# A plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

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.

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:

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

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