))^2. (8 \sin (t))^2 \]

\[ f(x,y)= \cos^2 t. 64 \sin^2 t \]

Using the **trigonometric** formula $ \sin^2(x) + \cos^2(x) = 1$,

$\cos^2(t)$ is equals to $1 – \sin^2(t)$.

Inserting the new $\cos^2(t)$ in $f(x,y)$:

\[f(x,y)= 1 -\sin^2(t) + 64 \sin^2(t) \]

\[f(x,y)= 1 + 63 \sin^2(t) \]

We know that the **range** of $\sin^2 (t)$ function is always between $0$ to $1$, that is:

\[ 0 \leq \sin^2(t) \leq 1 \]

\[ 0 \leq 63 \sin^2(t) \leq 63 \]

\[ 1 \leq 1+ 63\sin^2(t) \leq 64 \]

\[ 1 \leq f(x,y) \leq 64 \]

## Numerical Answer

**Part a**: **Maximum** and **minimum** value attained by the function $f(x,y) = xy$ along the **path** $ (cos(t), sin(t))$ is $\dfrac{-1}{2}$ and $\dfrac{1}{2}$.

**Part b: Maximum** and **minimum** value attained by the function $f(x,y = x^2 + y^2)$ along the **path** $ ( \cos(t), 8\sin (t))$ is $1$ and $64$.

## Example

Find the **maximum** and **minimum** range of the function $f$ along the path $c(t)$

**\[ -(b) \space f(x,y) = x^2 + y^2; \space c(t)= (\cos(t), 4 \sin(t)); \space 0 \leq t \leq 2 \pi \]**

Given, $f(x,y)= x^2+y^2$ and $c(t) = ( \cos(t), 4\sin (t));$ for $0 \leq t \leq 2 \pi$.

\[f(x,y)= x^2+y^2\]

\[f(x,y)= \cos^2 t. 16 \sin^2 t\]

Using the **trigonometric** formula $ \sin^2(x)+ \cos^2(x)=1$,

$\cos^2 (t)$ is equals to $1 – \sin^2 (t)$.

$f(x,y)$ becomes:

\[ f(x,y)=1 -\sin^2(t)+16 \sin^2 (t) \]

\[ f(x,y)=1+15 \sin^2(t) \]

**Range** of $\sin^2 (t)$ function is **between** $0$ to $1$, that is:

\[ 0 \leq \sin^2(t) \leq 1 \]

\[ 0 \leq 15 \sin^2(t) \leq 15 \]

\[ 1 \leq 1+ 15\sin^2(t) \leq 16 \]

\[1 \leq f(x,y) \leq 16\]