This question aims to develop an understanding of the **Pythagorean theorem** and basic rules of **differentiation**.

If we have a **right triangle**, then according to the **Pythagorean theorem** the **relation between its different sides** can be described mathematically with the help of the **following formula**:

\[ ( hypotenuse )^{ 2 } \ = \ ( base )^{ 2 } \ + \ ( perpendicular )^{ 2 } \]

The use of **differentiation** is explained as per its use in the following solution. We first develop the **starting function** using the **Pythagorean theorem**. Then we **differentiate** it to calculate the **required rate** of change.

## Expert Answer

**Given that:**

\[ \text{ Horizontal speed of plane } = \dfrac{ x }{ t } \ = \ 500 \ mi/h \]

\[ \text{ Distance of plane from the radar } = \ y \ = \ 2 \ mi \]

\[ \text{ Height of plane from the radar } = \ z \ = \ 1 \ mi \]

Given the situation described, we can** construct a triangle** such that the **Pythagorean theorem** is applied as follows:

\[ x^{ 2 } \ + \ ( 1 )^{ 2 } \ = \ y^{ 2 } \]

\[ x^{ 2 } \ + \ 1 \ = \ y^{ 2 } \ … \ … \ … \ ( 1 ) \]

**Substituting values:**

\[ x^{ 2 } \ + \ 1 \ = \ ( 2 )^{ 2 } \ = \ 4 \]

\[ x^{ 2 } \ = \ 4 \ – \ 1 \ = \ 3 \]

\[ x \ = \ \pm \sqrt{ 3 } \ mi \]

Since **distance cannot be negative:**

\[ x \ = \ + \sqrt{ 3 } \ mi \]

**Taking derivative of equation (1):**

\[ \dfrac{ d }{ dt } ( x^{ 2 } ) \ + \ \dfrac{ d }{ dt } ( 1 ) \ = \ \dfrac{ d }{ dt } ( y^{ 2 } ) \]

\[ 2 x \dfrac{ d x }{ d t } \ = \ 2 y \dfrac{ d y }{ d t } \]

\[ \dfrac{ d y }{ d t } \ = \ \dfrac{ x }{ y } \dfrac{ d x }{ d t } \ … \ … \ … \ ( 2 ) \]

**Substituting values:**

\[ \dfrac{ d y }{ d t } \ = \ \dfrac{ \sqrt{ 3 } }{ 2 } ( 500 ) \]

\[ \dfrac{ d y }{ d t } \ = \ 250 \sqrt{ 3 } \ mi/h \]

## Numerical Result

\[ \dfrac{ d y }{ d t } \ = \ 250 \sqrt{ 3 } \ mi/h \]

## Example

Suppose the **plane** described in the above question is **at a distance of 4 mi**. What will be the **rate of separation** in this case?

**Recall equation (1):**

\[ x^{ 2 } \ + \ 1 \ = \ y^{ 2 } \]

**Substituting values:**

\[ x^{ 2 } \ + \ 1 \ = \ ( 4 )^{ 2 } \ = \ 16 \]

\[ x^{ 2 } \ = \ 16 \ – \ 1 \ = \ 15 \]

\[ x \ = \ \pm \sqrt{ 15 } \ mi \]

Since **distance cannot be negative:**

\[ x \ = \ + \sqrt{ 15 } \ mi \]

**Recall equation (2):**

\[ \dfrac{ d y }{ d t } \ = \ \dfrac{ x }{ y } \dfrac{ d x }{ d t } \]

**Substituting values:**

\[ \dfrac{ d y }{ d t } \ = \ \dfrac{ \sqrt{ 15 } }{ 4 } ( 500 ) \]

\[ \dfrac{ d y }{ d t } \ = \ 125 \sqrt{ 15 } \ mi/h \]